ĐK: $x\leq 1$pt (1) $\Leftrightarrow 2y^{3}+y=(3-2x)\sqrt{1-x}=2(1-x)\sqrt{1-x}+\sqrt{1-x}$ $\Leftrightarrow y=\sqrt{1-x}$(2) TT:$\sqrt{2(1-x)+1}-\sqrt{1-x}-(2-x)=0$$\Leftrightarrow \sqrt{2(1-x)+1}-1-\sqrt{1-x}-(1-x)=0$$\Leftrightarrow \sqrt{1-x} \left[ \sqrt{1-x}(\frac{2}{\sqrt{2(1-x)+1}+1}-1)-1\right]=0$ ta có $\sqrt{2(1-x)+1}+1\geq2 \Rightarrow \frac{2}{\sqrt{...}+1}\leq 1$$\Rightarrow \frac{2}{\sqrt{..}+1}-1\leq0$$\Rightarrow$ phần gạch chân $<0$$\Leftrightarrow x=1 \Rightarrow (x;y)(1;0)$

ĐK: $x\leq 1$pt (1) $\Leftrightarrow 2y^{3}+y=(3-2x)\sqrt{1-x}=2(1-x)\sqrt{1-x}+\sqrt{1-x}$ $\Leftrightarrow y=\sqrt{1-x}$(2) TT:$\sqrt{2(1-x)+1}-\sqrt{1-x}-(2-x)=0$$\Leftrightarrow \sqrt{2(1-x)+1}-1-\sqrt{1-x}-(1-x)=0$$\Leftrightarrow \sqrt{1-x} \left[ \sqrt{1-x}(\frac{2}{\sqrt{2(1-x)+1}+1}-1)-1\right]=0$ ta có $\sqrt{2(1-x)+1}+1\geq2 \Rightarrow \frac{2}{\sqrt{...}+1}\leq 1$$\Rightarrow \frac{2}{\sqrt{..}+1}-1\leq0$$\Rightarrow$ \left[ ..... \right] $<0$$\Leftrightarrow x=1 \Rightarrow (x;y)=(1;0)$

pt (1) $\Leftrightarrow 2y^{3}+y=(3-2x)\sqrt{1-x}=2(1-x)\sqrt{1-x}+\sqrt{1-x}$ $\Leftrightarrow y=\sqrt{1-x}$(2) TT:$\sqrt{2(1-x)+1}-\sqrt{1-x}-(2-x)=0$$\Leftrightarrow \sqrt{2(1-x)+1}-1-\sqrt{1-x}-(1-x)=0$$\Leftrightarrow \sqrt{1-x} (( \sqrt{1-x}\frac{2}{\sqrt{2(1-x)+1}+1}-1)-1)=0$ta có$\sqrt{2(1-x)+1}+1\geq2 \Rightarrow \frac{2}{\sqrt{...}+1}\leq 1 $$\Rightarrow \frac{2}{\sqrt{}+1}-1\leq0$$ \Rightarrow $phần gạch chân$<0$$\Leftrightarrow x=1 \Rightarrow (x;y)(1;0)$

ĐK: $x\leq 1$pt (1) $\Leftrightarrow 2y^{3}+y=(3-2x)\sqrt{1-x}=2(1-x)\sqrt{1-x}+\sqrt{1-x}$ $\Leftrightarrow y=\sqrt{1-x}$(2) TT:$\sqrt{2(1-x)+1}-\sqrt{1-x}-(2-x)=0$$\Leftrightarrow \sqrt{2(1-x)+1}-1-\sqrt{1-x}-(1-x)=0$$\Leftrightarrow \sqrt{1-x} \left[ \sqrt{1-x}(\frac{2}{\sqrt{2(1-x)+1}+1}-1)-1\right]=0$ta có$\sqrt{2(1-x)+1}+1\geq2 \Rightarrow \frac{2}{\sqrt{...}+1}\leq 1$$\Rightarrow \frac{2}{\sqrt{..}+1}-1\leq0 $$\Rightarrow $ phần gạch chân $<0$$ \Leftrightarrow x=1 \Rightarrow (x;y)(1;0)$

(1) $\Leftrightarrow \frac{(x+y)^{2}-2xy}{xy}+\frac{2}{x+y}-\frac{1}{xy}=0$ $\Leftrightarrow \frac{(x+y)^{2}-1}{xy} + \frac{2}{x+y}-2=0$ $\Leftrightarrow \frac{(x+y-1)(x+y+1)}{xy} +\frac{2(1-x-y)}{xy}=0$ $\Leftrightarrow x+y=1$ or $x^{2}+y^{2}+x+y=0$ +) $x+y=1$$\Rightarrow 3x^{2}-4x-1=0 \Leftrightarrow x=\frac{2\pm \sqrt{7}}{3}$+) $x^{2}+y^{2}=-(x+y)$$\Rightarrow 1+2x-x^{2}=x^{2}+y^{2}+\frac{1}{x^{2}+y^{2}}\geq2$$\Leftrightarrow x=1$$\Rightarrow (x;y)=.....$

(1) $\Leftrightarrow \frac{(x+y)^{2}-2xy}{xy}+\frac{2}{x+y}-\frac{1}{xy}=0$ $\Leftrightarrow \frac{(x+y)^{2}-1}{xy} + \frac{2}{x+y}-2=0$ $\Leftrightarrow \frac{(x+y-1)(x+y+1)}{xy} +\frac{2(1-x-y)}{x+y}=0\Leftrightarrow x+y=1$or$x^{2}+y^{2}+x+y=0$+)$x+y=1 $$\Rightarrow 3x^{2}-4x-1=0 \Leftrightarrow x=\frac{2\pm \sqrt{7}}{3}$+) $x^{2}+y^{2}=-(x+y)$$ \Rightarrow 1+2x-x^{2}=x^{2}+y^{2}+\frac{1}{x^{2}+y^{2}}\geq2 $$\Leftrightarrow x=1$$ \Rightarrow (x;y)=.....$

ta có $P=\frac{(x+y)^{2}}{z^{2}+4(xy+yz+zx)}=\frac{(x+y)^{2}}{z^{2}+4z(x+y)+(x+y)^{2}}$ $=\frac{(\frac{x}{z}+\frac{y}{z})^{2}}{1+4(\frac{x}{z}+\frac{y}{z})+(\frac{x}{z}+\frac{y}{z})^{2}}$Đặt $t=\frac{x+y}{z} (t\in \left[ 1{;}4 \right])$ khi đó P$=\frac{t^{2}}{1+4t+t^{2}}$ ta cần cm $P\geq \frac{1}{6}$ $\Leftrightarrow 5t^{2}-4t-1\geq0 \Leftrightarrow (t-1)(5t+1)\geq0$ ( lđ vs $\forall t\in \left[ 1{;}4 \right]$)

ta có $P=\frac{(x+y)^{2}}{z^{2}+4(xy+yz+zx)}=\frac{(x+y)^{2}}{z^{2}+4z(x+y)+(x+y)^{2}}$ $=\frac{(\frac{x}{z}+\frac{y}{z})^{2}}{1+4(\frac{x}{z}+\frac{y}{z})+(\frac{x}{z}+\frac{y}{z})^{2}}$Đặt $t=\frac{x+y}{z} (t\in \left[ 1{;}4 \right])$ khi đó P$=\frac{t^{2}}{1+4t+t^{2}}$ ta cần cm $P\geq \frac{1}{6}$ $\Leftrightarrow 5t^{2}-4t-1\geq0 \Leftrightarrow (t-1)(5t+1)\geq0$ ( lđ vs $\forall t\in \left[ 1{;}4 \right]$) Dấu "=" $\Leftrightarrow x=y=1;z=2$

DK:.....pt $\Leftrightarrow x+4=(\sqrt{x+2}+1)(1+\frac{4(\sqrt[3]{2x+3}-1)}{2x+3-1}$ $\Leftrightarrow (x+4)(x+1)=(\sqrt{x+2}+1)(2\sqrt[3]{2x+3}+x-1)$ $\Leftrightarrow (x+4)(\sqrt{x+2}+1)(\sqrt{x+2}-1)=(\sqrt{x+2}+1)(2\sqrt[3]{2x+3}+x-1)$ $\Leftrightarrow (x+4)\sqrt{x+2}-x-4=2\sqrt[3]{2x+3}+x-1$ $\Leftrightarrow (\sqrt{x+2})^{3}+2\sqrt{x+2}=2x+3+2\sqrt[3]{2x+3}$ $\Leftrightarrow (\sqrt{x+2}-\sqrt[3]{2x+3})(....)=0$$\Leftrightarrow \sqrt{x+2}=\sqrt[3]{2x+3}$ Đến đây bạn giải tip nhé!!! :D

DK:.....+) $x=-1$ k là no of pt+) $x\neq -1$pt $\Leftrightarrow x+4=(\sqrt{x+2}+1)(1+\frac{4(\sqrt[3]{2x+3}-1)}{2x+3-1}$ $\Leftrightarrow (x+4)(x+1)=(\sqrt{x+2}+1)(2\sqrt[3]{2x+3}+x-1)$ $\Leftrightarrow (x+4)(\sqrt{x+2}+1)(\sqrt{x+2}-1)=(\sqrt{x+2}+1)(2\sqrt[3]{2x+3}+x-1)$ $\Leftrightarrow (x+4)\sqrt{x+2}-x-4=2\sqrt[3]{2x+3}+x-1$ $\Leftrightarrow (\sqrt{x+2})^{3}+2\sqrt{x+2}=2x+3+2\sqrt[3]{2x+3}$ $\Leftrightarrow (\sqrt{x+2}-\sqrt[3]{2x+3})(....)=0$$\Leftrightarrow \sqrt{x+2}=\sqrt[3]{2x+3}$ Đến đây bạn giải tip nhé!!! :D

(2) $\Leftrightarrow (y-1)^{2}=(x^{2}+y^{2})(1-x^{2}-y^{2})$$\Rightarrow x^{2}+y^{2}\leq 1 \Rightarrow |x| \leq 1$mà (x^{2}+y^{2})(1-x^{2}-y^{2}) \leq \frac{1}{4}\Rightarrow (y-1)^{2}\leq \frac{1}{4}\Leftrightarrow \frac{1}{2}\leq y\leq \frac{3}{2} \Rightarrow y>0, |x|\leq 1(\sqrt{x^{2}+1}+1) (x^{2}-y^{3}+3y-2)=(\sqrt{x^{2}+1}+1)(x^{2}-(y-1)^{2}(y+2))\leq (\sqrt{x^{2}+1}+1)x^{2} \leq (\sqrt{2}+1)x^{2}\leq 4x^{2}\Rightarrow x=0;y=1$

(2) $\Leftrightarrow (y-1)^{2}=(x^{2}+y^{2})(1-x^{2}-y^{2})$$\Rightarrow x^{2}+y^{2}\leq 1 \Rightarrow |x| \leq 1$mà $(x^{2}+y^{2})(1-x^{2}-y^{2})\leq \frac{1}{4}\Rightarrow (y-1)^{2}\leq \frac{1}{4}\Leftrightarrow \frac{1}{2}\leq y\leq \frac{3}{2} \Rightarrow y>0, |x|\leq 1(\sqrt{x^{2}+1}+1) (x^{2}-y^{3}+3y-2)=(\sqrt{x^{2}+1}+1)(x^{2}-(y-1)^{2}(y+2))\leq (\sqrt{x^{2}+1}+1)x^{2} \leq (\sqrt{2}+1)x^{2}\leq 4x^{2}\Rightarrow x=0;y=1$

(2) $\Leftrightarrow (y-1)^{2}=(x^{2}+y^{2})(1-x^{2}-y^{2})$$\Rightarrow x^{2}+y^{2}\leq 1 \Rightarrow |x| \leq 1$mà (x^{2}+y^{2})(1-x^{2}-y^{2})\leq \frac{1}{4}$ $\Rightarrow (y-1)^{2}\leq \frac{1}{4}$ $\Leftrightarrow \frac{1}{2}\leq y\leq \frac{3}{2} \Rightarrow y>0, |x|\leq 1$ $(\sqrt{x^{2}+1}+1) (x^{2}-y^{3}+3y-2)=(\sqrt{x^{2}+1}+1)(x^{2}-(y-1)^{2}(y+2))$ $\leq (\sqrt{x^{2}+1}+1)x^{2} \leq (\sqrt{2}+1)x^{2}\leq 4x^{2}$ $\Rightarrow x=0;y=1$

(2) $\Leftrightarrow (y-1)^{2}=(x^{2}+y^{2})(1-x^{2}-y^{2})$$\Rightarrow x^{2}+y^{2}\leq 1 \Rightarrow |x| \leq 1$mà (x^{2}+y^{2})(1-x^{2}-y^{2}) \leq \frac{1}{4}\Rightarrow (y-1)^{2}\leq \frac{1}{4}\Leftrightarrow \frac{1}{2}\leq y\leq \frac{3}{2} \Rightarrow y>0, |x|\leq 1(\sqrt{x^{2}+1}+1) (x^{2}-y^{3}+3y-2)=(\sqrt{x^{2}+1}+1)(x^{2}-(y-1)^{2}(y+2))\leq (\sqrt{x^{2}+1}+1)x^{2} \leq (\sqrt{2}+1)x^{2}\leq 4x^{2}\Rightarrow x=0;y=1$

http://thayquocvuong.com/uploads/news/de-thi/2006/da-toan-b-dh2006-03.jpgcâu 2

http://thayquocvuong.com/uploads/news/de-thi/2006/da-toan-b-dh2006-03.jpgcâu 2 phần IV

C2) ta có $3(x^{2}+y^{2}+z^{2}\geq (x+y+z)^{2} \Rightarrow x+y+z \leq \sqrt{3(x^{2}+y^{2}+z^{2}}$ $\Rightarrow P\leq \frac{xyz(\sqrt{3(x^{2}+y^{2}+z^{2}}+\sqrt{x^{2}+y^{2}+z^{2}})}{(x^{2}+y^{2}+z^{2})(xy+yz+zx)}$ $=\frac{(\sqrt{3}+1)xyz}{(xy+yz+zx)(\sqrt{x^{2}+y^{2}+z^{2}})}$mà$(xy+yz+zx)(\sqrt{x^{2}+y^{2}+z^{2}} \geq 3\sqrt[3]{x^{2}y^{2}z^{2}}\sqrt{3\sqrt[3]{x^{2}y^{2}z^{2}}}=3\sqrt{3}xyz\Rightarrow P\leq \frac{\sqrt{3}+1}{3\sqrt{3}}=\frac{3+\sqrt{3}}{9}$dấu '="$\Leftrightarrow a=b=c$

C2) ta có $3(x^{2}+y^{2}+z^{2})\geq (x+y+z)^{2} \Rightarrow x+y+z \leq \sqrt{3(x^{2}+y^{2}+z^{2})}\Rightarrow P\leq \frac{xyz(\sqrt{3(x^{2}+y^{2}+z^{2}}+\sqrt{x^{2}+y^{2}+z^{2}})}{(x^{2}+y^{2}+z^{2})2(xy+yz+zx)}$ =$\frac{(\sqrt{3}+1)xyz}{2(xy+yz+zx)\sqrt{x^{2}+y^{2}+z^{2}}}$mà$(xy+yz+zx)\sqrt{x^{2}+y^{2}+z^{2}} \geq 3\sqrt[3]{x^{2}y^{2}z^{2}}\sqrt{3\sqrt[3]{x^{2}y^{2}z^{2}}}=3\sqrt{3}xyz\Rightarrow P\leq \frac{\sqrt{3}+1}{2.3\sqrt{3}}=\frac{3+\sqrt{3}}{18}$dấu '="$\Leftrightarrow a=b=c$

C2)bđt $\Leftrightarrow \frac{\sqrt[3]{\frac{a}{b}}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} \leq1\Leftrightarrow \sum \frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+ \frac{1}{b} +\frac{1}{c})}}\leq1$ta có$\frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+frac{1}{b}+\frac{1}{c})}} =\sqrt[3]{\frac{1}{3}.\frac{a}{a+b+c}.\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}}\leq \frac{1}{3}(\frac{1}{3}+\frac{a}{a+b+c}+\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})$TT$\Rightarrow VT\leq \frac{1}{3}(1+1+1)=1$dấu '="$\Leftrightarrow a=b=c$

C2)bđt $\Leftrightarrow \frac{\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} \leq1$ $\Leftrightarrow \sum \frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+ \frac{1}{b} +\frac{1}{c})}}\leq1$ ta có $\frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} =\sqrt[3]{\frac{1}{3}.\frac{a}{a+b+c}.\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}}\leq \frac{1}{3}(\frac{1}{3}+\frac{a}{a+b+c}+\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})$TT$\Rightarrow VT\leq \frac{1}{3}(1+1+1)=1$dấu '="$\Leftrightarrow a=b=c$

C2)bđt $\Leftrightarrow \frac{\sqrt[3]{\frac{a}{b}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} \leq1\Leftrightarrow \sum \frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}}\leq1$ta có$\frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+frac{1}{b}+\frac{1}{c})}} =\sqrt[3]{\frac{1}{3}.\frac{a}{a+b+c}.\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}}\leq \frac{1}{3}(\frac{1}{3}+\frac{a}{a+b+c}+\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})$TT$\Rightarrow VT\leq \frac{1}{3}(1+1+1)=1$dấu '="$\Leftrightarrow a=b=c$

C2)bđt $\Leftrightarrow \frac{\sqrt[3]{\frac{a}{b}}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} \leq1\Leftrightarrow \sum \frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+ \frac{1}{b} +\frac{1}{c})}}\leq1$ta có$\frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+frac{1}{b}+\frac{1}{c})}} =\sqrt[3]{\frac{1}{3}.\frac{a}{a+b+c}.\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}}\leq \frac{1}{3}(\frac{1}{3}+\frac{a}{a+b+c}+\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})$TT$\Rightarrow VT\leq \frac{1}{3}(1+1+1)=1$dấu '="$\Leftrightarrow a=b=c$

ta có $x\sqrt{8y-5}\leq \sqrt{3} x\sqrt{\frac{8y-5}{3}} \leq \frac{\sqrt{3}}{2}x(\frac{8y-5}{3}+1)=\frac{\sqrt{3}}{3}x(4y-1)$ $\Rightarrow VT(1) \leq \frac{\sqrt{3}}{3}(8xy-x-y) \leq \frac{\sqrt{3}}{3}(2(x+y)^{2}-x-y)$ VP $\geq \sqrt[4]{24(\frac{1}{2}((x+y)^{2}+4))} =\sqrt[4]{12(x+y)^{2}+96}$ ta cần cm $\sqrt[4]{12(x+y)^{2}+96}\geq \frac{\sqrt{3}}{3} (2(x+y)^{2}-x-y)$ đặt $t=x+y$ $\Leftrightarrow \sqrt[4]{12t^{2}+96}\geq \frac{\sqrt{3}}{3}(2t^{2}-1)$ nâng lũy thừa bậc 4 $\Leftrightarrow (t-2) ( t^{7}+\frac{3t}{2}^{5}+\frac{5t}{2}^{4}+\frac{8t}{16}^{3}+\frac{81t}{8}^{2}+\frac{27t}{2}+2)\leq 0$ $\Leftrightarrow t\leq 2$ ta cần tìm đk này từ pt (2) vs t-x=y (2) $\Leftrightarrow 11x^{2}-6x(t-x)+3(t-x)=12x-4(t-x)$ $\Leftrightarrow 20x^{2}-(16+12t)x+3t^{2}+4=0$ $\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]$ $\Rightarrow x=y=1$

ta có $x\sqrt{8y-5}\leq \sqrt{3} x\sqrt{\frac{8y-5}{3}} \leq \frac{\sqrt{3}}{2}x(\frac{8y-5}{3}+1)=\frac{\sqrt{3}}{3}x(4y-1)$ $\Rightarrow VT(1) \leq \frac{\sqrt{3}}{3}(8xy-x-y) \leq \frac{\sqrt{3}}{3}(2(x+y)^{2}-x-y)$ VP $\geq \sqrt[4]{24(\frac{1}{2}((x+y)^{2}+4))} =\sqrt[4]{12(x+y)^{2}+96}$ ta cần cm $\sqrt[4]{12(x+y)^{2}+96}\geq \frac{\sqrt{3}}{3} (2(x+y)^{2}-x-y)$ đặt $t=x+y$ $\Leftrightarrow \sqrt[4]{12t^{2}+96}\geq \frac{\sqrt{3}}{3}(2t^{2}-1)$ nâng lũy thừa bậc 4 $\Leftrightarrow (t-2) ( t^{7}+\frac{3t}{2}^{5}+\frac{5t}{2}^{4}+\frac{8t}{16}^{3}+\frac{81t}{8}^{2}+\frac{27t}{2}+2)\leq 0$ $\Leftrightarrow t\leq 2$ ta cần tìm đk này từ pt (2) vs t-x=y (2) $\Leftrightarrow 11x^{2}-6x(t-x)+3(t-x)^{2}=12x-4(t-x)\Leftrightarrow 20x^{2}-(16+12t)x+3t^{2}+4t=0\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]\Rightarrow x=y=1$

ta có $x\sqrt{8y-5}\leq \sqrt{3} x\sqrt{\frac{8y-5}{3}} \leq \frac{\sqrt{3}}{2}x(\frac{8y-5}{3}+1)=\frac{\sqrt{3}}{3}x(4y-1)$ $\Rightarrow VT(1) \leq \frac{\sqrt{3}}{3}(8xy-x-y) \leq \frac{\sqrt{3}}{3}(2(x+y)^{2}-x-y)$ VP $\geq \sqrt[4]{24(\frac{1}{2}((x+y)^{2}+4))} =\sqrt[4]{12(x+y)^{2}+96}$ ta cần cm $\sqrt[4]{12(x+y)^{2}+96}\geq \frac{\sqrt{3}}{3} (2(x+y)^{2}-x-y)$ đặt $t=x+y$ $\Leftrightarrow \sqrt[4]{12t^{2}+96}\geq \frac{\sqrt{3}}{3}(2t^{2}-1)$ nâng lũy thừa bậc 4 $\Leftrightarrow (t-2) ( t^{7}+\frac{3t}{2}^{5}+\frac{5t}{2}^{4}+\frac{8t}{16}^{3}+\frac{81t}{8}^{2}+\frac{27t}{2}+2)\leq 0$ $\Leftrightarrow t\leq 2$ ta cần tìm đk này từ pt (2) vs t-x=y (2) $\Leftrightarrow 11x^{2}-6x(t-x)+3(t-x)=12x-4(t-x)$ $\Leftrightarrow 20x^{2}-(16+12y)x+3t^{2}+4=0\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]\Rightarrow x=y=1$

ta có $x\sqrt{8y-5}\leq \sqrt{3} x\sqrt{\frac{8y-5}{3}} \leq \frac{\sqrt{3}}{2}x(\frac{8y-5}{3}+1)=\frac{\sqrt{3}}{3}x(4y-1)$ $\Rightarrow VT(1) \leq \frac{\sqrt{3}}{3}(8xy-x-y) \leq \frac{\sqrt{3}}{3}(2(x+y)^{2}-x-y)$ VP $\geq \sqrt[4]{24(\frac{1}{2}((x+y)^{2}+4))} =\sqrt[4]{12(x+y)^{2}+96}$ ta cần cm $\sqrt[4]{12(x+y)^{2}+96}\geq \frac{\sqrt{3}}{3} (2(x+y)^{2}-x-y)$ đặt $t=x+y$ $\Leftrightarrow \sqrt[4]{12t^{2}+96}\geq \frac{\sqrt{3}}{3}(2t^{2}-1)$ nâng lũy thừa bậc 4 $\Leftrightarrow (t-2) ( t^{7}+\frac{3t}{2}^{5}+\frac{5t}{2}^{4}+\frac{8t}{16}^{3}+\frac{81t}{8}^{2}+\frac{27t}{2}+2)\leq 0$ $\Leftrightarrow t\leq 2$ ta cần tìm đk này từ pt (2) vs t-x=y (2) $\Leftrightarrow 11x^{2}-6x(t-x)+3(t-x)=12x-4(t-x)$ $\Leftrightarrow 20x^{2}-(16+12t)x+3t^{2}+4=0\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]\Rightarrow x=y=1$

ta có $x\sqrt{8y-5}\leq \sqrt{3} x\sqrt{\frac{8y-5}{3}} \leq \frac{\sqrt{3}}{2}x(\frac{8y-5}{3}+1)=\frac{\sqrt{3}}{3}x(4y-1)$ $\Rightarrow VT(1) \leq \frac{\sqrt{3}}{3}(8xy-x-y) \leq \frac{\sqrt{3}}{3}(2(x+y)^{2}-x-y)$ VP $\geq \sqrt[4]{24(\frac{1}{2}((x+y)^{2}+4))} =\sqrt[4]{12(x+y)^{2}+96}$ ta cần cm $\sqrt[4]{12(x+y)^{2}+96}\geq \frac{\sqrt{3}}{3} (2(x+y)^{2}-x-y)$ đặt $t=x+y$ $\Leftrightarrow \sqrt[4]{12t^{2}+96}\geq \frac{\sqrt{3}}{3}(2t^{2}-1)$ nâng lũy thừa bậc 4 $\Leftrightarrow (t-2) ( t^{7}+\frac{3t}{2}^{5}+\frac{5t}{2}^{4}+\frac{8t}{16}^{3}+\frac{81t}{8}^{2}+\frac{27t}{2}+2)\leq 0$ $\Leftrightarrow t\leq 2$ ta cần tìm đk này từ pt (2) vs t-x=y (2) $\Leftrightarrow 11x^{2}-6x9t-x)+3(t-x)=12x-4(t-x)\Leftrightarrow 20x^{2}-(16+12y)x+3t^{2}+4=0\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]\Rightarrow x=y=1$

ta có $x\sqrt{8y-5}\leq \sqrt{3} x\sqrt{\frac{8y-5}{3}} \leq \frac{\sqrt{3}}{2}x(\frac{8y-5}{3}+1)=\frac{\sqrt{3}}{3}x(4y-1)$ $\Rightarrow VT(1) \leq \frac{\sqrt{3}}{3}(8xy-x-y) \leq \frac{\sqrt{3}}{3}(2(x+y)^{2}-x-y)$ VP $\geq \sqrt[4]{24(\frac{1}{2}((x+y)^{2}+4))} =\sqrt[4]{12(x+y)^{2}+96}$ ta cần cm $\sqrt[4]{12(x+y)^{2}+96}\geq \frac{\sqrt{3}}{3} (2(x+y)^{2}-x-y)$ đặt $t=x+y$ $\Leftrightarrow \sqrt[4]{12t^{2}+96}\geq \frac{\sqrt{3}}{3}(2t^{2}-1)$ nâng lũy thừa bậc 4 $\Leftrightarrow (t-2) ( t^{7}+\frac{3t}{2}^{5}+\frac{5t}{2}^{4}+\frac{8t}{16}^{3}+\frac{81t}{8}^{2}+\frac{27t}{2}+2)\leq 0$ $\Leftrightarrow t\leq 2$ ta cần tìm đk này từ pt (2) vs t-x=y (2) $\Leftrightarrow 11x^{2}-6x(t-x)+3(t-x)=12x-4(t-x)\Leftrightarrow 20x^{2}-(16+12y)x+3t^{2}+4=0\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]\Rightarrow x=y=1$

$3\sqrt{2(3x+4)^{3}}\leq (x^{2}+2x-2)(\sqrt{x^{2}+2x-3}+19x+26).\sqrt{x+1}$

Giải bất phương trình$3\sqrt{2(3x+4)^{3}}\leq (x^{2}+2x-2)(\sqrt{x^{2}+2x-3}+19x+26).\sqrt{x+1}$

Phương trình mũ

Bất phương trình

$3\sqrt{2(3x+4)^{3}}\leq (x^{2}+2x-2)(\sqrt{x^{2}+2x-3}+19x+26).\sqrt{x+1}$

Giải bất phương trình$3\sqrt{2(3x+4)^{3}}\leq (x^{2}+2x-2)\sqrt{x^{2}+2x-3}+(19x+26).\sqrt{x+1}$

Phương trình mũ

Bất phương trình