pt $\Leftrightarrow 36x^3-36x^2+9x-1=0$Đặt $y=x-1/3\Rightarrow x=y+1/3$Ta có $36(y+ \frac{1}{3})^3-36(y+\frac{1}{3})^2+9(y+\frac{1}{3})-1=0\Leftrightarrow 36y^3-3y-\frac{2}{3}=0$Đặt $a,b$ tùy ý sao cho $\begin{cases}a+b=y \\ a \geq b \end{cases}$ $\Rightarrow 36(a+b)^3-3(a+b)= \frac{2}{3}\Rightarrow 36(a^3+b^3)+108ab(a+b)-3(a+b)=\frac{2}{3}$$\Rightarrow36(a^3+b^3)+(108ab-3)(a+b)=\frac{2}{3}(*)$Chọn $a,b$ sao cho $108ab-3=0$ hay $ab=\frac{1}{36}$$*\Leftrightarrow 36(a^3+b^3)=\frac{2}{3}\Rightarrow a^3+b^3=\frac{1}{54}$Ta có $\begin{cases}a^3+b^3=\frac{1}{54} \\ a^3b^3=\frac{1}{36^3}\end{cases}$Théo Vi-ét thì $a^3,b^3$ là 2 nghiệm của pt $X^2-\frac{1}{54}.X+\frac{1}{36^3}=0$mà $a \geq b\Rightarrow a^3=\frac{\frac{1}{54}+\frac{1}{36\sqrt{3}}}{2},b^3=\frac{\frac{1}{54}+\frac{1}{36\sqrt{3}}}{2}$$\Rightarrow x=y+\frac{1}{3}=a+b+\frac{1}{3}=\color{red}{\sqrt[3]{\frac{\frac{1}{54}+\frac{1}{36\sqrt{3}}}{2}}+\sqrt[3]{\frac{\frac{1}{54}+\frac{1}{36\sqrt{3}}}{2}}+\frac{1}{3}}$

pt $\Leftrightarrow 36x^3-36x^2+9x-1=0$Đặt $y=x-1/3\Rightarrow x=y+1/3$Ta có $36(y+ \frac{1}{3})^3-36(y+\frac{1}{3})^2+9(y+\frac{1}{3})-1=0\Leftrightarrow 36y^3-3y-\frac{2}{3}=0$Đặt $a,b$ tùy ý sao cho $\begin{cases}a+b=y \\ a \geq b \end{cases}$ $\Rightarrow 36(a+b)^3-3(a+b)= \frac{2}{3}\Rightarrow 36(a^3+b^3)+108ab(a+b)-3(a+b)=\frac{2}{3}$$\Rightarrow36(a^3+b^3)+(108ab-3)(a+b)=\frac{2}{3}(*)$Chọn $a,b$ sao cho $108ab-3=0$ hay $ab=\frac{1}{36}$$*\Leftrightarrow 36(a^3+b^3)=\frac{2}{3}\Rightarrow a^3+b^3=\frac{1}{54}$Ta có $\begin{cases}a^3+b^3=\frac{1}{54} \\ a^3b^3=\frac{1}{36^3}\end{cases}$Théo Vi-ét thì $a^3,b^3$ là 2 nghiệm của pt $X^2-\frac{1}{54}.X+\frac{1}{36^3}=0$mà $a \geq b\Rightarrow a^3=\frac{\frac{1}{54}+\frac{1}{36\sqrt{3}}}{2},b^3=\frac{\frac{1}{54}-\frac{1}{36\sqrt{3}}}{2}$$\Rightarrow x=y+\frac{1}{3}=a+b+\frac{1}{3}=\color{red}{\sqrt[3]{\frac{\frac{1}{54}+\frac{1}{36\sqrt{3}}}{2}}+\sqrt[3]{\frac{\frac{1}{54}-\frac{1}{36\sqrt{3}}}{2}}+\frac{1}{3}}$

PT$ \color{green}{\Leftrightarrow [(x+2)^2-1]+(x+3)^3+[(x+4)^4-1=0]}$$\color{green}{\Leftrightarrow (x+2+1)(x+2-1)+(x+3)^3+[(x+4)^2-1][(x+4)^2+1]=0}$$\color{green}{\Leftrightarrow (x+3)(x+1)+(x+3)^3+(x+4+1)(x+4-1)(x^2+8x+17)=0}$$\color{green}{\Leftrightarrow (x+3)[(x+1)+(x+3)^2]+(x+3)(x+5)(x^2+8x+17)=0}$$\color{green}{\Leftrightarrow (x+3)(x^2+7x+10)+(x+3)(x+5)(x^2+8x+17)=0}$$\color{green}{\Leftrightarrow (x+3)(x+5)(x+2)+(x+3)(x+5)(x^2+8x+17)=0}$$\color{green}{\Leftrightarrow (x+3)(x+5)(x^2+9x+19)=0}$Vậy phương trình có 4 nghiệm $\color{red}{x_1=-3,x_2=-5,x_{3,4}=\frac{-9\pm \sqrt{5}}{2}}$

PT$ \color{green}{\Leftrightarrow [(x+2)^2-1]+(x+3)^3+[(x+4)^4-1]=0}$$\color{green}{\Leftrightarrow (x+2+1)(x+2-1)+(x+3)^3+[(x+4)^2-1][(x+4)^2+1]=0}$$\color{green}{\Leftrightarrow (x+3)(x+1)+(x+3)^3+(x+4+1)(x+4-1)(x^2+8x+17)=0}$$\color{green}{\Leftrightarrow (x+3)[(x+1)+(x+3)^2]+(x+3)(x+5)(x^2+8x+17)=0}$$\color{green}{\Leftrightarrow (x+3)(x^2+7x+10)+(x+3)(x+5)(x^2+8x+17)=0}$$\color{green}{\Leftrightarrow (x+3)(x+5)(x+2)+(x+3)(x+5)(x^2+8x+17)=0}$$\color{green}{\Leftrightarrow (x+3)(x+5)(x^2+9x+19)=0}$Vậy phương trình có 4 nghiệm $\color{red}{x_1=-3,x_2=-5,x_{3,4}=\frac{-9\pm \sqrt{5}}{2}}$

Pt $\Leftrightarrow ( \frac{1}{x}+\frac{1}{x+7})+(\frac{1}{x+2}+\frac{1}{x+5})=(\frac{1}{x+1}+\frac{1}{x+6})+(\frac{1}{x+3}+\frac{1}{x+4})$$\Leftrightarrow\frac{2x+7}{x(x+7)}+\frac{2x+7}{(x+2)(x+5)}-\frac{2x+7}{(x+1)(x+6)}-\frac{2x+7}{(x+3)(x+4)}=0$$\Leftrightarrow(2x+7)[(\frac{1}{x^2+7}-\frac{1}{x^2+7x+6})+(\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+12)]}=0$Dễ thấy $\frac{1}{x^2+7}>\frac{1}{x^2+7x+6},\frac{1}{x^2+7x+10}>\frac{1}{x^2+7x+12}$$\Rightarrow(\frac{1}{x^2+7}-\frac{1}{x^2+7x+6})+(\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+12)}\neq 0$$\Rightarrow2x+7=0\Rightarrow \color{red}{x=\frac{-7}{2}}$

Điều kiện $\color{red}{x\neq 0,-1,-2,-3,-4,-5,-6,-7}$Pt $\Leftrightarrow ( \frac{1}{x}+\frac{1}{x+7})+(\frac{1}{x+2}+\frac{1}{x+5})=(\frac{1}{x+1}+\frac{1}{x+6})+(\frac{1}{x+3}+\frac{1}{x+4})$$\Leftrightarrow\frac{2x+7}{x(x+7)}+\frac{2x+7}{(x+2)(x+5)}-\frac{2x+7}{(x+1)(x+6)}-\frac{2x+7}{(x+3)(x+4)}=0$$\Leftrightarrow(2x+7)[(\frac{1}{x^2+7}-\frac{1}{x^2+7x+6})+(\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+12)]}=0$Dễ thấy $\frac{1}{x^2+7}>\frac{1}{x^2+7x+6},\frac{1}{x^2+7x+10}>\frac{1}{x^2+7x+12}$$\Rightarrow(\frac{1}{x^2+7}-\frac{1}{x^2+7x+6})+(\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+12)}\neq 0$$\Rightarrow2x+7=0\Rightarrow \color{red}{x=\frac{-7}{2}}$ (thõa đk)

Đặt $\color{red}{ x^2=y \geq 0}$Ta có $ \color{red}{ y^3-7y+\sqrt{6}=0\Rightarrow (y^3+\sqrt{6}y^2-y)-( \sqrt6y^2+6y-\sqrt{6})=0}$$\color{red}{\Rightarrow y(y^2+\sqrt{6}y-1)-\sqrt{6}(y^2+\sqrt{6}y-1)=0}$$\color{red}{\Rightarrow (y-\sqrt{6})(y^2+\sqrt{6}y-1)=0}$$\color{red}{\Rightarrow y=\sqrt{6}}$ hoặc $\color{red}{y^2+\sqrt{6}y-1=0(\bigstar)}$Phương trình $\color{red}{(\bigstar)}$ có 2 nghiệm $\color{red}{y=\frac{\pm\sqrt{10}-\sqrt{6}}{2}}$Ta có nghiệm $\color{red}{y=\frac{\sqrt{10}-\sqrt{6}}{2}}$ thõa vì nghiệm còn lại âm$\color{red}{\Rightarrow x_{1,2}=\pm \sqrt[4]{6},x_{3,4}=\pm \sqrt{\frac{\sqrt{10}-\sqrt{2}}{2}}}$

Đặt $\color{red}{ x^2=y \geq 0}$Ta có $ \color{red}{ y^3-7y+\sqrt{6}=0\Rightarrow (y^3+\sqrt{6}y^2-y)-( \sqrt6y^2+6y-\sqrt{6})=0}$$\color{red}{\Rightarrow y(y^2+\sqrt{6}y-1)-\sqrt{6}(y^2+\sqrt{6}y-1)=0}$$\color{red}{\Rightarrow (y-\sqrt{6})(y^2+\sqrt{6}y-1)=0}$$\color{red}{\Rightarrow y=\sqrt{6}}$ hoặc $\color{red}{y^2+\sqrt{6}y-1=0(\bigstar)}$Phương trình $\color{red}{(\bigstar)}$ có 2 nghiệm $\color{red}{y=\frac{\pm\sqrt{10}-\sqrt{6}}{2}}$Ta có nghiệm $\color{red}{y=\frac{\sqrt{10}-\sqrt{6}}{2}}$ thõa vì nghiệm còn lại âm$\color{red}{\Rightarrow x_{1,2}=\pm \sqrt[4]{6},x_{3,4}=\pm \sqrt{\frac{\sqrt{10}-\sqrt{6}}{2}}}$