1)Giải phương trình
$a)\sqrt[3]{2x+1} - \sqrt[3]{3x-2} = (2x-6)\sqrt{x-1}$
$b)\sqrt[3]{(x-2)^{2}} + \sqrt[3]{x^{2}-4} = 2\sqrt[3]{(x+2)^{2}}$

2)Cho $m = \sqrt{2} + \sqrt[3]{3}$.Lập một phương trình bậc 6 với hệ số nguyên nhận $m$ làm một nghiệm

3)Tìm Max $A = a^{3} + b^{3} + c^{3},$ biết $0 \leq c \leq b \leq a \leq 2$ và $a + b + c = 3$
   Tìm Min $B = (x-2)^{4} + (x-4)^{4} + 6(x-2)^{2}(x-4)^{2}$

4)Tìm $x, y, z$ thỏa mãn phương trình:
   $\sqrt{x-2000} + \sqrt{y-2000} + \sqrt{z-2000} = \frac{1}{2}(x+y+z) - 3000$
3a)
từ điều kiện suy ra $1\leq a\leq 2$
$A=a^3+b^3+c^3=a^3+(b+c)^3-3bc(b+c)\leq a^3+(b+c)^3=a^3+(3-a)^3=9(a^2-3a+3)=9\left ( (a-\frac{3}{2})^2+\frac{3}{4} \right )$
do $1\leq a\leq 2\Rightarrow 0\leq \left| {a-\frac{3}{2}} \right|\leq \frac{1}{2}\Rightarrow (a-\frac{3}{2})^2\leq \frac{1}{4}$
suy ra $A_{max}=9\Leftrightarrow (a;b;c)=(2;1;0)$
1b)
Đặt $a=\sqrt[3]{x-2},b=\sqrt[3]{x+2}$
Được pt:$a^2+ab=2b^2\Leftrightarrow (a-b)(a+2b)=0$
1a)
$\sqrt[3]{2x+1}-\sqrt[3]{3x-2}=(2x-6)\sqrt{x-1}$
ĐKXĐ: $x\geq 1\Rightarrow 2x+1>0,3x-2>0$
PT $\Leftrightarrow (2x-6)\sqrt{x-1}+\sqrt[3]{3x-2}-\sqrt[3]{2x+1}=0$
$\Leftrightarrow 2(x-3)\sqrt{x-1}+\frac{(3x-2)-(2x+1)}{(\sqrt[3]{2x+1})^2+\sqrt[3]{2x+1}\sqrt[3]{3x-2}+(\sqrt[3]{3x-2})^2}=0$
$\Leftrightarrow (x-3)(...)=0\Leftrightarrow x=3$ (do(...)>0)
đặt$4 x-2=a;x-4=b.$ta có :
$B=a^4+b^4+6a^2b^2.$vì $a-b=2$ nên $a^2+2ab+b^2=4$
$(a^2+b^2)^2+4a^2b^2-4ab(a^2+b^2)=16$
$\rightarrow B+4a^2.b^2-4ab(a^2+b^2)=16,mà 2ab=(a^2+b^2)-4$
thay vào :B=.................. tự tính tiếp

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