a) x1^2+x2^2=(x1+x2)^2-2x1x2

x1^3+x2^3=(x1+x2)(x1^2+x2^2-x1x2)

áp dụng viét thay vô

b) giải hệ pt

đenta>=0

x1+x2=-m

x1x2=m+3

và 2x1+3x2=5

c)thay x=-3 vào tìm ra m rồi thay m đó vô giải ra lại

d)áp dụng viét 

x1+x2=-m

x1x2=m+3

CT liên hệ ko phụ thuộc m là x1 +x2+x1x2=-m+m+3=3

sao ko ai làm hộ tôi vậy 

a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)

c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)

\(=\dfrac{3}{\sqrt{x}-2}\)

\(\frac{2x}{x+3\sqrt{x}+2}+\frac{5\sqrt{x}+1}{x+4\sqrt{x}+3}+\frac{\sqrt{x}+10}{x+5\sqrt{x}+6}\)

\(=\frac{2x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}+\frac{5\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}+10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{2x\left(\sqrt{x}+3\right)+\left(5\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+\left(\sqrt{x}+10\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{2\sqrt{x^3}+6x+5x+11\sqrt{x}+2+x+11\sqrt{x}+10}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{12x+22\sqrt{x}+2\sqrt{x^3}+12}{6x+11\sqrt{x}+\sqrt{x^3}+6}\)

\(=\frac{2\left(6x+11\sqrt{x}+\sqrt{x^3}+6\right)}{6x+11\sqrt{x}+\sqrt{x^3}+6}\)

\(=2\) (ko phụ thuộc vào biến ) (đpcm)

a,Ta có  \(x=4-2\sqrt{3}=\sqrt{3}^2-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)do \(\sqrt{3}-1>0\)

\(\Rightarrow A=\frac{1}{\sqrt{3}-1-1}=\frac{1}{\sqrt{3}-2}\)

b, Với \(x\ge0;x\ne1\)

 \(B=\left(\frac{-3\sqrt{x}}{x\sqrt{x}-1}-\frac{1}{1-\sqrt{x}}\right):\left(1-\frac{x+2}{1+\sqrt{x}+x}\right)\)

\(=\left(\frac{-3\sqrt{x}+x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x+\sqrt{x}+1-x-2}{x+\sqrt{x}+1}\right)\)

\(=\left(\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\right)\)

\(=\frac{\sqrt{x}-1}{x+\sqrt{x}+1}.\frac{x+\sqrt{x}+1}{\sqrt{x}-1}=1\)

Vậy biểu thức ko phụ thuộc biến x 

c, Ta có : \(\frac{2A}{B}\)hay \(\frac{2}{\sqrt{x}-1}\)để biểu thức nhận giá trị nguyên 

thì \(\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\left(\sqrt[3]{x}+1\right)^3-\left(\sqrt[3]{x}-1\right)^3-6\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x}+1\right)\\ =x+3\sqrt[3]{x^2}+3\sqrt[3]{x}+1-\left(x-3\sqrt[3]{x^2}+3\sqrt[3]{x}-1\right)-6\left(\sqrt[3]{x^2}-1\right)\\ =x+3\sqrt[3]{x^2}+3\sqrt[3]{x}+1-x+3\sqrt[3]{x^2}-3\sqrt[3]{x}+1-6\sqrt[3]{x^2}+6\\ =8\)

Lời giải:
Gọi biểu thức là $A$

\(A=(x+3\sqrt[3]{x^2}+3\sqrt[3]{x}+1)-(x-3\sqrt[3]{x^2}+3\sqrt[3]{x}-1)-6(\sqrt[3]{x^2}-1)\)

\(6\sqrt[3]{x^2}+2-6(\sqrt[3]{x^2}-1)=8\) là giá trị không phụ thuộc vào biến.