Bài 2: 

Ta có: \(x\left(x-4\right)-x^2+8=0\)

\(\Leftrightarrow x^2-4x-x^2+8=0\)

\(\Leftrightarrow-4x=-8\)

hay x=2

1)=x²-49-x²

=-49

2)=>x²-4x-x²+8=0

=>-4x+8=0

=>-4x=-8

=>x=2

\(A=\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}+\sqrt{13-4\sqrt{3}}-\sqrt{22+12\sqrt{2}}\)

\(=\left|2\sqrt{3}-3\sqrt{2}\right|+\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}+\sqrt{1^2}}-\sqrt{\left(3\sqrt{2}\right)^2+2.2.3\sqrt{2}+2^2}\)

\(=-2\sqrt{3}+3\sqrt{2}+\sqrt{\left(2\sqrt{3}-1\right)^2}-\sqrt{\left(3\sqrt{2}+2\right)^2}\)

\(=-2\sqrt{3}+3\sqrt{2}+\left|2\sqrt{3}-1\right|-\left|3\sqrt{2}+2\right|\)

\(=-2\sqrt{3}+3\sqrt{2}+2\sqrt{3}-1-3\sqrt{2}-2\)

\(=-3\)

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

\(A=\dfrac{x}{x-2}-\dfrac{x^2+x-2}{x^2-4}=\dfrac{x^2+2x-x^2-x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\)

\(A=\dfrac{x}{x-2}+\dfrac{x^2+x-2}{4-x^2}\left(x\ne\pm2\right).\)

\(A=\dfrac{x}{x-2}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x-2}-\dfrac{x-1}{x-2}=\dfrac{x-x+1}{x-2}=\dfrac{1}{x-2.}\)

A= \(\dfrac{x}{x-2}+\dfrac{x^2+x-2}{4-x^2}=\dfrac{-x\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}+\dfrac{x^2+x-2}{\left(2-x\right)\left(2+x\right)}=\dfrac{-2x-x^2+x^2+x-2}{\left(2-x\right)\left(2+x\right)}=\dfrac{-x-2}{\left(2-x\right)\left(2-x\right)}=\dfrac{-1\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}=\dfrac{-1}{2-x}\)

\(a,=-x^3+3x^2+2x-6\)

\(\left(x^2-2\right)\left(-x+3\right)\)

\(=-x^3+3x^2+2x-6\)

a: \(\left(1-cosx\right)\left(1+cosx\right)=1^2-cos^2x=sin^2x\)

b: \(tan^2x\left(2cos^2x+sin^2x-1\right)\)

\(=tan^2x\left(1-1+cos^2x\right)\)

\(=\dfrac{sin^2x}{cos^2x}\cdot cos^2x=sin^2x\)

c: \(sin^4x+cos^4x+2\cdot cos^2x\cdot sin^2x\)

\(=\left(sin^2x+cos^2x\right)^2\)

\(=1^2=1\)

\(A=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=\sqrt{3}\)

giúp minh