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

\(=\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)

\(=\left[\left(x+y\right)-\left(x-y\right)\right]^2=\left(x+y-x+y\right)^2=4y^2\)

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h như cái phần bỏ đi

chả có ai thèm dùng

a, \(A=\left(\frac{2}{x-2}-\frac{2}{x+2}\right)\frac{x^2+4x+4}{8}\)ĐK : \(x\ne\pm2\)

\(=\left(\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right)\frac{\left(x+2\right)^2}{8}\)

\(=\frac{2x+2-2x+2}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}=\frac{4}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}\)

\(=\frac{x+2}{2\left(x-2\right)}=\frac{x+2}{2x-4}\)

b, A = x hay 

\(\frac{x+2}{2x-4}=x\Leftrightarrow x+2=2x^2-4x\)

\(\Leftrightarrow5x+2-2x^2=0\)vô nghiệm 

tương tự với A = x/2 nhé ! 

a, \(M=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

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

\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)

\(=\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\frac{x^2-12-x}{\left(x-2\right)\left(x+3\right)}\)

\(=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-4}{x-2}\)

c, Đặt \(\frac{x-4}{x-2}=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)( thỏa mãn )

Thử : \(\frac{x-4}{x-2}=\frac{4-4}{4-2}=0\)