\(\left(2x+1\right)^2+\left(3x-1\right)^2+2\left(2x+1\right)\left(3x-1\right)\)

\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)

\(=\left[\left(2x+1\right)+\left(3x-1\right)\right]^2=\left(2x+3x+1-1\right)^2=25x^2\)

\(\left(2x+1\right)^2+\left(3x-1\right)^2+2\left(2x+1\right)\)\(\left(3x-1\right)\)

\(=\left(2x+1\right)^2+2\times\left(2x+1\right)\left(3x-1\right)\)\(+\left(3x-1\right)^2\)

\(=\left[\left(2x+1\right)+\left(3x-1\right)\right]^2\)

\(=\left(2x+1+3x-1\right)^2\)

\(=\left(5x\right)^2\)

\(=25x^2\)

Học tốt ~!

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

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

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

olm .vn

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 !