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

=>\(4x^2-4x+1=4x^2+6x\)

=>\(4x^2-4x+1-4x^2-6x=0\)

=>\(-10x+1=0=>x=\frac{1}{10}\)

                                                      Bài giải

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

\(\left(2x-1\right)^2=2x\left(2x-1+4\right)\)

\(\left(2x-1\right)^2=2x\left(2x-1\right)+8x\)

\(\left(2x-1\right)^2-2x\left(2x-1\right)=8x\)

\(\left(2x-1-2x\right)\left(2x-1\right)=8x\)

\(-2x+1=8x\)

\(8x+2x=1\)

\(10x=1\)

\(x=\frac{1}{10}\)

\(M=a+\frac{\left(2a+b\right)\left(2+b\right)-\left(2a-b\right)\left(2-b\right)}{4-b^2}-\frac{4a}{4-b^2}.\)

\(=a+\frac{4b\left(a+1\right)-4a}{4-b^2}\)

Ta có \(4ab+4b-4a=4\left[\frac{a^2}{a+1}+\frac{a}{a+1}-4a\right]=-12a\)

     \(4-b^2=4-\frac{a^2}{\left(a+1\right)^2}=\frac{4\left(a^2+2a+1\right)-a^2}{\left(a+1\right)^2}=\frac{3a^2+8a+4}{\left(a+1\right)^2}\)

\(\Rightarrow M=a+\frac{-12a\left(a+1\right)^2}{3a^2+8a+4}\)

\(=-\frac{9a^3+16a^2+8a}{3a^2+8a+4}\)

 \(M=a+\frac{2a+b}{2-b}-\frac{2a-b}{2+b}+\frac{4a}{b^2-4}\)

      \(=a-\frac{2a+b}{b-2}-\frac{2a-b}{2+b}+\frac{4a}{b^2-4}\)

      \(=a-\frac{\left(2a+b\right)\left(2+b\right)+\left(2a-b\right)\left(b-2\right)}{\left(b-2\right)\left(b+2\right)}+\frac{4a}{b^2-4}\) 

      \(=a-\frac{4b\left(a+1\right)}{b^2-4}+\frac{4a}{b^2-4}\)

      \(=a-\frac{4\frac{a}{a+1}\left(a+1\right)}{b^2-4}+\frac{4a}{b^2-4}\)

      \(=a-\frac{4a}{b^2-4}+\frac{4a}{b^2-4}\)

      \(=a\)

Hắc hắc :P Cứ làm từ từ sẽ thành công em ạ :D

\(=\frac{a+b+a-b}{a^2-b^2}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)

\(=\frac{2a\left(a^2+b^2\right)+2a\left(a^2-b^2\right)}{a^4-b^4}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)

\(=\frac{4a^3\left(a^4+b^4\right)+4a^3\left(a^4-b^4\right)}{a^8-b^8}+\frac{8a^7}{a^8+b^8}\)

\(=\frac{8a^7\left(a^8+b^8\right)+8a^7\left(a^8-b^8\right)}{\left(a^8-b^8\right)\left(a^8+b^8\right)}\)

\(=\frac{16a^{15}}{a^{16}-b^{16}}\)