Bài 1: Thực hiện phép tính
a) [ \(\frac{1}{\left(2x-y\right)^2}\) + \(\frac{2}{4x^2-y^2}\) + \(\frac{1}{\left(2x+y\right)^2}\) ] : \(\frac{16x}{4x^2+4xy+y^2}\)
b) \(\frac{3}{3x+3}\) + \(\frac{10}{5-5x}\) + \(\frac{5x-1}{x^2-1}\)
c) A = ( x\(^4\) - x\(^2\) + 2x - 1) : ( x\(^2\) + x - 1) - ( x\(^2\) - x )
Bài 1:
a) Ta có: \(\left(\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right):\frac{16x}{4x^2+4xy+y^2}\)
\(=\left(\frac{\left(2x+y\right)^2}{\left(2x-y\right)^2\cdot\left(2x+y\right)^2}+\frac{2\cdot\left(2x+y\right)\left(2x-y\right)}{\left(2x+y\right)^2\cdot\left(2x-y\right)^2}+\frac{\left(2x-y\right)^2}{\left(2x+y\right)^2\cdot\left(2x-y\right)^2}\right)\cdot\frac{\left(2x+y\right)^2}{16x}\)
\(=\frac{\left(2x+y+2x-y\right)^2}{\left(2x-y\right)^2\cdot\left(2x+y\right)^2}\cdot\frac{\left(2x+y\right)^2}{16x}\)
\(=\frac{\left(4x\right)^2}{\left(2x-y\right)^2}\cdot\frac{1}{16x}\)
\(=\frac{16x^2}{16x\cdot\left(2x-y\right)^2}\)
\(=\frac{x}{\left(2x-y\right)^2}\)
b) Ta có: \(\frac{3}{3x+3}+\frac{10}{5-5x}+\frac{5x-1}{x^2-1}\)
\(=\frac{1}{x+1}-\frac{2}{x-1}+\frac{5x-1}{x^2-1}\)
\(=\frac{x-1}{\left(x+1\right)\left(x-1\right)}-\frac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{5x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x-1-2\left(x+1\right)+5x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x-1-2x-2+5x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{4x-4}{\left(x-1\right)\left(x+1\right)}=\frac{4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{4}{x+1}\)
c) Ta có: \(A=\left(x^4-x^2+2x-1\right):\left(x^2+x-1\right)-\left(x^2-x\right)\)
\(=\frac{\left(x^2\right)^2-\left(x^2-2x+1\right)}{x^2+x-1}-x^2+x\)
\(=\frac{\left(x^2\right)^2-\left(x-1\right)^2}{x^2+x-1}-x^2+x\)
\(=\frac{\left(x^2-x+1\right)\left(x^2+x-1\right)}{x^2+x-1}-x^2+x\)
\(=x^2-x+1-x^2+x\)
=1
