Bài 5:
a: \(\frac{15x}{7y^3}\cdot\frac{2y^2}{x^2}=\frac{15x}{x^2}\cdot\frac{2y^2}{7y^3}=\frac{15}{x}\cdot\frac{2}{7y}=\frac{30}{7xy}\)
b: \(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)
\(=\frac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\frac{-2\left(x-2\right)}{x+2}=\frac{5\cdot\left(-2\right)}{4\cdot1}=\frac{-10}{4}=-\frac52\)
c: \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)
\(=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}\cdot\frac{3x}{2\left(1-2x\right)}\)
\(=\frac{3\left(1+2x\right)}{2\left(x+4\right)}\)
d: \(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)\)
\(=\frac{1-x\left(2-x\right)}{x\left(x+1\right)}:\frac{1+x^2-2x}{x}\)
\(=\frac{1-2x+x^2}{x\cdot\left(x+1\right)}\cdot\frac{x}{1-2x+x^2}=\frac{x}{x\left(x+1\right)}=\frac{1}{x+1}\)
Bài 4:
a: ĐKXĐ: x<>1; x<>-1; x<>-1/2
\(A=\left(\frac{1}{x-1}-\frac{x}{1-x^3}\cdot\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x+1\right)^2}{2x+1}=\frac{x+1}{x-1}\)
b: Khi x=1/2 thì \(A=\left(\frac12+1\right):\left(\frac12-1\right)=\frac32:\frac{-1}{2}=-3\)
Bài 2:
a: ĐKXĐ: \(x^3-8<>0\)
=>\(x^3<>8\)
=>x<>2
b: \(\frac{3x^2+6x+12}{x^3-8}\)
\(=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)
