Question

[Thanh Thủy]

B=\(\left(\frac{x^2-2}{x^2+2x}+\frac{1}{x+2}\right).\frac{x+1}{x-1}\)

a, CMR B= \(\frac{x+1}{x}\)v ới x#0,x#-2 ,x#1

b. tìm x để 2B=2x+5

Answer 1

[Trà My]

a)\(B=\left(\frac{x-2}{x^2+2x}+\frac{1}{x+2}\right).\frac{x+1}{x-1}=\left(\frac{x^2-2}{x^2+2x}+\frac{x}{x^2+2x}\right).\frac{x+1}{x-1}=\frac{x^2+x-2}{x^2+2x}.\frac{x+1}{x-1}\)

\(=\frac{x^2-x+2x-2}{x\left(x+2\right)}.\frac{x+1}{x-1}=\frac{x\left(x-1\right)+2\left(x-1\right)}{x\left(x+2\right)}.\frac{x+1}{x-1}=\frac{\left(x-1\right)\left(x+2\right)}{x\left(x+2\right)}.\frac{x+1}{x-1}=\frac{x+1}{x}\)

b)\(2B=2x+5\Leftrightarrow2.\frac{x+1}{x}=2x+5\Leftrightarrow\frac{2x+2}{x}=2x+5\Leftrightarrow2x+2=2x^2+5x\)

\(\Leftrightarrow0=2x^2+3x-2\Leftrightarrow2x^2+4x-x-2=0\Leftrightarrow2x\left(x+2\right)-\left(x+2\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x+2\right)=0\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-2\end{cases}}\)

Answer 2

[Thanh Thủy]

cảm ơn bạn nhé