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a, \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}=\frac{x+1}{2\left(x+3\right)}+\frac{3x+2}{x\left(x+3\right)}\)
\(=\frac{x^2+x}{2x\left(x+3\right)}+\frac{6x+4}{2x\left(x+3\right)}=\frac{x^2+7x+4}{2x\left(x+3\right)}\)
b, Sua de : \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)
\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{2x+6}{2x\left(x+3\right)}=\frac{1}{x}\)
a: \(E=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x^2-1+x+2-x^2}\)
\(=\dfrac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2}{x-1}\)
c: |2x+1|=5
=>2x+1=5 hoặc 2x+1=-5
=>x=-3(nhận) hoặc x=2(nhận)
Khi x=-3 thì \(E=\dfrac{\left(-3\right)^2}{-3-1}=-\dfrac{9}{4}\)
Khi x=2 thì \(E=\dfrac{2^2}{2-1}=4\)
b:
ĐKXĐ: \(x\notin\left\{0;2;-2\right\}\)
\(\left(\dfrac{4}{x^3-4x}+\dfrac{1}{x+2}\right):\left(\dfrac{x-2}{x^2+2x}-\dfrac{x}{2x+4}\right)\)
\(=\left(\dfrac{4}{x\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right):\left(\dfrac{x-2}{x\left(x+2\right)}-\dfrac{x}{2\left(x+2\right)}\right)\)
\(=\dfrac{4+x\left(x-2\right)}{x\left(x-2\right)\cdot\left(x+2\right)}:\dfrac{2\left(x-2\right)-x^2}{x\left(x+2\right)\cdot2}\)
\(=\dfrac{x^2-2x+4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{2x\left(x+2\right)}{-\left(x^2-2x+4\right)}\)
\(=\dfrac{-2}{x-2}\)
c:ĐKXĐ: x<>0
\(\left(x-\dfrac{3}{x}\right):\left(\dfrac{x^2+2x+1}{x}-\dfrac{2x+4}{x}\right)\)
\(=\dfrac{x^2-3}{x}:\dfrac{x^2+2x+1-2x-4}{x}\)
\(=\dfrac{x^2-3}{x}\cdot\dfrac{x}{x^2-3}\)
=1
a, \(B=\left(\frac{2x+1}{2x-1}+\frac{4}{1-4x^2}-\frac{2x-1}{2x+1}\right):\frac{x^2+2}{2x+1}\)
\(=\left(\frac{2x+1}{2x-1}+\frac{4}{\left(1-2x\right)\left(2x+1\right)}-\frac{2x-1}{2x+1}\right):\frac{x^2+2}{2x+1}\)
\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{4}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right):\frac{x^2+2}{2x+1}\)
\(=\left(\frac{4x^2+4x+1-4-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\right):\frac{x^2+2}{2x+1}\)
\(=\frac{8x-4}{\left(2x-1\right)\left(2x+1\right)}.\frac{2x+1}{x^2+2}=\frac{8x-4}{\left(2x-1\right)\left(x^2+2\right)}\)
b, Thay x = -1 ta được : \(\frac{9\left(-1\right)-4}{\left[2\left(-1\right)-1\right]\left[\left(-1\right)^2+2\right]}=-\frac{13}{-9}=\frac{13}{9}\)