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a: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)
\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{6\left(x+1\right)\cdot x\left(x+2\right)}{3\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x}{x-2}\)
a: \(A=\dfrac{x^2-8x+16-x^2+16}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-8\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-4x}{\left(x+4\right)\left(x-1\right)}\)
a: Khi x=3 thì \(A=\dfrac{3+2}{3-1}=\dfrac{5}{2}\)
b: \(B=\dfrac{x-1}{x}+\dfrac{2x+1}{x\left(x+1\right)}=\dfrac{x^2-1+2x+1}{x\left(x+1\right)}=\dfrac{x+2}{x+1}\)
\(P=A:B=\dfrac{x+2}{x-1}\cdot\dfrac{x+1}{x+2}=\dfrac{x+1}{x-1}\)
3: Để P>1/3 thì \(P-\dfrac{1}{3}>0\)
=>\(\Leftrightarrow3\left(x+1\right)-x+1>0\)
=>3x+3-x+1>0
=>2x+4>0
hay x>-2
a: ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};\dfrac{1}{2};-2\right\}\)
b: \(B=\dfrac{4x^2+4x+1-4-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{2x+1}{x+2}\)
\(=\dfrac{8x-4}{2x-1}\cdot\dfrac{1}{x+2}=\dfrac{4}{x+2}\)
a: \(A=\dfrac{x+2+x^2-2x+x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2}{x^2-4}\)