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a: \(A=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{\left(x-2\right)}+\dfrac{1}{x+2}\right):\dfrac{x^2-4+10-x^2}{x+2}\)
\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{6}=\dfrac{-1}{x-2}\)
b: |x|=1/2 khi x=1/2 hoặc x=-1/2
Khi x=1/2 thì \(A=\dfrac{-1}{\dfrac{1}{2}-2}=-1:\dfrac{-3}{2}=\dfrac{2}{3}\)
Khi x=-1/2 thì \(A=\dfrac{-1}{-\dfrac{1}{2}-2}=-1:\dfrac{-5}{2}=\dfrac{2}{5}\)
c: Để A=2 thì x-2=-1/2
hay x=3/2
d:Để A<0 thì x-2>0
hay x>2
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
\(A=[\dfrac{2}{\left(x+1\right)^3}.\dfrac{1+x}{x}+\left(\dfrac{1}{\left(x+1\right)^2}.\dfrac{1+x^2}{x^2}\right)].\dfrac{x^3}{x-1}=\left(\dfrac{2+2x}{x\left(x+1\right)^3}+\dfrac{1+x^2}{x^2}\right).\dfrac{x^3}{x-1}=\dfrac{2x+2x^2+\left(1+x^2\right)\left(x+1\right)}{x^2\left(x+1\right)^3}.\dfrac{x^3}{x-1}=\dfrac{2x\left(1+x\right)+\left(1+x^2\right)\left(x+1\right)}{x^2\left(x+1\right)^3}.\dfrac{x^3}{x-1}=\dfrac{\left(x+1\right)\left(2x+1+x^2\right)}{x^2\left(x+1\right)^3}.\dfrac{x^3}{x-1}=\dfrac{\left(x+1\right)^3}{x^2\left(x+1\right)^3}.\dfrac{x^3}{x-1}=\dfrac{x\left(x+1\right)}{x-1}=\dfrac{x^2+x}{x-1}\)
ý a có bn lm rồi, mk lm ý b,c thôi nhé
b/ A < 1 \(\Leftrightarrow\dfrac{x^2+x}{x-1}< 1\)
\(\Leftrightarrow x^2+x< x-1\)
\(\Leftrightarrow x^2+x-x+1< 0\)
\(\Leftrightarrow x^2+1< 0\)
\(\Leftrightarrow x^2< -1\) (vô lí)
Vậy k có gt nào của x t/m
c/ \(\dfrac{x^2+x}{x-1}=\dfrac{x^2+x-2+2}{x-1}=\dfrac{\left(x+2\right)\left(x-1\right)+2}{x-1}\)
\(=\dfrac{\left(x+2\right)\left(x-1\right)}{x-1}+\dfrac{2}{x-1}=x+2+\dfrac{2}{x-1}\)
Để A \(\in\) Z <=> \(\dfrac{2}{x-1}\in Z\Leftrightarrow x-1\inƯ\left(2\right)\)
\(\Leftrightarrow x-1=\left\{\pm1;\pm2\right\}\)
\(\Leftrightarrow x=\left\{-1;0;2;3\right\}\)
Vậy....
a: \(N=\left(\dfrac{\left(1-a\right)\left(a^2+a+1\right)}{1-a}-a\right)\cdot\dfrac{a^3-a^2-a+1}{-\left(a^2-1\right)}\)
\(=\left(a^2+1\right)\cdot\dfrac{a^2\left(a-1\right)-\left(a-1\right)}{-\left(a-1\right)\left(a+1\right)}\)
\(=-\left(a^2+1\right)\cdot\dfrac{\left(a-1\right)\left(a^2-1\right)}{\left(a-1\right)\left(a+1\right)}\)
\(=-\left(a^2+1\right)\cdot\left(a-1\right)\)
b: Để N<0 thì \(-\left(a^2+1\right)\left(a-1\right)< 0\)
\(\Leftrightarrow\left(a^2+1\right)\left(a-1\right)>0\)
=>a-1>0
hay a>1