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a/ \(A=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(A=x^3+8-\left[x^3+1+3x\left(x+1\right)\right]+3\left(x^2-1\right)\)
\(A=x^3+8-x^3-1-3x\left(x+1\right)+3x^2-3\)
\(A=-3x^2-3x+3x^2+4\)
\(A=4-3x\)
b/ Để \(\left|A\right|=A\)
=> \(A\ge0\)
<=> \(4-3x\ge0\)
<=> \(4\ge3x\)
<=> \(x\ge\frac{3}{4}\)
Vậy khi \(x\ge\frac{3}{4}\)thì \(\left|A\right|=A\).
a, \(A=\left(\frac{1}{x-2}+\frac{2x}{x^2-4}+\frac{1}{x+2}\right)\left(\frac{2}{x-1}\right)\)ĐK : \(x\ne1;\pm2\)
\(=\left(\frac{x+2+2x+x-2}{x^2-4}\right)\left(\frac{2}{x-1}\right)=\frac{4x}{x^2-4}.\frac{2}{x-1}=\frac{8x}{\left(x-1\right)\left(x-2\right)\left(x+2\right)}\)
b, bạn check lại đề bài nhé
A=(1x−2 +2xx2−4 +1x+2 )(2x−1 )ĐK : x≠1;±2
=(x+2+2x+x−2x2−4 )(2x−1 )=4xx2−4 .2x−1 =8x(x−1)(x−2)(x+2)
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
a, \(A=\left(\frac{1}{x-2}+\frac{2x}{x^2-4}+\frac{1}{x+2}\right).\left(\frac{2}{x}-1\right)\) \(\left(ĐK:x\ne\pm2\right)\)
\(A=\left(\frac{x+2}{\left(x-2\right)\left(x+2\right)}+\frac{2x}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right).\left(\frac{2}{x}-\frac{x}{x}\right)\)
\(A=\frac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}.\frac{2-x}{x}\)
\(A=\frac{4x}{\left(x-2\right)\left(x+2\right)}.\frac{-\left(x-2\right)}{x}\)
\(A=\frac{-4x.\left(x-2\right)}{\left(x-2\right)\left(x+2\right).x}\)
\(A=\frac{-4}{x+2}\)
b, \(A=\frac{-4}{x+2}=1\)
\(\rightarrow\frac{-4}{x+2}=\frac{x+2}{x+2}\)
\(\rightarrow-4=x+2\)
\(\rightarrow-6=x\)
a) ĐKXĐ của A là x\(\ne\pm2\); x\(\ne1\)
Ta có
A= \((\frac{1}{x-2}+\frac{2x}{x^2-4}+\frac{1}{x+2})\cdot\frac{2}{x-1}\)
A=\(\frac{x+2+2x+x-2}{\left(x-2\right)\cdot\left(x+2\right)}\cdot\frac{2}{x-1}\)
A=\(\frac{6x}{\left(x-1\right)\left(x-2\right)\left(x+2\right)}\)
a) Ta có: \(A=\left(\dfrac{1}{x-2}+\dfrac{2x}{x^2-4}+\dfrac{1}{x+2}\right)\cdot\left(\dfrac{2}{x}-1\right)\)
\(=\dfrac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{2-x}{x}\)
\(=\dfrac{4x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)}{x}\)
\(=\dfrac{-4}{x+2}\)
b) Để A=1 thì x+2=-4
hay x=-6(nhận)