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Bài 3:
a: ĐKXĐ: x<>2
b: \(M=\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}\)
c: Khi x=4001/2000 thì \(M=\dfrac{3}{\dfrac{4001}{2000}-2}=3:\dfrac{1}{2000}=6000\)
a) \(\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
\(\Leftrightarrow\left(\dfrac{x-2x-4+x-2}{x^2-4}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)
\(\Leftrightarrow\left(\dfrac{-6}{x^2-4}\right).\left(\dfrac{x+2}{6}\right)\Leftrightarrow\dfrac{1}{2-x}\)
b) với \(x^2=2x\Leftrightarrow x^2-2x+1-1=0\)
\(\Leftrightarrow\left(x-1\right)^2=1\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(KTM\right)\\x=0\left(TM\right)\end{matrix}\right.\)
Vậy với x=0 thì A = \(\dfrac{1}{2}\)
c) A nhận giá trị nguyên dương tức \(\dfrac{1}{2-x}\) nhận giá trị dương
Để A nhận giá trị dương thì
\(2-x\inƯ_{\left(1\right)}\)\(\Leftrightarrow2-x\in\left\{-1;1\right\}\)
| \(2-x\) | -1(loại) | 1 |
| \(x\) | 3(loại) | 1 |
Vậy x=1 thì A nhận giá trị dương
a) Rút gọn A
\(A=\left(\dfrac{1}{2-x}+\dfrac{1}{2+x}\right):\left(\dfrac{1}{2-x}-\dfrac{1}{2+x}\right)+\dfrac{2}{2+x}\)
ĐKXĐ : \(\left\{{}\begin{matrix}2-x\ne0\\2+x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.andx\ne0\)
Ta có : \(A=\left(\dfrac{2+x}{\left(2-x\right)\left(2+x\right)}+\dfrac{2-x}{\left(2-x\right)\left(2+x\right)}\right):\left(\dfrac{2+x}{\left(2-x\right)\left(2+x\right)}-\dfrac{2-x}{\left(2-x\right)\left(2+x\right)}\right)+\dfrac{2}{2+x}\)
\(A=\dfrac{4}{\left(2-x\right)\left(2+x\right)}.\dfrac{\left(2-x\right)\left(2+x\right)}{2+x}+\dfrac{2}{2+x}\)
\(A=\dfrac{4}{2+x}+\dfrac{2}{2+x}\)
\(A=\dfrac{6}{2+x}\)
Câu 3:
\(\Leftrightarrow3x^3-2x^2+6x^2-4x+9x-6>0\)
\(\Leftrightarrow\left(3x-2\right)\left(x^2+2x+3\right)>0\)
=>3x-2>0
=>x>2/3
Câu 1:
a: \(A=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{x+1+2x-2}{\left(x^2-1\right)}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{3x-1}{x^2-1}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{3x^2-x-3x^2+3}{x\left(x^2-1\right)}\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{-\left(x-3\right)}{x\left(x+2\right)}\)
\(=x-2+\dfrac{6x-3-x^2+3x}{x\left(x+2\right)}\)
\(=x-2+\dfrac{-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x\left(x^2-4\right)-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-4x-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-x^2+5x-3}{x\left(x+2\right)}\)
b: TH1: \(\left\{{}\begin{matrix}x^3-x^2+5x-3>0\\x\left(x+2\right)< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2< x< 2\\x>0.63\end{matrix}\right.\Leftrightarrow0.63< x< 2\)
TH2: \(\left\{{}\begin{matrix}x^3-x^2+5x-3< 0\\x\left(x+2\right)>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0.63\\\left[{}\begin{matrix}x>0\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< x< 0.63\\x< -2\end{matrix}\right.\)
ta có : \(\left(a-\dfrac{x^2+a^2}{x+a}\right).\left(\dfrac{2a}{x}-\dfrac{4a}{x-a}\right)\)
\(=\dfrac{-x^2-a^2+ax+a^2}{x+a}.\dfrac{2a\left(x-a\right)-4ax}{x\left(x-a\right)}\)
\(=\dfrac{-x^2+ax}{x+a}.\dfrac{2ax-2a^2-4ax}{x\left(x-a\right)}\)
\(=\dfrac{-x\left(x-a\right)}{x+a}.\dfrac{-2a^2-2ax}{x\left(x-a\right)}\)
\(=\dfrac{-x\left(x-a\right)}{x+a}.\dfrac{-2a\left(a+x\right)}{x\left(x-a\right)}=\dfrac{2a}{1}=2a\) vì a nguyên \(\Rightarrow2a\) nguyên (đpcm)