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a) ĐKXĐ: \(x\ne-2;x\ne2\), rút gọn:
\(A=\left[\frac{3\left(x-2\right)-2x\left(x+2\right)+2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right]\div\frac{2x-1}{4\left(x-2\right)}\)
\(A=\frac{3x-6-2x^2-4x+4x^2+6}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{4\left(x-2\right)}{2x-1}=\frac{4\left(2x^2-x\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4x\left(2x-1\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4}{x+2}\)
b) Ta có: \(\left|x-1\right|=3\Leftrightarrow\hept{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\left(n\right)\\x=-2\left(l\right)\end{cases}}}\)
=> Khi \(x=4\)thì \(A=\frac{4}{4+2}=\frac{4}{6}=\frac{2}{3}\)
c) \(A< 2\Leftrightarrow\frac{4}{x+2}< 2\Leftrightarrow4< 2x+4\Leftrightarrow0< 2x\Leftrightarrow x>0\)Vậy \(A< 2,\forall x>0\)
d) \(\left|A\right|=1\Leftrightarrow\left|\frac{4}{x+2}\right|=1\Leftrightarrow\hept{\begin{cases}\frac{4}{x+2}=1\\\frac{4}{x+2}=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\left(l\right)\\x=-6\left(n\right)\end{cases}}}\)Vậy \(\left|A\right|=1\)khi và chỉ khi x = -6
\(A=\left(\frac{2x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{5-x^2}{x+2}\right)\) ĐKXĐ : \(x\ne\pm2\)
\(A=\left(\frac{2x}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4}{x+2}+\frac{5-x^2}{x+2}\right)\)
\(A=\left(\frac{2x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4+5-x^2}{x+2}\right)\)
\(A=\frac{x-6}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{1}\)
\(A=\frac{x-6}{x-2}\)
a) \(ĐKXĐ:\hept{\begin{cases}x^3+1\ne0\\x^3-2x^2\ne0\\x+1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne2\end{cases}}\)(chỗ chữ và là do OLM thiếu ngoặc 4 cái nên mk để thế nha! trình bày thì kẻ thêm 1 ngoặc nưax)
\(Q=1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)
\(=1+\left[\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right]:\frac{x^2\left(x-2\right)}{x\left(x^2-x+1\right)}\)
\(=1+\frac{\left(x+1\right)+\left(x+1\right)-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\frac{4x-2x^2}{x+1}.\frac{1}{x\left(x-2\right)}\)
\(=1-\frac{2x\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}=1-\frac{2}{x+1}=\frac{x-1}{x+1}\)
b, Với \(x\ne0;x\ne-1;x\ne2\)Ta có:
\(|x-\frac{3}{4}|=\frac{5}{4}\)
*TH1:
\(x-\frac{3}{4}=\frac{5}{4}\Rightarrow x=2\)(ko thảo mãn)
*TH2:
\(x-\frac{3}{4}=-\frac{5}{4}\Rightarrow x=-\frac{1}{2}\)
\(\Rightarrow Q=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)
c,
\(Q=\frac{x-1}{x+1}=1-\frac{2}{x+1}\)
Để Q nguyên thì x+1 phải thuộc ước của 2!! tự làm tiếp dễ rồi!!
ĐK: \(x\ne\pm2\)
\( a)A = \left( {\dfrac{3}{{2x + 4}} + \dfrac{x}{{2 - x}} + \dfrac{{2{x^2} + 3}}{{{x^2} - 4}}} \right):\left( {\dfrac{{2x - 1}}{{4x - 8}}} \right)\\ = \dfrac{{3\left( {2 - x} \right) + 2x\left( {2 + x} \right) - 4{x^2} - 6}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}:\dfrac{{2x - 1}}{{4\left( {x - 2} \right)}}\\ = \dfrac{{ - 2{x^2} + x}}{{2\left( {x + 2} \right)}}.\dfrac{{4\left( {x - 2} \right)}}{{2x - 1}} = \dfrac{{2x}}{{x + 2}} \)
\(b)\dfrac{{2x}}{{x + 2}} - 2 = \dfrac{{2x - 2x + 4}}{{x + 2}} = \dfrac{4}{{x + 2}}\)
Ta có: \(4>0\) để \(\dfrac{4}{{x + 2}} < 0 \Rightarrow x + 2 < 0 \Rightarrow x < - 2\)
\(c)\left| {x - 1} \right| = 3 \Leftrightarrow \left[ \begin{array}{l} x - 1 = 3\\ x - 1 = - 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 4\left( {tm} \right)\\ x = - 2\left( {ktm} \right) \end{array} \right.\)
Với \(x = 4 \Rightarrow A = \dfrac{{2.4}}{{4 + 2}} = \dfrac{4}{3}\)
\( d)\left| A \right| = \left| {\dfrac{{2x}}{{x + 2}}} \right| = 1\\ T{H_1}:\dfrac{{2x}}{{x + 2}} = 1 \Leftrightarrow 2x = x + 2 \Leftrightarrow x = 2\\ T{H_2}:\dfrac{{2x}}{{x + 2}} = - 1 \Leftrightarrow 2x = - x - 2 \Leftrightarrow x = - \dfrac{2}{3} \)