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a) Ta có: A = \(\left(\frac{x}{x-1}+\frac{x}{x^2-1}\right):\left(\frac{2}{x^2}-\frac{2-x^2}{x^3+x^2}\right)\)
A = \(\left(\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2\left(x+1\right)}{x^2\left(x+1\right)}-\frac{2-x^2}{x^2\left(x+1\right)}\right)\)
A = \(\left(\frac{x^2+x+x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2x+2-2+x^2}{x^2\left(x+1\right)}\right)\)
A = \(\left(\frac{x^2+2x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x^2+2x}{x^2\left(x+1\right)}\right)\)
A = \(\frac{x\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x^2\left(x+1\right)}{x\left(x+2\right)}\)
A = \(\frac{x^2}{x+1}\)
b) ĐKXĐ: x \(\ne\)\(\pm\)1; x \(\ne\)0; x \(\ne\)-2
Ta có: A = 4
<=> \(\frac{x^2}{x+1}=4\)
<=> x2 = 4(x + 1)
<=> x2 - 4x - 4 = 0
<=>(x2 - 4x + 4) - 8 = 0
<=> (x - 2)2 = 8
<=> \(\orbr{\begin{cases}x-2=\sqrt{8}\\x-2=-\sqrt{8}\end{cases}}\)
<=> \(\orbr{\begin{cases}x=2\sqrt{2}+2\\x=2-2\sqrt{2}\end{cases}}\)(tm)
Vậy ...
c) Ta có: A < 0
<=> \(\frac{x^2}{x+1}< 0\)
Do x2 \(\ge\)0 => x + 1 < 0
=> x < -1
Vậy để A < 0 thì x < -1 và x khác -2
a) \(ĐKXĐ:x\ne\pm3\)
b) \(A=\left(\frac{x}{x+3}+\frac{3-x}{x+3}\cdot\frac{x^2+3x+9}{x^2-9}\right):\frac{3}{x+3}\)
\(\Leftrightarrow A=\left(\frac{x}{x+3}-\frac{\left(x-3\right)\left(x^2+3x+9\right)}{\left(x+3\right)\left(x^2-9\right)}\right):\frac{3}{x+3}\)
\(\Leftrightarrow A=\left(\frac{x}{x+3}-\frac{x^2+3x+9}{\left(x+3\right)^2}\right):\frac{3}{x+3}\)
\(\Leftrightarrow A=\frac{x^2+3x-x^2-3x-9}{\left(x+3\right)^2}:\frac{3}{x+3}\)
\(\Leftrightarrow A=\frac{-9\left(x+3\right)}{3\left(x+3\right)^2}\)
\(\Leftrightarrow A=\frac{-3}{x+3}\)
c) Tại \(x=-\frac{1}{2}\)
\(\Leftrightarrow A=\frac{-3}{-\frac{1}{2}+3}\)
\(\Leftrightarrow A=\frac{-6}{5}\)
d) Để \(A>0\)
\(\Leftrightarrow\frac{-3}{x+3}>0\)
\(\Leftrightarrow x+3< 0\)(Vì -3 < 0)
\(\Leftrightarrow x< -3\)
e) +) Với \(A>\frac{-1}{2}\)
\(\Leftrightarrow\frac{-3}{x+3}>-\frac{1}{2}\)
\(\Leftrightarrow-6>-x-3\)
\(\Leftrightarrow x>3\)(tm)
+) Với \(A< -\frac{1}{2}\)
\(\Leftrightarrow\frac{-3}{x+3}< -\frac{1}{2}\)
\(\Leftrightarrow-6< -x-3\)
\(\Leftrightarrow x< 3\)(chú ý : \(x\ne-3\))
+) Với \(A=-\frac{1}{2}\)
\(\Leftrightarrow-\frac{3}{x+3}=-\frac{1}{2}\)
\(\Leftrightarrow x+3=6\)
\(\Leftrightarrow x=3\)(ktm)
Vậy \(\orbr{\begin{cases}A>-\frac{1}{2}\\A< -\frac{1}{2}\end{cases}}\)
a)\(M=\left(\frac{x^3+1}{x+1}-x\right):\left(1-\frac{1}{x}\right)\left(ĐKXĐ:x\ne-1;0\right)\)
\(M=\left[\frac{\left(x+1\right)\left(x^2-x+1\right)}{x+1}-x\right]:\left(\frac{x-1}{x}\right)\)
\(M=\left(x^2-x+1-x\right).\frac{x}{x-1}\)
\(M=\left(x-1\right)^2.\frac{x}{x-1}\)
\(M=x\left(x-1\right)\)
b)Ta có:\(\left|A\right|-A=0\)
\(\Leftrightarrow\left|x\left(x-1\right)\right|-x\left(x-1\right)=0\)
\(\Leftrightarrow\left|x^2-x\right|-x^2+x=0\)
\(TH1:x^2-x-x^2+x=0\)
\(\Leftrightarrow0x=0\)
\(\Rightarrow x\)vô số nghiệm
\(TH2:-\left(x^2-x\right)-x^2+x=0\)
\(\Leftrightarrow x-x^2-x^2+x=0\)
\(\Leftrightarrow2x=0\)
\(\Rightarrow x=0\)
c)Để M < \(-\frac{1}{2}\) ta có:
\(x\left(x-1\right)< -\frac{1}{2}\)
\(\Leftrightarrow x^2-x< -\frac{1}{2}\)
\(\Leftrightarrow x^2-x+\frac{1}{2}< 0\)
\(\Leftrightarrow x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{1}{4}< 0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{1}{4}< 0\)
Vậy ko có x nào TM để A < -1/2