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a) ĐKXĐ: \(x\ne2\); x \(\ne\)-2
Ta có: P = \(\left(\frac{2+x}{x-2}+\frac{2}{x+3}-\frac{x^2+5x}{x^2-4}\right):\left(1-\frac{x+1}{x+2}\right)\)
P = \(\left(\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x^2+5x}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x+2-x-1}{x+2}\right)\)
P = \(\left(\frac{x^2+4x+4+2x-4-x^2-5x}{\left(x-2\right)\left(x+2\right)}\right):\frac{1}{x+2}\)
P = \(\frac{x}{\left(x-2\right)\left(x+2\right)}\cdot\left(x+2\right)\)
P = \(\frac{x}{x-2}\) (đk: x khác 2)
b) Ta có: x2 - 2x = 0
=> x(x - 2) = 0
=> \(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\left(tm\right)\\x=2\end{cases}}\)
Vì biểu thức P x \(\ne\)2 => x = 0=> P = \(\frac{0}{0-2}=0\)
\(C=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{1}{1-x}+\frac{2}{x+1}+\frac{-5x}{\left(1-x\right)\left(x+1\right)}\right):\frac{1-2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\left(\frac{x+1+2-2x-5x}{\left(1-x\right)\left(x+1\right)}\right):\frac{1-2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\left(\frac{3-6x}{\left(1-x\right)\left(x+1\right)}\right).\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\frac{3\left(1-2x\right)}{\left(1-x\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\frac{3\left(x-1\right)}{1-x}\)
nho k nha
b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
ĐKXĐ : \(x\ne\pm2;x\ne0;x\ne3\)
\(A=\left(\frac{4x}{2+x}+\frac{8x^2}{4-x^2}\right):\left(\frac{x-1}{x^2-2x}-\frac{2}{x}\right)\)
\(=\frac{4x\left(2-x\right)+8x^2}{\left(2-x\right)\left(2+x\right)}:\frac{x-1-2\left(x-2\right)}{x\left(x-2\right)}\)
\(=\frac{8x-4x^2+8x^2}{\left(2-x\right)\left(2+x\right)}:\frac{x-1-2x+4}{x\left(x-2\right)}\)
\(=\frac{8x+4x^2}{\left(2-x\right)\left(2+x\right)}:\frac{3-x}{x\left(x-2\right)}\)
\(=\frac{8x+4x^2}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(x-2\right)}{3-x}\) \(=\frac{4x\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{x-3}\)
\(=\frac{4x^2}{x-3}\)
\(A< 0\Leftrightarrow\frac{4x^2}{x-3}< 0\Leftrightarrow x-3< 0\) ( do \(4x^2>0\) )
\(\Leftrightarrow x< 3\)
Vậy :........
a) ĐKXĐ : x \(\ne-2;x\ne1;x\ne0\)
\(A=\left(\frac{x}{x+2}-\frac{4}{x^2+2x}\right):\left(\frac{x^2-2x+1}{x^2-x}\right)=\left(\frac{x}{x+2}-\frac{4}{x\left(x+2\right)}\right):\left(\frac{\left(x-1\right)^2}{x\left(x-1\right)}\right)\)
\(=\frac{x^2-4}{x\left(x+2\right)}:\frac{x-1}{x}=\frac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)}.\frac{x}{x-1}=\frac{x-2}{x}.\frac{x}{x-1}=\frac{x-2}{x-1}\)
b) Để A > 1
=> \(\frac{x-2}{x-1}>1\)
=> \(\frac{x-2}{x-1}-1>0\Rightarrow\frac{-1}{x-1}>0\Rightarrow x-1< 0\Rightarrow x< 1\)
Vậy để A > 1 thì x < 1 và x \(\ne-2;x\ne1;x\ne0\)
c) Ta có \(A=\frac{x-2}{x-1}=\frac{x-1-1}{x-1}=1-\frac{1}{x-1}\)
Để A \(\inℤ\Rightarrow\frac{1}{x-1}\inℤ\Rightarrow1⋮x-1\Rightarrow x-1\inƯ\left(1\right)\Rightarrow x-1\in\left\{1;-1\right\}\)
Khi x - 1 = 1 => x = 2( tm)
Khi x - 1 =-1 => x = 0 (loại)
Vậy x = 2 thì A nguyên
\(A=\left(\frac{x^2-16}{x-4}+1\right):\left(\frac{x-2}{x-3}+\frac{x+3}{x+1}+\frac{x+2-x^2}{x^2-2x-3}\right)\)
\(=\left(x+5\right):\left(\frac{\left(x-2\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}+\frac{x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)
\(=\left(x+5\right):\left(\frac{x^2+x-2x-2+x^2-9+x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)
\(=\left(x+5\right):\left(\frac{x^2-9}{\left(x-3\right)\left(x+1\right)}\right)\)
\(=\left(x+5\right):\left(\frac{x+3}{x+1}\right)=\frac{x+3}{\left(x+5\right)\left(x+1\right)}\)
ĐKXĐ : \(\hept{\begin{cases}x-1\ne0\\x+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}}\)
a) \(A=\left(\frac{x+2}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\)
\(\Leftrightarrow A=\left[\frac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{5\left(x-1\right)}{2x}\)
\(\Leftrightarrow A=\frac{x^2+3x+2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\frac{5\left(x-1\right)}{2x}\)
\(\Leftrightarrow A=\frac{x^2+3x+2-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{5\left(x-1\right)}{2x}\)
\(\Leftrightarrow A=\frac{5\left(5x-1\right)\left(x-1\right)}{2x\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow A=\frac{5\left(5x-1\right)}{2x\left(x+1\right)}\)
b) Xét x = -1 không thỏa mãn ĐKXĐ nên ta xét x = 3
Thay x = 3 vào A ta có :
\(A=\frac{5\left(5\cdot3-1\right)}{2\cdot3\cdot\left(3+1\right)}=\frac{35}{12}\)
c) Để A = 2 thì :
\(\frac{5\left(5x-1\right)}{2x\left(x+1\right)}=2\)
\(\Leftrightarrow4x\left(x+1\right)=5\left(5x-1\right)\)
\(\Leftrightarrow4x^2+4x=25x-5\)
\(\Leftrightarrow4x^2+4x-25x+5=0\)
\(\Leftrightarrow4x^2-21x+5=0\)
\(\Leftrightarrow4x^2-20x-x+5=0\)
\(\Leftrightarrow4x\left(x-5\right)-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{4}\end{cases}}\)( thỏa mãn ĐKXĐ )
Vậy....
\(A=\left(\dfrac{1}{x-2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\cdot\dfrac{2-x}{x}\)
\(=\dfrac{x+2+2x+x-2}{-\left(2-x\right)\left(x+2\right)}\cdot\dfrac{2-x}{x}\)
\(=\dfrac{4x}{-\left(x+2\right)\cdot x}=\dfrac{-4}{x+2}\)
\(A=\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right)\div\left(\frac{x^2-2x}{x^3-x^2+x}\right)\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
\(=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right)\div\left(\frac{x\left(x-2\right)}{x\left(x^2-x+1\right)}\right)\)
\(=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right)\div\frac{x-2}{x^2-x+1}\)
\(=\left(\frac{x+1+x+1-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}\right)\times\frac{x^2-x+1}{x-2}\)
\(=\frac{-2x^2+4x}{\left(x+1\right)\left(x^2-x+1\right)}\times\frac{x^2-x+1}{x-2}\)
\(=\frac{-2x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{-2x}{x+1}\)
b) \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)
<=> \(\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(loai\right)\\x=-\frac{1}{2}\left(nhan\right)\end{cases}}\)
Với x = -1/2 => \(A=\frac{-2\cdot\left(-\frac{1}{2}\right)}{-\frac{1}{2}+1}=2\)
c) Để A ∈ Z thì \(\frac{-2x}{x+1}\)∈ Z
=> -2x ⋮ x + 1
=> -2x - 2 + 2 ⋮ x + 1
=> -2( x + 1 ) + 2 ⋮ x + 1
Vì -2( x + 1 ) ⋮ ( x + 1 )
=> 2 ⋮ x + 1
=> x + 1 ∈ Ư(2) = { ±1 ; ±2 }
| x+1 | 1 | -1 | 2 | -2 |
| x | 0 | -2 | 1 | -3 |
Các giá trị trên đều tm \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
Vậy x ∈ { -3 ; -2 ; 0 ; 1 }