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B1: ĐXXĐ: \(x\ne\pm2;x\ne-1\)
\(=\left(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\left(\dfrac{x-2-2x-2+x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}:\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x+1\right)}{-6\left(x+2\right)}=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}\)
b, \(A=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}>0\)
\(\Leftrightarrow2x+2>0\) (vì \(3\left(x+2\right)^2\ge0\forall x\))
\(\Leftrightarrow x>-1\).
-Vậy \(x\in\left\{x\in Rlx>-1;x\ne2\right\}\) thì \(A>0\).
a: ĐKXĐ: x<>1; x<>2; x<>3
\(K=\left(\dfrac{x^2}{\left(x-2\right)\left(x-3\right)}+\dfrac{x^2}{\left(x-1\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+2x^2+1-x^2}\)
\(=\dfrac{x^3-x^2+x^3-3x^2}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}\)
\(=\dfrac{2x^3-4x^2}{\left(x-2\right)}\cdot\dfrac{1}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{2x^2\left(x-2\right)}{\left(x-2\right)\left(x^4+x^2+1\right)}=\dfrac{2x^2}{x^4+x^2+1}\)
b:
a) Ta có: \(A=\left(\dfrac{2}{x+2}-\dfrac{1}{x-3}+\dfrac{5-x}{x^2-x-6}\right)\cdot\left(x-\dfrac{6}{x-1}\right)\)
\(=\left(\dfrac{2\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}-\dfrac{x+2}{\left(x-3\right)\left(x+2\right)}+\dfrac{5-x}{\left(x-3\right)\left(x+2\right)}\right)\cdot\dfrac{x\left(x-1\right)-6}{x-1}\)
\(=\dfrac{2x-6-x-2+5-x}{\left(x+2\right)\left(x-3\right)}\cdot\dfrac{x^2-x-6}{x-1}\)
\(=\dfrac{-3}{x-1}\)
a) \(-ĐKXĐ:x\ne\pm2;1\)
Rút gọn : \(A=\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)
\(=\left(\frac{1}{x+2}+\frac{-2}{x-2}+\frac{x}{x^2-4}\right).\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\left[\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{\left(x-2\right)\left(x+2\right)}\right]\)\(.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\left[\frac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\right].\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)\(=\frac{x+1}{\left(x+2\right)^2}\)
b) \(A>0\Leftrightarrow\frac{x+1}{\left(x+2\right)^2}>0\Leftrightarrow\orbr{\begin{cases}x+1< 0;\left(x+2\right)^2< 0\left(voly\right)\\x+1>0;\left(x+2\right)^2>0\end{cases}}\)
\(\Leftrightarrow x>1;x>-2\Leftrightarrow x>1\)
Vậy với mọi x thỏa mãn x>1 thì A > 0
c) Ta có : \(x^2+3x+2=0\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
Vậy x = -1;-2
Lời giải:
ĐK: $x\neq \pm 2; x\neq 0$
a)
\(A=\left[\frac{x+2}{(x+2)(x-2)}+\frac{2x}{(x-2)(x+2)}+\frac{x-2}{(x-2)(x+2)}\right].\frac{2-x}{x}=\frac{x+2+2x+x-2}{(x-2)(x+2)}.\frac{-(x-2)}{x}\)
\(=\frac{4x}{(x-2)(x+2)}.\frac{-(x-2)}{x}=\frac{-4}{x+2}\)
b) Để $A=1\Leftrightarrow \frac{-4}{x+2}=1$
$\Leftrightarrow x+2=-4$
$\Leftrightarrow x=-6$ (thỏa ĐKXĐ)
Vậy $x=-6$
ĐKXĐ: \(x\notin\left\{-1;2;-2\right\}\)
a) Ta có: \(A=\left(\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\dfrac{2x^2+4x-1}{x^3+1}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)
\(=\left(\dfrac{\left(x+1\right)^2}{x^2-x+1}-\dfrac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)
\(=\left(\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{\left(x-2\right)\left(x+2\right)}{3x\left(x+2\right)}\)
\(=\dfrac{x^3+3x^2+3x+1-2x^2-4x+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}:\dfrac{x-2}{3x}\)
\(=\dfrac{x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\dfrac{3x}{x-2}\)
\(=\dfrac{3x}{x-2}\)
b) Để A nguyên thì \(3x⋮x-2\)
\(\Leftrightarrow3x-6+6⋮x-2\)
mà \(3x-6⋮x-2\)
nên \(6⋮x-2\)
\(\Leftrightarrow x-2\inƯ\left(6\right)\)
\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(x\in\left\{3;1;4;0;5;-1;8;-4\right\}\)
Kết hợp ĐKXĐ, ta được:
\(x\in\left\{3;1;4;0;5;8;-4\right\}\)
Vậy: Để A nguyên thì \(x\in\left\{3;1;4;0;5;8;-4\right\}\)
