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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
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 \(A>0\).
Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
a)Có A=\(\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)(ĐKXĐ \(x\ne2,-2,-1\))
=\(\left(\frac{2-x}{\left(2-x\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}-\frac{x}{\left(2-x\right)\left(2+x\right)}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)
=\(\frac{2-x+2x+4-x}{\left(2-x\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
=\(\frac{6\left(2-x\right)\left(x+1\right)}{6\left(2-x\right)\left(x+2\right)^2}\)
=\(\frac{x+1}{\left(x+2\right)^2}\)
b)Có A=\(\frac{x+1}{\left(x+2\right)^2}\)
Để A>0 <=> x+1>0 <=>x>-1
c) Có x2+3x+2=0
<=> x2+2x+x+2=0
<=> x(x+2)+(x+2)=0
<=>(x+1)(x+2)=0
<=> x=-1 hoặc x=-2
\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{x^3+1}\)
\(A=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3-3x}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x^3-x^2+x-3-3x+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{1}{x^3+1}\)
a) điều kiện \(x\ne\pm2\)
\(A=\left(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x-6}{4-x^2}\right):\dfrac{1}{3x-2x^2-6}\)
\(A=\left(\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x-6}{x^2-4}\right):\dfrac{1}{3x-2x^2-6}\)
\(A=\left(\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x-6}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{4\left(x-2\right)+2\left(x+2\right)-\left(5x-6\right)}{\left(x+2\right)\left(x-2\right)}:\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{4x-8+2x+4-5x+6}{\left(x+2\right)\left(x-2\right)}:\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{x+2}{\left(x+2\right)\left(x-2\right)}:\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{1}{x-2}.\dfrac{3x-2x^2-6}{1}=\dfrac{3x-2x^2-6}{x-2}\)
b) ta có : \(3x-2x^2-6=-2x^2+3x-6=-\left(2x^2-3x+6\right)\)
\(=\left(\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\dfrac{3}{2\sqrt{2}}+\left(\dfrac{3}{2\sqrt{2}}\right)^2\right)+\dfrac{39}{8}\)
\(=\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2+\dfrac{39}{8}\ge\dfrac{39}{8}>0\)
\(\Rightarrow A\le0\) \(\Leftrightarrow x-2\le0\) (mà đk : \(x\ne2\) \(\Rightarrow x-2\ne0\))
vậy \(A\le0\Leftrightarrow A< 0\) \(\Leftrightarrow x-2< 0\Leftrightarrow x< 2\) vậy \(x< 2\)
\(A=\left(\dfrac{1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{6x+3}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right):\left(x+2\right)\)\(A=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-x+1\right)\left(x+2\right)}\)
a) \(A=\left\{{}\begin{matrix}x\ne-1;-2\\\dfrac{1}{x^2-x+1}\end{matrix}\right.\)
b)
\(A>1;\dfrac{1}{x^2-x+1}>1\Leftrightarrow x^2-x< 0\Leftrightarrow0< x< 1\)
\(P=\dfrac{1}{x^2-x+1}.\dfrac{x^3-x^2+x}{\left(x+1\right)^2}=\dfrac{x}{\left(x+1\right)^2}\)
x>0 => P >0 đang tìm Giá trị LN => chỉ xét P>0 <=> x>0
\(\dfrac{1}{P}=\dfrac{\left(x+1\right)^2}{x}=x+2+\dfrac{1}{x}\)
áp co si hai số dương x ; 1/x
\(\dfrac{1}{P}\ge2.\sqrt{x.\dfrac{1}{x}}+2=4\Rightarrow P\le\dfrac{1}{4}\)
đẳng thức khi x =1/x => x=1 thỏa mãn đk của x
\(MaxP=\dfrac{1}{4}\)
a: \(A=\dfrac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\dfrac{-6}{\left(x+2\right)}\cdot\dfrac{-\left(x+1\right)}{6\left(x+2\right)}=\dfrac{\left(x+1\right)}{\left(x+2\right)^2}\)
b: A>0
=>x+1>0
=>x>-1
c: x^2+3x+2=0
=>(x+1)(x+2)=0
=>x=-2(loại) hoặc x=-1(loại)
Do đó: Khi x^2+3x+2=0 thì A ko có giá trị