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\(A=\frac{1}{x+2}+\frac{1}{x-2}+\frac{x^2+1}{x^2-4}\)
\(=\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}\)
Với \(\forall x\in\left[-2;2\right]\) thì \(\left(x-2\right)\left(x+2\right)< 0\Rightarrow\frac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}< 0\Rightarrow A< 0\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne\frac{1}{2}\\x\ne\pm1\end{cases}}\)
\(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(\Leftrightarrow A=\frac{-x-1+2x-2+5-x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Leftrightarrow A=\frac{2}{1-2x}\)
b) Để |A| = A
\(\Leftrightarrow A>0\)
\(\Leftrightarrow\frac{2}{1-2x}>0\)
Vì 2 > 0
\(\Leftrightarrow1-2x>0\)
\(\Leftrightarrow1>2x\)
\(\Leftrightarrow x< \frac{1}{2}\)
Vậy để \(\left|A\right|=A\Leftrightarrow x< \frac{1}{2}\)
\(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\left(x\ne\pm1;x\ne\frac{1}{2}\right)\)
A=(1/x-2 - (2x/(2-x)(2+x) - 1/2+x) ) *(2-x)/x
=(1/x-2 - x^2+5x-2/(2-x)(2+x))*2-x/x
=(-x^3-4x^2+12x/(x-2)(2-x)(2+x))*2-x/x
= - x(x-2)(x+6)(2-x)/x(x-2)(2-x)(2+x)
= - x+6/x+2
\(A=\frac{\left|x-1\right|+\left|x\right|-x}{3x^2+4x+1}=\frac{1-x-x-x}{3x^2+3x+x+1}=\frac{1-3x}{\left(x+1\right)\left(3x+1\right)}\)
\(B=\frac{\left|2x-1\right|+x}{3x^2-22x+7}=\frac{1-2x+x}{3x^2-21x-x+7}=\frac{1-x}{\left(x-7\right)\left(3x-1\right)}\)
Để \(A\)có nghĩa thì \(x^3-3x-2\ne0\)
\(\Rightarrow\left(x^3-x\right)-\left(2x-2\right)\ne0\)
\(\Rightarrow x\left(x^2-1\right)-2\left(x-1\right)\ne0\)
\(x\left(x+1\right)\left(x-1\right)-2\left(x-1\right)\ne0\)
\(\left(x^2+x-2\right)\left(x-1\right)\ne0\)
\(\Rightarrow\left[x^2-1+x-1\right]\left(x-1\right)\ne0\)
\(\left[\left(x-1\right)\left(x+1\right)+\left(x-1\right)\right]\left(x-1\right)\ne0\)
\(\left(x-1\right)^2\left(x+2\right)\ne0\)
\(\Rightarrow x\ne1;-2\)
Vậy...
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1) ĐKXĐ : \(\left\{{}\begin{matrix}x^3-1\ne0\\x^3+x\ne0\\x^2+x\ne0\\3x+\left(x-1\right)^2\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x-1\ne0\\x\left(x^2+1\right)\ne0\\x\left(x+1\right)\ne0\\x^2+x+1\ne0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x-1\ne0\\x\ne0\\x+1\ne0\\\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne0\\x\ne-1\\\left(x+\frac{1}{2}\right)^2\ne-\frac{3}{4}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne\pm1\\x\ne0\end{matrix}\right.\)
2) Ta có : \(P=\left(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}+\frac{1}{x-1}\right):\frac{x^2+x}{x^3+x}\)
=> \(P=\left(\frac{x^2-2x+1}{3x+x^2-2x+1}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right):\frac{x^2+x}{x^3+x}\)
=> \(P=\left(\frac{\left(x-1\right)^2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\frac{x^2+x}{x^3+x}\)
=> \(P=\left(\frac{\left(x-1\right)^3-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\frac{x^2+x}{x^3+x}\)
=> \(P=\left(\frac{x^3-3x^2+3x-1-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\frac{x\left(x+1\right)}{x\left(x^2+1\right)}\)
=> \(P=\left(\frac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\frac{x+1}{x^2+1}\)
=> \(P=\left(\frac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right):\frac{x+1}{x^2+1}\)
=> \(P=1:\frac{x+1}{x^2+1}=\frac{x^2+1}{x+1}\)
- Thay P = 0 vào phương trình trên ta được :\(\frac{x^2+1}{x+1}=0\)
=> \(x^2+1=0\)
=> \(x^2=-1\) ( Vô lý )
Vậy phương trình vô nghiệm .
3) Ta có : \(\left|P\right|=1\)
=> \(\left|\frac{x^2+1}{x+1}\right|=1\)
=> \(\frac{x^2+1}{\left|x+1\right|}=1\)
=> \(\left|x+1\right|=x^2+1\)
TH1 : \(x+1\ge0\left(x\ge-1\right)\)
=> \(x+1=x^2+1\)
=> \(x^2=x\)
=> \(x=1\) ( TM )
TH2 : \(x+1< 0\left(x< -1\right)\)
=> \(-x-1=x^2+1\)
=> \(x^2+1+1+x=0\)
=> \(x^2+\frac{1}{2}x.2+\frac{1}{4}+\frac{7}{4}=0\)
=> \(\left(x+\frac{1}{2}\right)^2=-\frac{7}{4}\) ( Vô lý )
Vậy giá trị của x thỏa mãn là x = 1 .
1. A = -4 phần x+2
2. 2x^2 + x = 0 => x = 0 hoặc x = -1/2
Với x = 0 thì A = -2
Với x = -1/2 thì A = -8/3
3. A = 1/2 => -4 phần x + 2 = 1/2
<=> -8 = x + 2
<=> x = -10
4. A nguyên dương => A > 0
=> -4 phần x + 2 > 0
Do -4 < 0 nên -4 phần x + 2 > 0 khi x + 2 < 0
=> x < -2