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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
1.A=\(\frac{x^4-2x^2+1}{x^3-3x-2}\)
A có nghĩa \(\Leftrightarrow x^3-3x-2\ne0\Leftrightarrow\left(x+1\right)^2\left(x-2\right)\ne0\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
2 .A = \(\frac{x^4-2x^2+1}{x^3-3x-2}\)=\(\frac{\left(x^2-1\right)^2}{\left(x+1\right)^2\left(x-2\right)}=\frac{\left(x+1\right)^2\left(x-1\right)^2}{\left(x+1\right)^2\left(x-2\right)}=\frac{\left(x-1\right)^2}{x-2}\)
A<1\(\Rightarrow\frac{\left(x-1\right)^2}{x-2}-1< 0\Rightarrow\frac{x^2-2x+1-x+2}{x-2}< 0\)
\(\Rightarrow\frac{x^2-3x+3}{x-2}< 0\Rightarrow x-2< 0\)vì \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy x<2 thỏa mãn yêu cầu A<1
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 .
ĐKXĐ: \(x^3-3x-2\ne0\)
\(\Leftrightarrow x^3-x-2x-2\ne0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)-2\left(x+1\right)\ne0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x-2\right)\ne0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)^2\ne0\)
hay \(x\notin\left\{2;-1\right\}\)
\(A=\dfrac{x^4-2x^2+1}{x^3-3x-2}=\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{\left(x-2\right)\cdot\left(x+1\right)^2}=\dfrac{\left(x-1\right)^2}{x-2}\)
Để A<1 thì \(A-1< 0\)
\(\Leftrightarrow\dfrac{x^2-2x+1-x+2}{x-2}< 0\)
\(\Leftrightarrow\dfrac{x^2-3x+3}{x-2}< 0\)
=>x-2<0
hay x<2
Vậy: \(\left\{{}\begin{matrix}x< 2\\x< >-1\end{matrix}\right.\)
1.
<=> 7 - 2x - 4 = -x - 4
<=> -2x + x = -4 -7 + 4
<=> -x = -7
<=> x = 7
Vậy S = { 7 }
2.
<=> \(\frac{2\left(3x-1\right)}{6}\)= \(\frac{3\left(2-x\right)}{6}\)
<=> 2( 3x - 1 ) = 3( 2 - x )
<=> 6x -2 = 6 - 3x
<=> 6x + 3x = 6 + 2
<=> 9x = 8
<=> x = \(\frac{8}{9}\)
Vậy S = \(\left\{\frac{8}{9}\right\}\)
3.
<=> \(\frac{6x+10}{3}-\frac{x}{2}=5-\frac{3x+3}{4}\)
<=> \(\frac{4\left(6x+10\right)}{12}-\frac{6x}{12}=\frac{60}{12}-\frac{3\left(3x+3\right)}{12}\)
<=> 4( 6x + 10 ) - 6x = 60 - 3( 3x + 3 )
<=> 24x + 40 - 6x = 60 - 9x -9
<=> 18x + 40 = 51 - 9x
<=> 18x + 9x = 51 - 40
<=> 27x = 11
<=> x = \(\frac{11}{27}\)
Vậy S = \(\left\{\frac{11}{27}\right\}\)
<=>
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...
x khác 1 , x khác -2