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7 tháng 8 2020
Bài làm:
PT:
đkxđ: \(x\ne0;x\ne2\)
Ta có: \(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}+\frac{x-2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+2x=2+x-2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(vl\right)\\x+1=0\end{cases}}\Rightarrow x=-1\)
BPT:
Ta có: \(\frac{x+1}{2}-x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{x+1}{2}-x-\frac{1}{2}\le0\)
\(\Leftrightarrow\frac{x+1-2x-1}{2}\le0\)
\(\Leftrightarrow\frac{-x}{2}\le0\)
\(\Rightarrow-x\le0\)
\(\Rightarrow x\ge0\)
MN
7 tháng 8 2020
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)
\(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)
\(\Leftrightarrow\frac{2}{x\left(x-2\right)}+\frac{1}{x}-\frac{x+2}{x-2}=0\)
\(\Leftrightarrow\frac{2+x-2-x^2-2x}{x\left(x-2\right)}=0\)
\(\Leftrightarrow-x^2-x=0\)
\(\Leftrightarrow-x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}}\)
Vậy \(S=\left\{-1\right\}\)
b) \(\frac{x+1}{2}-x\le\frac{1}{2}\)
\(\Leftrightarrow x+1-2x-1\le0\)
\(\Leftrightarrow-x\le0\)
\(\Leftrightarrow x\ge0\)
Vậy \(x\ge0\)
LQ
1
14 tháng 2 2018
\(\frac{x-1}{x^2-x+1}-\frac{x+1}{x^2+x+1}=\frac{10}{x\left(x^4+x+1\right)}\)
\(\Leftrightarrow\frac{x\left(x-1\right)\left(x^2+x+1\right)-x\left(x+1\right)\left(x^2+x+1\right)-10}{x\left(x^4+x^2+1\right)}=0\)
\(\Rightarrow x\left(x^3-1\right)-x\left(x^3+1\right)-10=0\)
\(\Leftrightarrow x^4-x-x^4-x-10=0\)
\(\Leftrightarrow-2x-10=0\)
\(\Leftrightarrow x=-5\)
KC
3
11 tháng 3 2018
\(\left(\frac{1}{x-2}-\frac{1}{x+2}\right)+\left(\frac{1}{x-1}-\frac{1}{x+1}\right)=0\)
\(\frac{x+2-x+2}{x^2-4}+\frac{x+1-x+1}{x^2-1}=0\)
\(\frac{4}{x^2-4}+\frac{2}{x^2-1}=0\)
\(4x^2-4+2x^2-8=0\)
\(6x^2-12=0\)
\(x^2=2\)
\(x=\sqrt{2}\)
11 tháng 3 2018
ĐKXĐ: x
≠-2,-1,1,2
Ta có :
\(\frac{1}{x-1}+\frac{1}{x-2}=\frac{1}{x+1}+\frac{1}{x+2}\)
<=> \(\frac{1}{x-1}-\frac{1}{x+1}=\frac{1}{x+2}-\frac{1}{x-2}\)
<=>\(\frac{2}{x^2-1}=\frac{-4}{x^2-4}\)
<=> \(2x^2-8=-4x^2+4\)
<=> \(6x^2=12\)
<=> \(x^2=2\)
<=>\(\hept{\begin{cases}x=\sqrt{2}\left(TMĐK\right)\\x=-\sqrt{2}\left(TMĐK\right)\end{cases}}\)
Vậy pt trên có tập nghiệm S={\(\sqrt{2},-\sqrt{2}\)}
k mk nha mn
CB
0
3 tháng 8 2017
ĐK \(x\ne0\)
Ta có \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow\frac{x\left(x+1\right)\left(x^2-x+1\right)}{x\cdot\left(x^4+x^2+1\right)}-\frac{x\left(x-1\right)\left(x^2+x+1\right)}{x\left(x^4+x^2+1\right)}=\frac{3}{x\left(x^4+x^2+1\right)}\)
\(\Rightarrow\left(x^2+x\right)\left(x^2-x+1\right)-\left(x^2-x\right)\left(x^2+x+1\right)=3\)
\(\Leftrightarrow x^4-x^3+x^2+x^3-x^2+x-x^4-x^3-x^2+x^3+x^2+x=3\)
\(\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\left(tm\right)\)
Vậy \(x=\frac{3}{2}\)
25 tháng 3 2020
\(\frac{x-1}{x+1}-\frac{x^2+x-2}{x+1}=\frac{x+1}{x-1}-x-2\)
<=> \(\frac{x-1}{x+1}-\frac{\left(x-1\right)\left(x+2\right)}{x+1}=\frac{x+1}{x-1}-x-2\)
<=> \(\frac{x-1-\left(x-1\right)\left(x+1\right)}{x+1}=\frac{x+1}{x-1}-x-2\)
<=> \(\frac{-\left(x-1\right)\left(x+2-1\right)}{x+1}=\frac{x+1}{x-1}-x-2\)
<=> -(x - 1) = \(\frac{x+1}{x-1}\) - x - 2
<=> 1 - x = \(\frac{x+1}{x-1}\) - x - 2
<=> 1 = \(\frac{x+1}{x-1}\) - x - 2
<=> x - 1 = x + 1 - 2(x - 1)
<=> x - 1 = -x + 3
<=> x = 3 - x - 1
<=> x = 2 - x
<=> x + x = 2
<=> 2x = 2
<=> x = 1
HN
1
24 tháng 3 2020
\(\frac{2}{x^3-x^2-x+1}=\frac{3}{1-x^2}-\frac{1}{x+1}\)
<=> \(\frac{2}{\left(x^2-1\right)\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}+\frac{1}{x+1}=0\)
<=> \(\frac{2}{\left(x-1\right)^2\left(x+1\right)}+\frac{3\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}+\frac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}=0\)
<=> \(2+3x-3+x^2-2x+1=0\)
<=> x2 + x = 0
<=> x(x + 1) = 0
<=> \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
Vậy S = {0; -1}
NV
16 tháng 7 2016
ĐKXĐ: \(x\ne\left\{0;-1;-2;-3;-4;-5;-6;-7\right\}\)
\(\frac{1}{x}+\frac{1}{x+2}+\frac{1}{x+5}+\frac{1}{x+7}=\frac{1}{x+1}+\frac{1}{x+3}+\frac{1}{x+4}+\frac{1}{x+6}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{x+7}+\frac{1}{x+2}+\frac{1}{x+5}=\frac{1}{x+1}+\frac{1}{x+6}+\frac{1}{x+3}+\frac{1}{x+4}\)
\(\Rightarrow\frac{x+7+x}{x\left(x+7\right)}+\frac{x+5+x+2}{\left(x+2\right)\left(x+5\right)}=\frac{x+6+x+1}{\left(x+1\right)\left(x+6\right)}+\frac{x+4+x+3}{\left(x+3\right)\left(x+4\right)}\)
\(\Rightarrow\frac{2x+7}{x^2+7x}+\frac{2x+7}{x^2+7x+10}=\frac{2x+7}{x^2+7x+6}+\frac{2x+7}{x^2+7x+12}\)
\(\Rightarrow\left(2x+7\right)\left(\frac{1}{x^2+7x}+\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+6}-\frac{1}{x^2+7x+12}\right)=0\)
mà \(\frac{1}{x^2+7x}+\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+6}-\frac{1}{x^2+7x+12}\ne0\)
=> 2x + 7 = 0 => x = -7/2
Vậy x = -7/2
ĐKXĐ: \(x\ne\pm1\)
Ta có: \(\frac{x-1}{x+1}-\frac{x^2+x-2}{x+1}=\frac{x+1}{x-1}-x-2\)
=> \(\left(x-1\right)^2-\left(x^2+x-2\right)\left(x-1\right)=\left(x+1\right)^2-x\left(x^2-1\right)-2\left(x^2-1\right)\)
<=> x2 - 2x + 1 - x^3 + 3x - 2 = x2 + 2x + 1 - x3 + x - 2x2 + 2
<=> -x3 + x2 + x - 1 = -x3 - x2 + 3x + 3
<=> -x3 + x2 + x - 1 + x3 + x2 - 3x - 3 = 0
<=> 2x2 - 2x - 4 = 0
<=> x2 - x - 2 = 0
<=> x2 - 2x + x - 2 = 0
<=> (x + 1)(x - 2) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
Vậy S = {-1; 2}
kl lại. \(\orbr{\begin{cases}x=-1\left(ktm\right)\\x=2\end{cases}}\)
Vậy S = {2}