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7 tháng 10 2021
\(=\left(1-x\right)\left(6a+2a^2\right)=2a\left(3+2a\right)\left(1-x\right)\\ 2,=\left(x-5\right)\left(x-3-2\right)=\left(x-5\right)^2\)
TA
2
1 tháng 3 2021
1) ĐKXĐ: \(x\notin\left\{1;-2\right\}\)
Ta có: \(\dfrac{2x}{x-1}-\dfrac{1}{x+2}=2\)
\(\Leftrightarrow\dfrac{2x\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-1\right)\left(x+2\right)}=\dfrac{2\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)
Suy ra: \(2x^2+4x-x+1=2\left(x^2+x-2\right)\)
\(\Leftrightarrow2x^2+3x+1=2x^2+2x-4\)
\(\Leftrightarrow2x^2+3x+1-2x^2-2x+4=0\)
\(\Leftrightarrow x+5=0\)
hay x=-5(thỏa ĐK)
Vậy: S={-5}
2) ĐKXĐ: \(x\notin\left\{5;-5\right\}\)
Ta có: \(\dfrac{x}{x^2-25}-\dfrac{1-x}{x-5}=\dfrac{1}{x+5}\)
\(\Leftrightarrow\dfrac{x}{\left(x-5\right)\left(x+5\right)}+\dfrac{\left(x-1\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(x+x^2+5x-x-5=x-5\)
\(\Leftrightarrow x^2+5x-5-x+5=0\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
Vậy: S={0;-4}
1 tháng 3 2021
a/ ĐKXĐ : \(x\ne1;-2\)
\(\dfrac{2x}{x-1}-\dfrac{1}{x+2}=2\)
\(\Leftrightarrow\dfrac{2x\left(x+2\right)-\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=2\)
\(\Leftrightarrow2x^2+3x-x+1=2x^2+4x-2x-4\)
\(\Leftrightarrow2x+1=2x-4\)
\(\Leftrightarrow1=-4\left(loại\right)\)
Vậy...
b/ĐKXĐ : \(x\ne\pm5\)
\(\dfrac{x}{x^2-25}-\dfrac{1-x}{x-5}=\dfrac{1}{x+5}\)
\(\Leftrightarrow\dfrac{x}{\left(x-5\right)\left(x+5\right)}+\dfrac{\left(x-1\right)\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}\)
\(\Leftrightarrow x+x^2+5x-x-5=x-5\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy...
14 tháng 12 2021
\(a,=\dfrac{x^3-\left(x-1\right)\left(x^2+x+1\right)}{1-x}=\dfrac{x^3-x^3+1}{1-x}=\dfrac{1}{1-x}\\ b,=\dfrac{2x+x^2+3x+2+2-x}{\left(x+2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x+2\right)^2}=1\)
TQ
0
NN
1
28 tháng 3 2021
ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x-3}{x+1}=\dfrac{x^2}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(x^2-4x+3-x^2=0\)
\(\Leftrightarrow-4x=-3\)
hay \(x=\dfrac{3}{4}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{3}{4}\right\}\)
DK
1
15 tháng 4 2020
\(\frac{2}{x-2}-\frac{3}{x+2}=\frac{x+1}{x^2-4}\left(x\ne\pm2\right)\)
\(\Leftrightarrow\frac{2}{x-2}-\frac{3}{x+2}-\frac{x+1}{x^2-4}=0\)
\(\Leftrightarrow\frac{2}{x-2}-\frac{3}{x+2}-\frac{x+1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x+1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x+4-3x+6-x-1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{-2x-9}{\left(x-2\right)\left(x+2\right)}=0\)
=> -2x-9=0
<=> -2x=9
<=> \(x=\frac{-9}{2}\left(tmđk\right)\)
MT
6
7 tháng 9 2021
\(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)
\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)
\(\Leftrightarrow-x^2+6x-3=-x^2+3x+1\)
\(\Leftrightarrow3x=4\)
hay \(x=\dfrac{4}{3}\)
NT
7 tháng 9 2021
\(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)
\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)
\(\Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\)
DT
1
28 tháng 12 2021
\(a,\Leftrightarrow x^2-2x-x^2+1=0\\ \Leftrightarrow-2x+1=0\Leftrightarrow x=\dfrac{1}{2}\\ b,\Leftrightarrow\left(2x-1-x-4\right)\left(2x-1+x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(3x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
14 tháng 12 2021
\(=\dfrac{2x^2-x-x-1+2-x^2}{x-1}=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\)
\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\left(ĐKXĐ:x\ne0;x\ne2\right)\)
\(\Rightarrow x^2+2x-x+2-2=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(KTMĐK\right)\\x=-1\left(TMĐK\right)\end{cases}}\)
\(\Rightarrow S=\left\{-1\right\}\)
\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\left(x\ne0;x\ne2\right)\)
<=> \(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
=> x2 + 2x - x + 2 = 2
<=> x2 + x = 0
<=> x( x + 1 ) = 0
<=> x = 0 (ktm) hoặc x = -1(tm)
Vậy ...