Giải PT: \(\dfrac{2x-1}{3}\) - \(\dfrac{x-1}{2}\) + \(\dfrac{x+1}{6}\) = 1
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25 tháng 2 2021
`1/(3-x)-1/(x+1)=x/(x-3)-(x-1)^2/(x^2-2x-3)(x ne -1,3)`
`<=>(-x-1)/(x^2-2x-3)-(x-3)/(x^2-2x-3)=(x^2+x)/(x^2-2x-3)-(x-1)^2/(x^2-2x-3)`
`<=>-x-1-x+3=x^2+x-x^2+2x-1`
`<=>-2x+2=3x-1`
`<=>5x=3`
`<=>x=3/5`
Vậy `S={3/5}`
`1/(x-2)-6/(x+3)=6/(6-x^2-x)(x ne 2,-3)`
`<=>(x+3)/(x^2+x-6)-(6x-12)/(x^2+x-6)+6/(x^2+x-6)=0`
`<=>x+3-6x+12+6=0`
`<=>-5x+21=0`
`<=>x=21/5`
Vậy `S={21/5}`
25 tháng 2 2021
a) ĐKXĐ: \(x\notin\left\{3;-1\right\}\)
Ta có: \(\dfrac{1}{3-x}-\dfrac{1}{x+1}=\dfrac{x}{x-3}-\dfrac{\left(x-1\right)^2}{x^2-2x-3}\)
\(\Leftrightarrow\dfrac{-1\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}-\dfrac{x-3}{\left(x+1\right)\left(x-3\right)}=\dfrac{x\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}-\dfrac{x^2-2x+1}{\left(x-3\right)\left(x+1\right)}\)
Suy ra: \(-x-1-x+3=x^2+x-x^2+2x-1\)
\(\Leftrightarrow3x-1=-2x+2\)
\(\Leftrightarrow3x+2x=2+1\)
\(\Leftrightarrow5x=3\)
hay \(x=\dfrac{3}{5}\)(nhận)
Vậy: \(S=\left\{\dfrac{3}{5}\right\}\)
22 tháng 4 2021
a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow x^2-2x+12-8-x^2=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
22 tháng 4 2021
b) Ta có: \(\left|2x+6\right|-x=3\)
\(\Leftrightarrow\left|2x+6\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Vậy: S={-3}
HH
2
25 tháng 1 2022
\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-\dfrac{x}{4}\)
\(\Leftrightarrow\dfrac{4x}{12}-\dfrac{6\left(2x+1\right)}{12}=\dfrac{2x}{12}-\dfrac{3x}{12}\)
\(\Leftrightarrow4x-6\left(2x+1\right)=2x-3x\)
\(\Leftrightarrow4x-12x-6=-x\)
\(\Leftrightarrow4x-12x-6+x=0\)
\(\Leftrightarrow-7x-6=0\)
\(\Leftrightarrow x=-\dfrac{6}{7}\)
14 tháng 2 2022
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x-3\right)\left(x+1\right)}\)
\(\Leftrightarrow x^2+x+x^2-3x=4x\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
=>x=0(nhận) hoặc x=3(loại)
14 tháng 2 2022
đk : x khác -1 ; 3
\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)=4x\Leftrightarrow2x^2-2x-4x=0\)
\(\Leftrightarrow2x^2-6x=0\Leftrightarrow2x\left(x-3\right)=0\Leftrightarrow x=0;x=3\left(ktm\right)\)
S
22 tháng 3 2021
a, 3x - 7 = 0
<=> 3x = 7
<=> x = 7/3
b, 8 - 5x = 0
<=> -5x = -8
<=> x = 8/5
c, 3x - 2 = 5x + 8
<=> -2x = 10
<=> x = -5
22 tháng 3 2021
e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)
TA
1
11 tháng 9 2021
Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8+20x-12=0\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
\(\dfrac{2x-1}{3}-\dfrac{x-1}{2}+\dfrac{x+1}{6}=1\)
\(\Leftrightarrow\dfrac{2\left(2x-1\right)-3\left(x-1\right)+x+1-6}{6}=0\)
\(\Leftrightarrow4x-2-3x+3+x+1-6=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)