a. ĐKXĐ: \(x\ne2\).
\(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)
⇔\(\dfrac{1}{x-2}+\dfrac{3x-6}{x-2}=\dfrac{3-x}{x-2}\)
⇔\(1+3x-6=3-x\)
⇔\(4x-8=0\)
⇔\(x=2\) (không thỏa mãn)
-Vậy S=∅.
b. ĐKXĐ: \(x\ne-1\)
\(\dfrac{5x}{2x+2}+1=-\dfrac{6}{x+1}\)
⇔\(\dfrac{5x}{2\left(x+1\right)}+1=-\dfrac{6}{x+1}\)
⇔\(\dfrac{5x}{2\left(x+1\right)}+\dfrac{2\left(x+1\right)}{2\left(x+1\right)}=-\dfrac{12}{2\left(x+1\right)}\)
⇔\(5x+2\left(x+1\right)=-12\)
⇔\(5x+2x+2+12=0\)
⇔\(7x+14=0\)
⇔\(x=-2\) (thỏa mãn).
-Vậy \(S=\left\{-2\right\}\)
a, \(\Leftrightarrow\dfrac{1}{x-2}+\dfrac{3.\left(x-2\right)}{x-2}=\dfrac{3-x}{x-2}\\ \Leftrightarrow1+3x-6=3-x\)
\(\Leftrightarrow3x+x=3-1+6\\ \Leftrightarrow4x=8\\ \Leftrightarrow x=\dfrac{8}{4}=2\\ Vậy.S=\left\{2\right\}\)
b, \(\Leftrightarrow\)\(\dfrac{5x}{2x+2}+\dfrac{2x+2}{2x+2}=\dfrac{-6.2}{2.\left(x+1\right)}\)
\(\Leftrightarrow5x+2x+2=-12\\ \Leftrightarrow7x=-12-2\\ \Leftrightarrow7x=-14\\ \Leftrightarrow x=-\dfrac{14}{7}=-2\\ Vậy.S=\left\{-2\right\}\)
a) \(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\Leftrightarrow\dfrac{1}{x-2}+\dfrac{3\left(x-2\right)}{x-2}+\dfrac{-3+x}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3x-6+x-3}{x-2}=0\Leftrightarrow\dfrac{4x-8}{x-2}=0\Leftrightarrow\dfrac{4\left(x-2\right)}{x-2}=0\)
\(\Leftrightarrow4=0\) ( vô lý ) => PT vô nghiệm
b)\(\dfrac{5x}{2x+2}+1=-\dfrac{6}{x+1}\Leftrightarrow\dfrac{5}{2\left(x+1\right)}+\dfrac{2\left(x+1\right)}{2\left(x+1\right)}+\dfrac{12}{2\left(x+1\right)}=0\)
\(\Leftrightarrow\dfrac{5x+2x+2+12}{2\left(x+1\right)}=0\Leftrightarrow\dfrac{10x+14}{2\left(x+1\right)}=0\Leftrightarrow x=\dfrac{-7}{5}\)
a) ĐKXĐ: `x ne 2`
`1/(x-2) + 3 = (3-x)/(x-2)`
`<=> 1/(x-2) - (3-x)/(x-2) = -3`
`<=> (1-3+x)/(x-2) = -3`
`<=> (-2+x)/(x-2) =-3`
`<=> 1 = -3` (vô lí)
Vậy phương trình vô nghiệm
b)ĐKXĐ: `x ne 2`
`(5x)/(2x+2) + 1 = -6/(x+1)`
`<=>(5x)/(2x+2) + 1 = -12/(2x+2)`
`<=> 5x/(2x+2) + 12/(2x+2) =-1`
`<=> (5x+12)/(2x+2) = -1`
`=> 5x + 12 = -1(2x+2)`
`<=> 5x+12 = -2x - 2`
`<=> 7x = -14`
`<=> x= -2` (thỏa mãn ĐKXĐ)
Vậy tập nghiệm của phương trình là `S={ -2}`
