\(\dfrac{2}{x+1}-\dfrac{3}{x-1}=5\)
\(\Leftrightarrow5\left(x+1\right)\left(x-1\right)=2\left(x-1\right)-3\left(x+1\right)\)
\(\Leftrightarrow5x^2-5=2x-2-3x-3\)
\(\Leftrightarrow5x^2-5=-x-5\)
\(\Leftrightarrow5x^2+x=0\)
=>x(5x+1)=0
=>x=0 hoặc x=-1/5
`2/[x+1]-3/[x-1]=5` `ĐK: x \ne +-1`
`<=>[2(x-1)-3(x+1)]/[(x-1)(x+1)]=[5(x-1)(x+1)]/[(x-1)(x+1)]`
`=>2x-2-3x-3=5x^2-5`
`<=>5x^2+x=0`
`<=>x(5x+1)=0`
`<=>x=0` hoặc `x=-1/5`
(t/m) (t/m)
Vậy `S={0;-1/5}`
đề bài :\(\dfrac{2}{x+1}-\dfrac{3}{x-1}=5\)
\(ĐK:x\ne1;x\ne-1\)
\(\Leftrightarrow>\dfrac{2x-2-3x-1}{\left(x+1\right)\left(x-1\right)}=\dfrac{5\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow2x-2-3x-3=5x^2-5\)
\(\Leftrightarrow-x-5=5x^2+x\)
\(\Leftrightarrow x\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\)
vậy \(S=\left\{0,-\dfrac{1}{5}\right\}\)
\(\dfrac{2}{x+1}-\dfrac{3}{x-1}=5\left(ĐKXĐ:x\ne1;x\ne-1\right)\)
\(\Leftrightarrow\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=5\)
\(\Leftrightarrow\dfrac{2x-2-3x-3}{\left(x+1\right)\left(x-1\right)}=5\)
\(\Leftrightarrow\dfrac{-x-5}{x^2-1}=5\)
\(\Leftrightarrow-x-5=5\left(x^2-1\right)\)
\(\Leftrightarrow5x^2-5+x+5=0\)
\(\Leftrightarrow5x^2+x=0\)
\(\Leftrightarrow x\left(5x+1\right)=0\)
\(\Leftrightarrow x=0\) hoặc 5x+1=0
<=>x=0 hoặc x=\(\dfrac{-1}{5}\)(Thỏa mãn)
Vậy \(x\in\left\{0;\dfrac{-1}{5}\right\}\)
2x+1−3x−1=52x+1-3x-1=5 ĐK:x≠±1ĐK:x≠±1
⇔2(x−1)−3(x+1)(x−1)(x+1)=5(x−1)(x+1)(x−1)(x+1)⇔2(x-1)-3(x+1)(x-1)(x+1)=5(x-1)(x+1)(x-1)(x+1)
⇒2x−2−3x−3=5x2−5⇒2x-2-3x-3=5x2-5
⇔5x2+x=0⇔5x2+x=0
⇔x(5x+1)=0⇔x(5x+1)=0
⇔x=0⇔x=0 hoặc x=−15x=-15
(t/m) (t/m)
Vậy S={0;−15}
