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1: \(\dfrac{x-3}{x+1}=\dfrac{x^2}{x^2-1}\)
=>(x-3)(x-1)=x^2
=>x^2=x^2-4x+3
=>-4x+3=0
=>x=3/4
2: \(\dfrac{5}{3x+2}=2x-1\)
=>(2x-1)(3x+2)=5
=>6x^2+4x-3x-2-5=0
=>6x^2+x-7=0
=>6x^2+7x-6x-7=0
=>(6x+7)(x-1)=0
=>x=1hoặc x=-7/6
ĐKXĐ ; \(x\ne\pm1\)
Ta có : \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2+3}{1-x^2}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}+\dfrac{-x^2-3}{x^2-1}=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2-x^2-3=0\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1-x^2-3=0\)
\(\Leftrightarrow-x^2+4x-3=0\)
\(\Leftrightarrow-x^2+3x+x-3=0\)
\(\Leftrightarrow-x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=1\left(L\right)\end{matrix}\right.\)
=> X = 3
Vậy ..
a: 3x-2=2x-3
=>x=-1
b: 2x+3=5x+9
=>-3x=6
=>x=-2
c: 5-2x=7
=>2x=-2
=>x=-2
d: 10x+3-5x=4x+12
=>5x+3=4x+12
=>x=9
e: 11x+42-2x=100-9x-22
=>9x+42=78-9x
=>18x=36
=>x=2
f: 2x-(3-5x)=4(x+3)
=>2x-3+5x=4x+12
=>7x-3=4x+12
=>3x=15
=>x=5
1: \(\dfrac{x}{3}-\dfrac{2x+1}{2}=x-6-x\)
=>2x-3(2x+1)=-36
=>2x-6x-3=-36
=>-4x=-33
=>x=33/4
2: \(3x-15=2x\left(x-5\right)\)
=>(x-5)(2x-3)=0
=>x=3/2 hoặc x=5
3: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)
=>x(x+3)+(x-2)(x+1)=2x(x+1)
=>x^2+3x+x^2+x-2-2x^2-2x=0
=>2x-2=0
=>x=1
\(\dfrac{3}{x}=\dfrac{5}{x-3}\left(x\ne0;x\ne3\right)\) đề như thế này phải ko ạ?
suy ra: \(3\left(x-3\right)=5x\\ < =>3x-9=5x\\ < =>3x-5x=9\\ < =>-2x=9\\ < =>x=-\dfrac{9}{2}\left(tm\right)\)