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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) 3-4x\(\ge\)11
\(4x\le3-11=-8\)
\(x\le-2\)
( câu b bn ghi rõ đề bài đc ko ?)
a) \(3-2x>4\)
\(\Leftrightarrow-2x>1\)
\(\Leftrightarrow x< \frac{-1}{2}\)
b) \(\frac{2}{3-x}-\frac{9}{3+x}=\frac{1}{2}\)ĐKXĐ : \(x\pm3\)
\(\Leftrightarrow\frac{-4\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}-\frac{18\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{\left(x-3\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow-4x-13-18x+54=x^2-9\)
\(\Leftrightarrow x^2+22x-50=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot11+11^2-171=0\)
\(\Leftrightarrow\left(x+11\right)^2=\left(\pm\sqrt{171}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{171}-11\\x=-\sqrt{171}-11\end{cases}}\)( thỏa )
Vậy....
\(a,\)\(3-2x>4\)
\(\Rightarrow-2x>1\)
\(\Rightarrow x< \frac{-1}{2}\)
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
3x+5<14
<=>3x<9
<=>x<3
vậy S=(x\(\in\)R/x<3)