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a) \(A=\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{8x}{3-6x}\left(ĐK:x\ne\pm\frac{1}{2}\right)\)
\(=\frac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}:\frac{8x}{3\left(1-2x\right)}\)
\(=\frac{4x^2+4x+1-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\cdot\frac{3\left(1-2x\right)}{8x}\)
\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}\cdot\frac{-3\left(2x-1\right)}{8x}\)
\(=\frac{-3}{2x+1}\)
b) Với mọi x thuộc ĐKXĐ mà \(A=-\frac{3}{4031}\Leftrightarrow\frac{-3}{2x+1}=\frac{-3}{4031}\Leftrightarrow2x+1=4031\Leftrightarrow x=2015\left(tm\right)\)
Vậy x=2015 thì \(A=-\frac{3}{4031}\)
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a, \(ĐKXĐ:x\ne2\)
\(\frac{1}{x-2}+3=\frac{x-3}{2-x}\)
\(\Leftrightarrow\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}=\frac{3-x}{x-2}\)
\(\Rightarrow1+3x-6=3-x\)
\(\Leftrightarrow1+3x-6-3+x=0\)
\(\Leftrightarrow4x-8=0\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\left(ktm\right)\)
vậy x thuộc tập hợp rỗng
b, \(ĐKXĐ:x\ne\pm1\)
\(\frac{x}{x-1}-\frac{2x}{x^2-1}=0\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Rightarrow x^2+x-2x=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x-1=0\Rightarrow x=1\left(ktm\right)\end{cases}}\)
vậy x = 0
c, \(ĐKXĐ:x\ne\pm\frac{1}{2}\)
\(\frac{8x^2}{3\left(1-4x^2\right)}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
\(\Leftrightarrow\frac{8x^2}{3\left(1-2x\right)\left(2x+1\right)}=\frac{2x}{3\left(2x-1\right)}-\frac{1+8x}{4\left(2x+1\right)}\)
\(\Leftrightarrow\frac{32x^2}{12\left(1-2x\right)\left(2x+1\right)}=\frac{-8x\left(2x+1\right)}{12\left(1-2x\right)\left(2x+1\right)}-\frac{3\left(1+8x\right)\left(1-2x\right)}{12\left(1-2x\right)\left(2x+1\right)}\)
\(\Rightarrow32x^2=-16x^2-8x-3+6x-24x+48x\)
\(\Leftrightarrow48x^2=22x-3\)
\(\Leftrightarrow48x^2-22x+3=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\Leftrightarrow\frac{8x^2}{3\left(1-2x\right)\left(1+2x\right)}=\frac{2x}{3\left(2x-1\right)}-\frac{1+8x}{4\left(1+2x\right)}\left(1\right)\)
Điều kiện : \(x\ne\frac{1}{2};\frac{-1}{2}\)
\(\left(1\right)\Leftrightarrow\frac{8x^2.4}{12\left(1-2x\right)\left(1+2x\right)}=\frac{-2x\left(1+2x\right).4}{12\left(1-2x\right)\left(1+2x\right)}-\frac{3\left(1+8x\right)\left(1-2x\right)}{12\left(1+2x\right)\left(1-2x\right)}\)
=> 32x2 = -8x(1+2x) - 3(1+8x)(1-2x)
<=> 32x2 = -8x - 16x2 + (-3-24x)(1-2x)
<=> 32x2 = -16x2 -8x -3 + 6x - 24x + 48x2
<=> -26x = 3
<=> x= -3/26 (nhận)
Vậy tập nghiệm \(S=\left\{\frac{-3}{26}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
ĐKXĐ: $x\neq \pm \frac{1}{2}$
PT \(\Leftrightarrow \frac{8x^2}{3(1-4x^2)}=\frac{2x}{3(2x-1)}-\frac{8x+1}{4(2x+1)}=\frac{8x(2x+1)-3(8x+1)(2x-1)}{12(2x-1)(2x+1)}\)
\(\Leftrightarrow \frac{8x^2}{3(1-4x^2)}=\frac{-32x^2+26x+3}{12(4x^2-1)}\)
\(\Leftrightarrow \frac{8x^2}{3(1-4x^2)}=\frac{32x^2-26x-3}{12(1-4x^2)}\)
\(\Leftrightarrow 32x^2=32x^2-26x-3\)
\(\Leftrightarrow 26x+3=0\Rightarrow x=-\frac{3}{26}\) (t/m)
Vậy.........
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d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{8x}{3-6x}\)
\(=\frac{8}{4x^2-1}.\frac{8x}{3-6x}\)
\(=\frac{8.8}{\left(4x^2-1\right).\left(3-6x\right)}\)
\(=\frac{64}{\left(4x^2-1\right)\left(3-6x\right)}\)