\(\Leftrightarrow\frac{8x^2}{3\left(1-2x\right)\left(1+2x\right)}=\frac{2x}{3\left(2x-1\right)}-\frac{1+8x}{4\left(1+2x\right)}\left(1\right)\)

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 \(S=\left\{\frac{-3}{26}\right\}\)

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.........

giup minh voi cac bạn

a,\(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{3+6x}{16x^2-1}\)

ĐKXĐ: x≠1/4, x≠-1/4

⇔\(-\frac{3}{4x-1}=\frac{2}{4x+1}-\frac{3+6x}{16x^2-1}\)

⇔\(\frac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\frac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\frac{3+6x}{16x^2-1}\)

⇒-12x-3=8x-2-3-6x

⇔8x-6x+12x=-3+2+3

⇔14x=2

⇔x=1/7(tmđk)

Vậy phương trình có nghiệm là x=1/7

b, \(\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\) (2)

ĐKXĐ: x≠0, x≠2

(2)⇔\(\frac{2\left(5-x\right)}{2.4x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4.\left(x-1\right)}{4.2x\left(x-2\right)}+\frac{x}{8.x\left(x-2\right)}\)

⇒10-2x+7x-14=4x-4+x

⇔-2x+7x-4x-x=-4-10+14

⇔0x=0

⇔ x∈R

Vậy phương trình có nghiệm là x∈R và x≠0, x≠2

c, \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\) (3)

ĐKXĐ: x≠0

(3)⇒x(x+1)(x²-x+1)-x(x-1)(x²+x+1)=3

⇔x⁴+x-x⁴+x=3

⇔2x=3

⇔x=3/2(tmđk)

Vậy phương trình có nghiệm là x=3/2

a) ĐKXĐ: x khác +2

\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)

<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)

<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)

<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22

<=> x^2 - 7x - 2 = 2x - 22

<=> x^2 - 7x - 2 - 2x + 22 = 0

<=> x^2 - 9x + 20 = 0

<=> (x - 4)(x - 5) = 0

<=> x - 4 = 0 hoặc x - 5 = 0

<=> x = 4 hoặc x = 5

làm nốt đi 

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, \(ĐKXĐ:x\ne\pm\frac{1}{2}\)

\(\frac{8x^2}{3\left(1-4x^2\right)}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)

\(\Leftrightarrow\frac{8x^2}{3\left(1-2x\right)\left(2x+1\right)}=\frac{2x}{3\left(2x-1\right)}-\frac{1+8x}{4\left(2x+1\right)}\)

\(\Leftrightarrow\frac{32x^2}{12\left(1-2x\right)\left(2x+1\right)}=\frac{-8x\left(2x+1\right)}{12\left(1-2x\right)\left(2x+1\right)}-\frac{3\left(1+8x\right)\left(1-2x\right)}{12\left(1-2x\right)\left(2x+1\right)}\)

\(\Rightarrow32x^2=-16x^2-8x-3+6x-24x+48x\)

\(\Leftrightarrow48x^2=22x-3\)

\(\Leftrightarrow48x^2-22x+3=0\)

Chuyển hết sang vế phải quy đồng ta được:

\(\frac{16x^2+4x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)}{6\left(2x-1\right)\left(2x+1\right)}=0\)

\(\Leftrightarrow\frac{16x^2+8x^2+4x-48x^2+6x+1}{6\left(2x+1\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow24x^2-10x-1=0\Leftrightarrow\left(12x+1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{12}\\x=\frac{1}{2}\end{cases}}\)