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7 tháng 12 2016

http://olm.vn/hoi-dap/question/772794.html

co vẽ bạn hỏi thật

\(A=\frac{1}{3x-2}+-\frac{4}{3x+2}-\frac{3x-6}{\left(2-3x\right)\left(2+3x\right)}\) DK \(x\ne\) +-2/3;

A=\(\frac{\left(3x-2\right)-4\left(3x-2\right)-2\left(x-3\right)}{\left(2-3x\right)\left(2+3x\right)}=\)\(\frac{3x-2-12x+8-2x+6}{\left(2-3x\right)\left(2+3x\right)}=\frac{12-11x}{\left(2-3x\right)\left(2+3x\right)}=\frac{11x-2}{\left(3x-2\right)\left(3x+2\right)}\)

7 tháng 12 2016

k nhanh thế thì giải tiếp (dk x khac +-2/3

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

\(\frac{\left(3x+2\right)-4\left(3x-2\right)+\left(3x-6\right)}{\left(3x-2\right)\left(3x+2\right)}=\frac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\frac{4-6x}{\left(3x-2\right)\left(3x+2\right)}\)\(=\frac{-2\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}=-\frac{2}{\left(3x+2\right)}\)

\(A=-\frac{2}{3x+2}\)

ĐKXĐ : \(x\ne\pm\frac{2}{3}\)

Suy ra : \(\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)

\(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)

\(\Leftrightarrow-6x=-16\)

\(\Leftrightarrow x=\frac{8}{3}\left(TM\right)\)

Vậy pt có tập nghiệm là \(S=\left\{\frac{8}{3}\right\}\)

18 tháng 4 2017

\(1.\frac{7x-3}{x-1}=\frac{2}{3}\)   ( \(x\ne1\))

\(\Leftrightarrow\frac{3\left(7x-1\right)}{3\left(x-1\right)}=\frac{2\left(x-1\right)}{3\left(x-1\right)}\)

\(\Rightarrow3\left(7x-3\right)=2\left(x-1\right)\)

\(\Leftrightarrow21x-9=2x-2\)

\(\Leftrightarrow19x=7\)

\(\Leftrightarrow x=\frac{7}{19}\)

\(2.\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\)

\(\Leftrightarrow\frac{\left(5x-1\right)\left(3x-1\right)}{\left(3x+2\right)\left(3x-1\right)}=\frac{\left(5x-7\right)\left(3x+2\right)}{\left(3x-1\right)\left(3x+2\right)}\)

\(\Rightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)

\(\Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\)

\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)

\(\Leftrightarrow\left(15x^2-15x^2\right)+\left(-8x+11x\right)=-14-1\)

\(\Leftrightarrow3x=-15\)

\(\Leftrightarrow x=-5\)

\(3.\frac{1-x}{x+1}+3=\frac{2x+3}{3x-1}\)

\(\Leftrightarrow\frac{\left(1-x\right)\left(3x-1\right)}{\left(x+1\right)\left(3x-1\right)}+\frac{3\left(x+1\right)\left(3x-1\right)}{\left(x+1\right)\left(3x-1\right)}=\frac{\left(2x+3\right)\left(x+1\right)}{\left(3x-1\right)\left(0+1\right)}\)

\(\Rightarrow\left(1-x\right)\left(3x-1\right)+3\left(x+1\right)\left(3x-1\right)=\left(2x+3\right)\left(x+1\right)\)

\(\Leftrightarrow3x-1-3x^2+x+3\left(3x^2-x+3x-1\right)=2x^2+2x+3x+3\)

\(\Leftrightarrow3x-1-3x^2+x+9x^2-3x+9x-3=2x^2+2x+3x+3\)

\(\Leftrightarrow6x^2+10x-4=2x^2+5x+3\)

\(\Leftrightarrow\left(6x^2-2x^2\right)+\left(10x-5x\right)=7\)

\(\Leftrightarrow4x^2+5x-7=0\)

\(\Leftrightarrow\left(2x\right)^2+4x.\frac{5}{4}+\frac{16}{25}+\frac{191}{25}=0\)

\(\Leftrightarrow\left(2x+\frac{5}{4}\right)^2-\frac{191}{25}=0\)

\(\left(2x+\frac{5}{4}\right)^2>0\)

\(\Rightarrow\left(2x+\frac{5}{4}\right)^2+\frac{191}{25}>0\)

=> PT vô nghiệm 

\(4.\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)

\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{x^2-4}+\frac{\left(9x+4\right)\left(x-2\right)}{x^2-4}=\frac{2\left(3x-2\right)+1}{x^2-4}\)

\(\Rightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3\left(3x-2\right)+1\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-2x+1\)

\(\Leftrightarrow3x^2-25x-6=3x^2-2x+1\)

\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(-25x+2x\right)+\left(-6-1\right)=0\)

\(\Leftrightarrow-23x-7=0\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=\frac{-7}{23}\)

\(5.\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)

\(\Leftrightarrow\frac{\left(3x+2\right)^2}{9x^2-4}-\frac{6\left(3x-2\right)}{9x^2-4}=\frac{9x^2}{9x^2-4}\)

\(\Rightarrow\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)

\(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)

\(\Leftrightarrow\left(9x^2-9x^2\right)+\left(12x-18x\right)+\left(4+12\right)=0\)

\(\Leftrightarrow-6x+16=0\)

\(\Leftrightarrow-6x=-16\)

\(\Leftrightarrow x=\frac{16}{6}\)

\(6.1+\frac{1}{x+2}=\frac{12}{8-x^3}\)

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

\(\Rightarrow\left(x+2\right)\left(8-x^3\right)+1\left(8-x^3\right)=12\left(x+2\right)\)

\(\Leftrightarrow8x+x^4+16+2x^3+8-x^3=12x+24\)

\(\Leftrightarrow x^4+\left(2x^3-x^3\right)+\left(8x-12x\right)+\left(16-24\right)=0\)

\(\Leftrightarrow x^4+x^3-4x-8=0\)

\(\Leftrightarrow\left(x^4-4x\right)+\left(x^3-8\right)=0\)

Đến đấy mk tắc r xl bạn nhé 

21 tháng 2 2017

Giúp zs ạ

21 tháng 2 2017

3x+2/3x-2 - 6/2+3x = 9x2/9x2-4 (đkxđ :3/2 , -3/2)

<=>3x+2/(3x-2).(3x+2) - 6/(3x-2).(3x+2) = 9x2/(3x-2).(3x+2)

=>3x+2 - 6 = 9x2

<=>3x-9x2=6+2

<=>3x-9x2=8

<=>3(x-3x)=8

<=>x-3x=8/3

<=>2x=8/3

<=>x=8/3 / 2

<=>x=4/3(thoải mãn)

vậy phương trình có nghiệm x = 4/3

25 tháng 4 2020

\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\) (1)

đk: \(x\ne\pm\frac{2}{3}\)

(1)\(\Leftrightarrow\frac{\left(3x+2\right)^2-6\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}=\frac{9x^2}{9x^2-4}\)

\(\Leftrightarrow\frac{9x^2-6x+16}{\left(3x-2\right)\left(3x+2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Rightarrow9x^2-6x+16=9x^2\)

\(\Leftrightarrow16-6x=0\)

\(\Leftrightarrow x=\frac{8}{3}\)(thỏa mãn đkxđ)

vậy:...................

25 tháng 4 2020

ĐKXĐ: \(x\ne\frac{2}{3};x\ne-\frac{2}{3}\)

\(\Leftrightarrow\frac{3x+2}{3x-2}-\frac{6}{3x+2}-\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}=0\)

\(\Leftrightarrow\frac{\left(3x+2\right)^2-6\left(3x-2\right)-9x^2}{\left(3x-2\right)\left(3x+2\right)}=0\)

\(\Leftrightarrow9x^2+12x+4-18x+12-9x^2=0\)

\(\Leftrightarrow16-6x=0\)

\(\Leftrightarrow6x=16\)

\(\Leftrightarrow x=\frac{8}{3}\left(TM\right)\)

Vậy \(S=\left\{\frac{8}{3}\right\}\)

15 tháng 4 2020
https://i.imgur.com/5GwnHAD.jpg
15 tháng 4 2020

a, 2x-6=5x-9

=>2x-5x=-9+6

=>-3x=-3

=>x=1

b, 4x2-6x=0

=> 2x(2x-3)=0

=>x=0 hoặc x=\(\frac{3}{2}\)

c,

\(\frac{4+3x}{3}=\frac{x^2+1}{x}\\ =>\frac{4x+3x^2-3x^2-3}{x}=0\\ =>\frac{4x-3}{x}=0\\ =>4x-3=0\\ =>x=\frac{3}{4}\)

d,

\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x}{9x^2-4}\\ =>\left(3x+2\right)^2-6\left(3x-2\right)=9x\\ =>9x^2+12x+4-18x-12-9x=0\\ =>9x^2-15x+16=0\\ =>x=..\)

22 tháng 4 2020

Bài làm

a) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x-4}\)

\(\Leftrightarrow\frac{3x+2}{3x-2}-\frac{6}{3x+2}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\frac{(3x+2)\left(3x+2\right)}{(3x-2)\left(3x+2\right)}-\frac{6\left(3x-2\right)}{(3x+2)\left(3x-2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Rightarrow\left(3x+2\right)^2-\left(18x-12\right)=9x^2\)

\(\Leftrightarrow9x^2+12x+4-18x+12x-9x^2=0\)

\(\Leftrightarrow6x+4=0\)

\(\Leftrightarrow x=-\frac{4}{6}\)

\(\Leftrightarrow x=-\frac{2}{3}\)

Vậy x = -2/3 là nghiệm.

23 tháng 4 2020

@Tao Ngu :))@ 9x-4 không tách thành (3x+4)(3x-4) được đâu bạn. Chỗ đó phải là: 9x2-4

Bài thiếu đkxđ của x \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)

27 tháng 2 2020

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

13 tháng 2 2018

      ĐKXĐ:   \(x\ne\pm\frac{2}{3}\)

             \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)

\(\Leftrightarrow\)\(\frac{\left(3x+2\right)^2}{\left(3x-2\right)\left(3x+2\right)}-\frac{6\left(3x-2\right)}{\left(2+3x\right)\left(3x-2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Rightarrow\)\(\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)

\(\Leftrightarrow\)\(9x^2+12x+4-18x+12-9x^2=0\)

\(\Leftrightarrow\)\(6x=16\)

\(\Leftrightarrow\)\(x=\frac{8}{3}\)  (t/m ĐKXĐ)

Vậy...

31 tháng 12 2018

a) \(=\frac{3x+2}{\left(3x+2\right).\left(3x-2\right)}-\frac{12x-8}{\left(3x+2\right).\left(3x-2\right)}-\frac{-3x+6}{\left(3x-2\right).\left(3x+2\right)}\)

\(b,\frac{x^2+1}{\left(x-1\right).\left(x^2+1\right)}-\frac{x.\left(x^2-1\right).\left(x-1\right)}{\left(x-1\right).\left(x^2+1\right)}.\left(\frac{1}{\left(x-1\right)^2}-\frac{1}{\left(x+1\right).\left(x-1\right)}\right)\)

p/s: hướng dấn cách tách thoy, tự làm nha~~lazy

31 tháng 12 2018

a )

\(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{3x-6}{4-9x^2}=0\)

\(\Leftrightarrow\frac{\left(3x+2\right)-4.\left(3x-2\right)}{9x^2-4}=\frac{3x-6}{4-9x^2}\) ( * ) 

Đkxđ : \(\hept{\begin{cases}9x^2-4\ne0\\4-9x^2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne\pm\sqrt{\frac{4}{9}}\\x\ne\pm\sqrt{\frac{4}{9}}\end{cases}}\Leftrightarrow x\ne\pm\frac{2}{3}\)

( * ) => \(\left(4-9x^2\right).\left[\left(3x+2\right)+\left(-12x+8\right)\right]=\left(9x^2-4\right).\left(3x-6\right)\)

\(\Leftrightarrow\left(4-9x^2\right).\left(-9x+10\right)=\left(9x^2-4\right).\left(3x-6\right)\)

\(\Leftrightarrow-36x+40+81x^3-90x^2=27x^3-54x^2-12x+24\)

\(\Leftrightarrow54x^3-36x^2-24x+16=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\left(loai\right)\\x=-\frac{2}{3}\left(loai\right)\end{cases}}\)

Vậy : phương trình vô nghiệm