\(\frac{3x+2}{3x-2}\)-\(\frac{6}{2+3x}\)=\(\frac{9x^2}{9x^2-4}\)
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KR
2
5 tháng 5 2019
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 \(S=\left\{\frac{8}{3}\right\}\)
HS
2
SF
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:...................
LT
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 \(S=\left\{\frac{8}{3}\right\}\)
LM
2
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
\(\frac{1}{\left(3x-2\right)}-\frac{4}{\left(3x+2\right)}+\frac{3x-6}{\left(3x\right)^2-2^2}\left(MSC=\left(3x\right)^2-2^2=\left(3x-2\right)\left(3x+2\right)\right)\)
\(\frac{\left(3x+2\right)-4\left(3x-2\right)+\left(3x-6\right)}{\left(3x-2\right)\left(3x+2\right)}=\frac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\frac{4-6x}{\left(3x-2\right)\left(3x+2\right)}\)\(=\frac{-2\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}=-\frac{2}{\left(3x+2\right)}\)
\(A=-\frac{2}{3x+2}\)
A
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,
\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x}{9x^2-4}\\ =>\left(3x+2\right)^2-6\left(3x-2\right)=9x\\ =>9x^2+12x+4-18x-12-9x=0\\ =>9x^2-15x+16=0\\ =>x=..\)
NV
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
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 \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)
NV
1
NH
20 tháng 3 2020
\(\frac{2x-8}{6}-\frac{3x-1}{4}=\frac{9x-2}{8}+\frac{3x-1}{12}\)
\(\Leftrightarrow\frac{2x-8}{6}-\frac{3x-1}{4}-\frac{9x-2}{8}-\frac{3x-1}{12}=0\)
\(\Leftrightarrow\frac{4\left(2x-8\right)}{24}-\frac{6\left(3x-1\right)}{24}-\frac{3\left(9x-2\right)}{24}-\frac{2\left(3x-1\right)}{24}=0\)
\(\Leftrightarrow\frac{8x-32-18x+6-27x+6-6x+2}{24}=0\)
\(\Leftrightarrow\frac{-43x-18}{24}=0\)
\(\Rightarrow-43x-18=0\)
\(\Leftrightarrow-43x=18\)
\(\Leftrightarrow x=\frac{-18}{43}\)
Vậy...
N
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
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...