K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

27 tháng 2 2020

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

Câu 3: Giải các phương trình sau bằng cách đưa về dạng ax+b=0 1. a, \(\frac{5x-2}{3}=\frac{5-3x}{2}\); b, \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\) c, \(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\); d, \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\) e, \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\); f, 4 (0,5-1,5x)=\(\frac{5x-6}{3}\) g, \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\); h, \(\frac{x+4}{5}.x+4=\frac{x}{3}-\frac{x-2}{2}\) i,...
Đọc tiếp

Câu 3: Giải các phương trình sau bằng cách đưa về dạng ax+b=0

1. a, \(\frac{5x-2}{3}=\frac{5-3x}{2}\); b, \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)

c, \(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\); d, \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)

e, \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\); f, 4 (0,5-1,5x)=\(\frac{5x-6}{3}\)

g, \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\); h, \(\frac{x+4}{5}.x+4=\frac{x}{3}-\frac{x-2}{2}\)

i, \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\); k, \(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)

m, \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\); n, \(\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+1\right).\frac{1}{3}\left(x+2\right)\)

p, \(\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x\); q, \(\frac{2+x}{5}-0,5x=\frac{1-2x}{4}+0,25\)

r, \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\); s, \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{6}\)

t, \(\frac{2x-8}{6}.\frac{3x+1}{4}=\frac{9x-2}{8}+\frac{3x-1}{12}\); u, \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{3}+\frac{2x-1}{12}\)

v, \(\frac{5x-1}{10}+\frac{2x+3}{6}=\frac{x-8}{15}-\frac{x}{30}\); w, \(\frac{2x-\frac{4-3x}{5}}{15}=\frac{7x\frac{x-3}{2}}{5}-x+1\)

17

Đây là những bài cơ bản mà bạn!

29 tháng 3 2020

bạn ấy muốn thách xem bạn nào đủ kiên nhẫn đánh hết chỗ này

ĐKXĐ : \(\hept{\begin{cases}x-2\ne0\\3-4x\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne2\\x\ne\frac{3}{4}\end{cases}}}\)

\(\frac{5}{x-2}+\frac{6}{3-4x}=0\)

\(\frac{5\left(3-4x\right)}{\left(x-2\right)\left(3-4x\right)}+\frac{6\left(x-2\right)}{\left(3-4x\right)\left(x-2\right)}=0\)

\(15-20x+6x-12=0\)

\(3-14x=0\Leftrightarrow14x=3\Leftrightarrow x=\frac{3}{14}\)theo ĐKXĐ : x thỏa mãn 

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