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, \(\frac{x+4}{5}.x+4=\frac{x}{3}-\frac{x-2}{2}\)
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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, \(\frac{2x-\frac{4-3x}{5}}{15}=\frac{7x\frac{x-3}{2}}{5}-x+1\)
Bài 1 :
a, Ta có : \(3x-1=2x+4\)
=> \(3x-2x=4+1\)
=> \(x=5\)
Vậy phương trình có tập nghiệm \(S=\left\{5\right\}\)
b, Ta có : \(5x-2=0\)
=> \(5x=2\)
=> \(x=\frac{2}{5}\)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{2}{5}\right\}\)
c, Ta có : \(7x-4=3x+12\)
=> \(7x-3x=12+4\)
=> \(4x=16\)
=> \(x=4\)
Vậy phương trình có tập nghiệm \(S=\left\{4\right\}\)
d, Ta có : \(\frac{x-1}{2}+\frac{3x+2}{4}=\frac{x-7}{12}\)
=> \(\frac{6\left(x-1\right)}{12}+\frac{3\left(3x+2\right)}{12}=\frac{x-7}{12}\)
=> \(6\left(x-1\right)+3\left(3x+2\right)=x-7\)
=> \(6x-6+9x+6=x-7\)
=> \(6x+9x-x=6-7-6\)
=> \(14x=-7\)
=> \(x=-\frac{1}{2}\)
Vậy phương trình có tập nghiệm \(S=\left\{-\frac{1}{2}\right\}\)
Bài 2 :
a, ĐKXĐ : \(\left\{{}\begin{matrix}x^2-2x+1\ne0\\x-1\ne0\end{matrix}\right.\)
=> \(x-1\ne0\)
=> \(x\ne1\)
- Ta có : \(\left(\frac{x+1}{x^2-2x+1}+\frac{1}{x-1}\right):\frac{x}{x-1}-\frac{2}{x-1}\)
= \(\left(\frac{x+1}{\left(x-1\right)^2}+\frac{x-1}{\left(x-1\right)^2}\right):\frac{x}{x-1}-\frac{2}{x-1}\)
= \(\left(\frac{2x}{\left(x-1\right)^2}\right):\frac{x}{x-1}-\frac{2}{x-1}\)
= \(\left(\frac{2x}{\left(x-1\right)^2}\right)\left(\frac{x-1}{x}\right)-\frac{2}{x-1}\)
= \(\frac{x}{x-1}-\frac{2}{x-1}\)
= \(\frac{x-2}{x-1}\)
cảm ơn