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, \(\frac{2x-\frac{4-3x}{5}}{15}=\frac{7x\frac{x-3}{2}}{5}-x+1\)
a) \(5x-7=3x+9\)
\(\Rightarrow5x-3x=9+7\)
\(\Rightarrow2x=16\)
\(\Rightarrow x=16:2\)
\(\Rightarrow x=8\)
Vậy \(x=8.\)
b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{2}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{2}{5}.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{2}{5}-\frac{1}{2}\\x=\left(-\frac{2}{5}\right)-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{10};-\frac{9}{10}\right\}.\)
c) \(5x-\left|9-7x\right|=3\)
\(\Rightarrow\left|9-7x\right|=5x-3\)
\(\Rightarrow\left[{}\begin{matrix}9-7x=5x-3\\9-7x=3-5x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}9+3=5x+7x\\9-3=-5x+7x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}12=12x\\6=2x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=12:12\\x=6:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{1;3\right\}.\)
d) \(-5+\left|3x-1\right|+6=\left|-4\right|\)
\(\Rightarrow-5+\left|3x-1\right|+6=4\)
\(\Rightarrow-5+\left|3x-1\right|=4-6\)
\(\Rightarrow-5+\left|3x-1\right|=-2\)
\(\Rightarrow\left|3x-1\right|=\left(-2\right)+5\)
\(\Rightarrow\left|3x-1\right|=3.\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=3\\3x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=4\\3x=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4:3\\x=\left(-2\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{4}{3};-\frac{2}{3}\right\}.\)
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