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\(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\)
=> \(\frac{3}{2}x-\frac{2}{5}-\frac{1}{3}x=-\frac{1}{4}\)
=> \(\frac{3}{2}x-\frac{1}{3}x-\frac{2}{5}=-\frac{1}{4}\)
=> \(\frac{3}{2}x-\frac{1}{3}x=-\frac{1}{4}+\frac{2}{5}\)
=> \(\frac{9}{6}x-\frac{2}{6}x=-\frac{5}{20}+\frac{8}{20}\)
=> \(\frac{7}{6}x=\frac{3}{20}\)
=> \(x=\frac{3}{20}:\frac{7}{6}=\frac{3}{20}\cdot\frac{6}{7}=\frac{3}{10}\cdot\frac{3}{7}=\frac{9}{70}\)
\(-\frac{4}{3}\left[x-\frac{1}{4}\right]=\frac{3}{2}\left[2x-1\right]\)
=> \(-\frac{4}{3}x-\left[-\frac{1}{3}\right]=3x-\frac{3}{2}\)
=> \(-\frac{4}{3}x+\frac{1}{3}=3x-\frac{3}{2}\)
=> \(-\frac{4}{3}x+\frac{1}{3}-3x=-\frac{3}{2}\)
=> \(-\frac{4}{3}x-3x+\frac{1}{3}=-\frac{3}{2}\)
=> \(-\frac{4}{3}x-\frac{3}{1}x=-\frac{3}{2}-\frac{1}{3}\)
=> \(-\frac{4}{3}x-\frac{9}{3}x=-\frac{9}{6}-\frac{2}{6}\)
=> \(-\frac{13}{3}x=-\frac{11}{6}\)
=> \(x=-\frac{11}{6}:\left[-\frac{13}{3}\right]=-\frac{11}{6}\cdot\left[-\frac{3}{13}\right]=-\frac{11}{2}\cdot\left[-\frac{1}{13}\right]=\frac{11}{26}\)
Ta có: \(\frac{2-x}{4}=\frac{3x-1}{3}\)
\(\Leftrightarrow\left(2-x\right).3=\left(3x-1\right).4\)
\(\Leftrightarrow6-3x=12x-4\)
\(\Leftrightarrow-3x-12x=-4-6\)
\(\Leftrightarrow-15x=-10\)
\(\Leftrightarrow x=\frac{2}{3}\)
Vậy ...
\(\frac{2-x}{4}=\frac{3x-1}{3}\)
\(3.\left(2-x\right)=4.\left(3x-1\right)\)
\(6-3x=12x-4\)
\(-3x-12x=-4-6\)
\(-15x=-10\)
\(x=\frac{2}{3}\)
Vậy \(x=\frac{2}{3}\)
b) \(\left||3x+1|+3\right|=2\)
Mà \(\left|3x+1\right|\ge0\)nên \(\left|3x+1\right|+3\ge3\)
Vậy biểu thức trong dấu GTTĐ luôn dương
\(\Rightarrow\left|3x+1\right|+3=2\)
\(\Rightarrow\left|3x+1\right|=-1\)(vô lí)
Vậy pt vô nghiệm
a) \(\left|2x-1\right|-4=5\)
\(\Leftrightarrow\left|2x-1\right|=5+4\)
\(\Leftrightarrow\left|2x-1\right|=9\)
\(\Leftrightarrow2x-1=\pm9\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=9\\2x-1=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)
c) \(\left|3x-2\right|=4-2x\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=4-2x\\-\left(3x-2\right)=4-2x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=-2\end{cases}}\)
d) \(\left|1-3x\right|=1+2x\)
\(\Leftrightarrow\orbr{\begin{cases}1-3x=1+2x\\-\left(1-3x\right)=1+2x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
b) \(\sqrt{2x-3}-7=4\)
\(\sqrt{2x-3}=11\)
\(\left(\sqrt{2x-3}\right)^2=11^2\)
\(2x-3=121\)
\(2x=124\)
\(x=62\)
c) \(\sqrt{3x-2}+7=0\)
\(\sqrt{3x-2}=-7\)
\(\Rightarrow x=\varnothing\)
bạn Hoàng Thanh Huyền ơi! cảm ơn đã là giúp nhưng phần a) bạn làm đến dong thứ 3 thì mk bt làm r nhưng mũ 2 phải chia ra hai trường hợp chứ :))
a) \(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\)
=> \(\frac{3}{2}x-\frac{2}{5}-\frac{1}{3}x+\frac{1}{4}=0\)
=> \(\left(\frac{3}{2}-\frac{1}{3}\right)x+\left(-\frac{2}{5}+\frac{1}{4}\right)=0\)
=> \(\frac{7}{6}x-\frac{3}{20}=0\)
=> \(\frac{7}{6}x=\frac{3}{20}\)
=> \(x=\frac{3}{20}:\frac{7}{6}=\frac{3}{20}\cdot\frac{6}{7}=\frac{9}{70}\)
b) \(2x-\frac{2}{3}=7x+\frac{2}{3}-1\)
=> \(2x-\frac{2}{3}=7x-\frac{1}{3}\)
=> \(2x-\frac{2}{3}-7x+\frac{1}{3}=0\)
=> (2x - 7x) + (-2/3 + 1/3) = 0
=> -5x - 1/3 = 0
=> -5x = 1/3
=> x = -1/15
\(2x -10 - [3x - 13 - (3 - 5x) - 4] =7\)
\(2x -10 - 3x + 13 + 3 - 5x + 4 =7\)
\(2x-3x-5x=10-13-3-4+7\)
\(-6x=-3\)
\(\frac{1}{2}\)
\(x = \) \(\frac{1}{2}\)