Tìm x biết:
\(\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\left|-\frac{3}{4}\right|\)
K
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6 tháng 2 2018
a, => |5/3.x| = 1/6
=> 5/3.x = -1/6 hoặc 5/3.x = 1/6
=> x = -1/10 hoặc x = 1/10
Tk mk nha
31 tháng 3 2017
\(\Rightarrow\frac{3}{4}x+5-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}+3\)+3
\(\Rightarrow\left(\frac{3}{4}x-\frac{2}{3}x-\frac{1}{6}x\right)+\left(5+4-1\right)=\frac{1}{3}x+\left(4-\frac{1}{3}+3\right)\)
=>\(\frac{-1}{12}x+8=\frac{1}{3}x+\frac{20}{3}\)\(\Rightarrow\frac{-1}{12}x+8-\frac{1}{3}x=\frac{20}{3}\)
\(\Rightarrow\left(\frac{-1}{12}-\frac{1}{3}\right)x+8=\frac{20}{3}\)
\(\Rightarrow\frac{-5}{12}x+8=\frac{20}{3}\Rightarrow\frac{-5}{12}x=\frac{20}{3}-8\)
\(\Rightarrow\frac{-5}{12}x=\frac{-4}{3}\Rightarrow x=\frac{-4}{3}:\frac{-5}{12}=\frac{16}{5}\)
12 tháng 5 2016
\(\frac{4}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)=\frac{4}{3}.\frac{-1}{3}=\frac{-4}{9}\)
k nha
MH
22 tháng 8 2016
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{3\left(x-1\right)\left(3x+3\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{4}{\left(3x-1\right)\left(3x+3\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)(Vì 3x + 3 lớn hơn 3x - 1 là 4 đơn vị)
\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{x+1-1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{x}{3x+3}=\frac{3}{10}\)
\(\Rightarrow10x=3.\left(3x+3\right)\)
\(\Rightarrow10x=9x+9\)
\(\Rightarrow x=9\)
Vậy...
31 tháng 1 2016
1
2(\(\frac{3}{4}\)-5x)=\(\frac{4}{5}\)-3x
=> \(\frac{6}{4}-10x=\frac{4}{5}-3x\)
=>\(-10x+3x=\frac{4}{5}-\frac{6}{4}\)
=> \(x=\frac{1}{10}\)
2 .
\(\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)
=>\(\frac{3}{2}-1+4x=\frac{2}{3}-7x\)
=>\(11x=\frac{1}{6}\)
=>x=\(\frac{1}{66}\)
3.
\(3\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
=>\(\frac{3}{2}-3x+\frac{1}{3}=\frac{7}{6}-x\)
=>\(-2x=\frac{-2}{3}\)
=>\(\frac{1}{3}\)
4. câu 4 ko hiểu bạn ơi
S
24 tháng 5 2016
a)
\(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)
\(\frac{9}{12}x-\frac{8}{12}x-\frac{2}{12}x=4+3-3-4+1\)
\(\left(\frac{9}{12}-\frac{8}{12}-\frac{2}{12}\right)x=1\)
\(-\frac{1}{12}x=1\)
\(x=-12\)
b)
\(\frac{x+1}{3}=\frac{x-2}{4}\)
\(\Rightarrow4\left(x+1\right)=3\left(x-2\right)\)
\(4x+4=3x-6\)
\(4x-3x=-6-4\)
\(x=-10\)
\(\left|\frac{3}{4}x-\frac{3}{40}\right|-\frac{3}{4}=\left|-\frac{3}{4}\right|\)
\(\Leftrightarrow\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\frac{3}{4}\)
\(\Leftrightarrow\left|\frac{3}{4}x-\frac{3}{4}\right|=\frac{3}{4}+\frac{3}{4}\)
\(\Leftrightarrow\left|\frac{3}{4}x-\frac{3}{4}\right|=\frac{3}{2}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\frac{3}{4}x-\frac{3}{4}=\frac{3}{2}\\3x-\frac{3}{4}=-\frac{3}{2}\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}3x=\frac{9}{4}\\3x=-\frac{3}{4}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{4}\\x=-\frac{1}{4}\end{array}\right.\)
Vậy ........................
\(\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\left|-\frac{3}{4}\right|\)
\(\Rightarrow\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\frac{3}{4}\)
\(\Rightarrow\left|\frac{3}{4}x-\frac{3}{4}\right|=\frac{3}{4}+\frac{3}{4}\)
\(\Rightarrow\left|\frac{3}{4}x-\frac{3}{4}\right|=\frac{3}{2}\)
\(\Rightarrow\left|\frac{3}{4}.\left(x-1\right)\right|=\frac{3}{2}\)
\(\Rightarrow\left|\frac{3}{4}\right|.\left|x-1\right|=\frac{3}{2}\)
\(\Rightarrow\left|x-1\right|=\frac{3}{2}:\frac{3}{4}\)
\(\Rightarrow\left|x-1\right|=\frac{3}{2}.\frac{4}{3}=2\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-1=2\\x-1=-2\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-1\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=3\\x=-1\end{array}\right.\)