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\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)
\(\left|x-\frac{1}{3}\right|=2\)
=> \(x-\frac{1}{3}=2\) hoặc \(x-\frac{1}{3}=-2\)
x = \(\frac{7}{3}\) x = \(\frac{-5}{3}\)
Vậy x = \(\frac{7}{3}\)hoặc x = \(\frac{-5}{3}\)
a) \(=10\frac{1}{4}\cdot\frac{-5}{3}-8\frac{1}{4}\cdot\frac{-5}{3}-5=\left(10\frac{1}{4}-8\frac{1}{4}\right)\cdot\frac{-5}{3}-5\)
\(=\left(\frac{41}{4}-\frac{33}{4}\right)\cdot\frac{-5}{3}-5=2\cdot\frac{-5}{3}-5\)\(=\frac{-10}{3}-\frac{15}{3}=\frac{-25}{3}\)
b)\(=\frac{5}{7}+1+\frac{2}{7}+\frac{2^{10}\cdot\left(2^3\right)^3}{\left(2^2\right)^9}\)
\(=\frac{5}{7}+\frac{2}{7}+1+\frac{2^{10}\cdot2^9}{2^{27}}\)
\(=1+1+\frac{1}{2^8}=2+\frac{1}{256}=\frac{512}{256}+\frac{1}{256}=\frac{513}{256}\)
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{4}\right|=4x.\)
Điều kiện \(4x\ge0\)nên
\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}=4x\)
\(\Leftrightarrow3x+\frac{13}{12}=4x\)
\(\Leftrightarrow4x-3x=\frac{13}{12}\)
\(\Leftrightarrow x=\frac{13}{12}\)
\(\frac{4}{5}-|x-\frac{1}{6}|=\frac{2}{3}\)
\(\Rightarrow|x-\frac{1}{6}|=\frac{2}{15}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{6}=\frac{2}{15}\\x-\frac{1}{6}=-\frac{2}{15}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{1}{30}\end{cases}}\)
Vậy.....
a) \(\left|2-x\right|+x=-3\\ \Rightarrow\left|2-x\right|=-3-x\left(ĐK:-3-x\ge0\right)\\ \Rightarrow\left[{}\begin{matrix}2-x=-3-x\\2-x=3+x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-x=-3-2\\-x-x=3-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0=-5\left(\text{vô lí}\right)\\-2x=1\end{matrix}\right.\Rightarrow x=\frac{-1}{2}\left(ktm\text{ }-3-x\ge0\right)\)
Vậy \(x\in\varnothing\)
b) \(\left|x-1\right|+1=2x-3\\ \Rightarrow\left|x-1\right|=2x-4\left(ĐK:2x-4\ge0\right)\\ \Rightarrow\left[{}\begin{matrix}x-1=2x-4\\x-1=-2x+4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-x=4-1\\x+2x=1+4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\left(t/m\right)\\3x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\left(t/m\right)\\x=\frac{5}{3}\left(ktm\right)\end{matrix}\right.\)
Vậy x = 3
c) \(\left|\frac{4}{3}x-\frac{4}{3}+\frac{1}{2}\right|=\left|2x-2+\frac{1}{3}\right|\\ \Rightarrow\left[{}\begin{matrix}\frac{4}{3}x-\frac{4}{3}+\frac{1}{2}=2x-2+\frac{1}{3}\\\frac{4}{3}x-\frac{4}{3}+\frac{1}{2}=-2x+2-\frac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-\frac{4}{3}x=2-\frac{1}{3}-\frac{4}{3}+\frac{1}{2}\\\frac{4}{3}x+2x=\frac{4}{3}-\frac{1}{2}+2-\frac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=\frac{5}{6}\\\frac{10}{3}x=\frac{5}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{5}{4}\\x=\frac{3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{5}{4};\frac{3}{4}\right\}\)
a, \(-\frac{187}{70}\)
b,\(\frac{27}{70}\)
c,\(\frac{53}{14}\)
d,\(\frac{27}{4}\)
e,1
f,\(\frac{23}{4}\)
g,-1
i,6
k,315
l,\(\frac{9}{2}\)
\(\frac{8}{16}+\frac{5}{\sqrt{2}}.\frac{2}{5}-\frac{3}{4}\)= \(\frac{1}{2}+\sqrt{2}-\frac{3}{4}=\sqrt{2}-\frac{1}{4}\)