BT1: Tìm x, biết:
4) \(\text{|}0,4x-25\%\text{|}-4=3\)
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Giải:
\(\left|\left|0,4x-25\%\right|-4\right|=3\)
\(\left|\left|\dfrac{2}{5}x-\dfrac{1}{4}\right|-4\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|\dfrac{2}{5}x-\dfrac{1}{4}\right|-4=3\\\left|\dfrac{2}{5}x-\dfrac{1}{4}\right|-4=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left|\dfrac{2}{5}x-\dfrac{1}{4}\right|=7\\\left|\dfrac{2}{5}x-\dfrac{1}{4}\right|=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{2}{5}x-\dfrac{1}{4}=7\\\dfrac{2}{5}x-\dfrac{1}{4}=-7\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{2}{5}x-\dfrac{1}{4}=1\\\dfrac{2}{5}x-\dfrac{1}{4}=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{2}{5}x=\dfrac{29}{4}\\\dfrac{2}{5}x=-\dfrac{27}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{2}{5}x=\dfrac{5}{4}\\\dfrac{2}{5}x=-\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{145}{8}\\x=-\dfrac{135}{8}\end{matrix}\right.\\\left[{}\begin{matrix}x=\dfrac{25}{8}\\x=-\dfrac{15}{8}\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
Chúc bạn học tốt!
\(\Leftrightarrow\left[{}\begin{matrix}\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|-3=-4\\\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|-3=4\end{matrix}\right.\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{4}=7\\\dfrac{1}{2}x-\dfrac{1}{4}=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{29}{4}\\\dfrac{1}{2}x=-\dfrac{27}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{2}\\x=-\dfrac{27}{2}\end{matrix}\right.\)
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{4}\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{4}\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{4}\right|\ge0\)
\(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{4}=4x\)
\(\Leftrightarrow3x+1=4x\)
\(\Leftrightarrow x=1\left(tm\right)\)
Vậy ..
\(x+\left|\dfrac{1}{2}-\dfrac{1}{3}\right|=\left|\dfrac{-2}{3}-\dfrac{1}{4}\right|\)
\(x+\left|\dfrac{1}{6}\right|=\left|\dfrac{-11}{12}\right|\)
\(x+\dfrac{1}{6}=\dfrac{11}{12}\)
\(x=\dfrac{11}{12}-\dfrac{1}{6}\)
\(x=\dfrac{3}{4}\)
Vậy ...
\(\dfrac{1}{2}\)| \(\dfrac{1}{3}x\)- \(\dfrac{1}{4}\)| - \(\dfrac{1}{5}\)= \(\dfrac{1}{6}\)
=> \(\dfrac{1}{2}\)| \(\dfrac{1}{3}x\) - \(\dfrac{1}{4}\)| = \(\dfrac{11}{30}\)
=> | \(\dfrac{1}{3}x\)- \(\dfrac{1}{4}\)| = \(\dfrac{11}{15}\)
=> \(\left[{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{4}=\dfrac{11}{15}\\\dfrac{1}{3}x-\dfrac{1}{4}=\dfrac{-11}{15}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\dfrac{1}{3}x=\dfrac{59}{60}\\\dfrac{1}{3}x=\dfrac{-29}{60}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{59}{20}\\x=\dfrac{-29}{20}\end{matrix}\right.\)
Chúc bạn học tốt !
\(\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{5}\right|+\left|x+\dfrac{1}{15}\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{5}\right|\ge0\\\left|x+\dfrac{1}{15}\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{5}\right|+\left|x+\dfrac{1}{15}\right|\ge0\)
\(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+\dfrac{1}{3}+x+\dfrac{1}{5}+x+\dfrac{1}{15}=4x\)
\(\Leftrightarrow3x+1=4x\)
\(\Leftrightarrow x=1\)
Vậy ..
d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+3y-2z}{\dfrac{1}{2}+3\cdot\dfrac{1}{3}-2\cdot\dfrac{1}{4}}=\dfrac{36}{1}=36\)
Do đó: x=18; y=12; z=9
a) Thay x + 3y - 2z vào biểu thức ta có:
\(\dfrac{x - 1}{3} = \dfrac{3(y + 2)}{3 . 4} = \dfrac{2(z - 2)}{2 . 3}\) = \(\dfrac{x - 1}{3} = \dfrac{3x + 6}{12} = \dfrac{2z - 4}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhua ta có:
\(\dfrac{x - 1}{3} = \dfrac{3y + 6}{12} = \dfrac{2z - 4}{6} = \dfrac{x - 1}{3}+ \dfrac{3y + 6}{12} -\dfrac{2z - 4}{6}\)
=\(\dfrac{x - 1 + 3y + 6 - 2z + 4}{3 + 12 -6} \) = \(\dfrac{(x + 3y - 2z) + ( -1 + 6 +4)}{3 + 12 - 6} \)
=\(\dfrac{36 + 9}{9}\) = 5
=> \(\dfrac{x - 1}{3} =\) 5 => x - 1 = 5.3 =15 => x = 5+1 = 6
=>
=>
Vậy ...
(Bạn dựa theo cách này và lm những bài tiếp nhé!)
Ta có: \(20\%x-x+\dfrac{1}{5}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-4}{5}x=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{15}{20}-\dfrac{4}{20}=\dfrac{11}{20}\)
\(\Leftrightarrow x=\dfrac{11}{20}:\dfrac{-4}{5}=\dfrac{11}{20}\cdot\dfrac{5}{-4}=\dfrac{-55}{80}=\dfrac{-11}{16}\)
Vậy: \(x=-\dfrac{11}{16}\)
\(\left|0,4x-25\%\right|-4=3\\ \left|\dfrac{2}{5}x-\dfrac{1}{4}\right|=3+4\\ \left|\dfrac{2}{5}x-\dfrac{1}{4}\right|=7\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{2}{5}x-\dfrac{1}{4}=7\\\dfrac{2}{5}x-\dfrac{1}{4}=-7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{2}{5}x=\dfrac{29}{4}\\\dfrac{2}{5}x=-\dfrac{27}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{145}{8}\\x=-\dfrac{135}{8}\end{matrix}\right.\)
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