KL
1
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Khách
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18 tháng 10 2017
\(\left|0,4x-25\%\right|-4=3\\ \left|\dfrac{2}{5}x-\dfrac{1}{4}\right|=3+4\\ \left|\dfrac{2}{5}x-\dfrac{1}{4}\right|=7\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{2}{5}x-\dfrac{1}{4}=7\\\dfrac{2}{5}x-\dfrac{1}{4}=-7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{2}{5}x=\dfrac{29}{4}\\\dfrac{2}{5}x=-\dfrac{27}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{145}{8}\\x=-\dfrac{135}{8}\end{matrix}\right.\)
Vậy ...
30 tháng 5 2022
\(\Leftrightarrow\left[{}\begin{matrix}\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|-3=-4\\\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|-3=4\end{matrix}\right.\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{4}=7\\\dfrac{1}{2}x-\dfrac{1}{4}=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{29}{4}\\\dfrac{1}{2}x=-\dfrac{27}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{2}\\x=-\dfrac{27}{2}\end{matrix}\right.\)
26 tháng 9 2017
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{4}\right|=4x\)
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 ..
18 tháng 10 2017
\(\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{5}\right|+\left|x+\dfrac{1}{15}\right|=4x\)
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 ..
LH
26 tháng 9 2017
\(\dfrac{1}{2}\)| \(\dfrac{1}{3}x\)- \(\dfrac{1}{4}\)| - \(\dfrac{1}{5}\)= \(\dfrac{1}{6}\)
=> \(\dfrac{1}{2}\)| \(\dfrac{1}{3}x\) - \(\dfrac{1}{4}\)| = \(\dfrac{11}{30}\)
=> | \(\dfrac{1}{3}x\)- \(\dfrac{1}{4}\)| = \(\dfrac{11}{15}\)
=> \(\left[{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{4}=\dfrac{11}{15}\\\dfrac{1}{3}x-\dfrac{1}{4}=\dfrac{-11}{15}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\dfrac{1}{3}x=\dfrac{59}{60}\\\dfrac{1}{3}x=\dfrac{-29}{60}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{59}{20}\\x=\dfrac{-29}{20}\end{matrix}\right.\)
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LD
0
13 tháng 8 2017
\(4x^2-12x-y^2-3=0\)
\(\Rightarrow4x^2-12x-y^2+9-12=0\)
\(\Rightarrow\left(2x-3\right)^2-\left(y^2+12\right)=0\)
Lập bảng xét dấu:v
b tương tự
MH
1
17 tháng 8 2015
xy + yx = 99 => 11.(x + y) = 99 => x + y = 9
0,xy(x) - 0,yx(y) = 0,4(5)
=> xy,(x) - yx,(y) = 45, (5)
=> xy + 0,(x) - yx - 0, (y) = 45 + 0,(5)
=> (xy - yx) + x/9 - y/9 = 45 + 5/9
=> 9(x - y) + (x - y)/ 9 = 410/9
=> (9 + 1/9). (x - y) = 410/9 => x - y = 5
Mà x + y = 9 nên (x + y) + (x - y) = 9 + 5 = 14 => 2x = 14 => x = 7
=> y = 2
Vậy....
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 ...
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