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DD
1 tháng 12 2019
1,5-3|5-2x|=12014-17/2
1,5-3|5-2x|=1-17/2
1,5-3|5-2x|=-15/2
-3|5-2x|=-15/2-3/2
-3|5-2x|=-9
|5-2x|=3
TH1:5-2x=3
2x=2
x=1
TH2:5-2x=-3
2x=8
2x=4
Vậy x=1 và x=4
8 tháng 7 2017
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
PT
Phạm Trần Hoàng Anh
CTVHS
18 tháng 8 2020
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
=> \(\frac{2}{3}x=-\frac{29}{70}\)
=> \(x=-\frac{29}{70}:\frac{2}{3}\)
=> \(x=-\frac{29}{70}.\frac{3}{2}\)
=> \(x=-\frac{87}{140}\)
b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)
=> \(-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)
=> \(-\frac{21}{13}x=-\frac{3}{3}\)
=> \(-\frac{21}{13}x=1\)
=> \(x=1:\left(-\frac{21}{13}\right)\)
=> \(x=-\frac{13}{21}\)
c) \(\left|x-1,5\right|=2\)
=> \(\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.=>\left[{}\begin{matrix}x=2+1,5\\x=-2+1,5\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)(T/M)
d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
=> \(\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
=> \(=>\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{1}{2}-\frac{3}{4}\\x=-\frac{1}{2}-\frac{3}{4}\end{matrix}\right.=>\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{matrix}\right.\)(T/M)
HỌC TỐT 
LT
18 tháng 8 2020
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
\(\Leftrightarrow\frac{2}{3}x=-\frac{29}{70}\)
\(\Leftrightarrow x=-\frac{29}{70}:\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{87}{140}\)
b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)
\(\Leftrightarrow-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)
\(\Leftrightarrow-\frac{21}{13}x=-1\)
\(\Leftrightarrow x=-1:\left(-\frac{21}{13}\right)\)
\(\Leftrightarrow x=\frac{13}{21}\)
c) \(\left|x-1,5\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Leftrightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=\frac{5}{4}\end{matrix}\right.\)
1 tháng 10 2020
+) \(5\frac{2}{3}x+1\frac{2}{3}=4\frac{1}{2}\Leftrightarrow\frac{17}{3}x+\frac{5}{3}=\frac{9}{2}\Leftrightarrow\frac{17}{3}x=\frac{17}{6}\Leftrightarrow x=\frac{1}{2}\)
+) \(\frac{x}{27}=\frac{-2}{9}\Leftrightarrow x=\frac{-2}{9}.27=-6\)
+) \(\left|x+1,5\right|=2\Leftrightarrow\orbr{\begin{cases}x+1,5=2\\x+1,5=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0,5\\x=-3,5\end{cases}}}\)
+) \(A=\left|x-1004\right|-\left|x+1003\right|\)
Ta có BĐT \(\left|x\right|-\left|y\right|\le\left|x-y\right|,\)dấu "=" xảy ra khi và chỉ khi x,y cùng dấu hay \(xy\ge0\)
Áp dụng: \(A=\left|x-1004\right|-\left|x+1003\right|\le\left|x-1004-x-1003\right|=\left|-2007\right|=2007\)
Vậy \(maxA=2007\Leftrightarrow\left(x-1004\right)\left(x+1003\right)\ge0\Leftrightarrow\orbr{\begin{cases}x\ge1004\\x\le-1003\end{cases}}\)
21 tháng 1 2017
a) \(\frac{3}{4}+\frac{1}{4}.x=\frac{1}{2}+\frac{1}{2}x\)
\(\Rightarrow3.\frac{1}{4}+\frac{1}{4}.x=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow\frac{1}{4}.\left(x+3\right)=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow\frac{x+1}{x+3}=\frac{1}{4}:\frac{1}{2}=\frac{1}{2}\)\(\Rightarrow\left(x+1\right).2=x+3\Rightarrow2x+2=x+3\)
\(\Rightarrow2x-x=3-2\Rightarrow x=1\)
vay x=1
G
3 tháng 3 2018
Câu 2 đây:
\(|x^2+|x-1||=x^2+2\)
\(\Rightarrow\orbr{\begin{cases}x^2+\left|x-1\right|=x^2+2\\x^2+\left|x-1\right|=-x^2-2\left(l\right)\end{cases}}\)
\(\Rightarrow\left|x-1\right|=2\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
KT
3 tháng 3 2018
a) \(M=\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+0,5}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)
\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{2}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\frac{2012}{2013}\)
\(=\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\right):\frac{2012}{2013}\)
\(=\left(\frac{2}{7}-\frac{2}{7}\right):\frac{2012}{2013}\)
\(=0\)
GC
1
19 tháng 8 2019
Lời giải :
Do \(VT\ge0\forall x;y\)nên ta có hệ :
\(\hept{\begin{cases}\frac{2}{3}-\frac{1}{2}+\frac{3}{4}x=0\\1,5-\frac{11}{17}+\frac{23}{13}y=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{-2}{9}\\y=\frac{-377}{782}\end{cases}}\)
Vậy...
\(\frac{-2}{1,5}=\frac{-2.2}{1,5.2}=\frac{-4}{3}=\frac{x-1}{3}\)
\(\Rightarrow x-1=-4\)\(\Rightarrow x=-3\)
Vậy \(x=-3\)