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LN
1
13 tháng 9 2018
a)\(\frac{X}{5}=\frac{5}{6}+\frac{-19}{30}\)
\(\frac{X}{5}=\frac{1}{5}\)
Vậy \(X=1\)
b)\(X-\frac{41}{5}=\frac{-2}{3}\)
\(X=\frac{-2}{3}+\frac{41}{5}\)
\(X=\frac{113}{15}\)
Vậy \(X=\frac{113}{15}\)
c)\(\frac{31}{5}-X=\frac{11}{3}+\frac{7}{10}\)
\(\frac{31}{5}-X=\frac{131}{30}\)
\(X=\frac{31}{5}-\frac{131}{30}\)
\(X=\frac{11}{6}\)
Vậy \(X=\frac{11}{6}\)
d)\(\frac{9}{X}=\frac{2}{5}+\frac{-7}{20}\)
\(\frac{9}{X}=\frac{1}{20}\)
\(X=9:\frac{1}{20}\)
\(X=180\)
Vậy \(X=180\)
Hc tốt
HN
2
PH
16 tháng 5 2022
a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)
PH
16 tháng 5 2022
b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)
1 tháng 4 2020
\(\text{a, 3(x+1)+4x=10}\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow7x+3=10\)
\(\Rightarrow7x=10-3=7\)
\(\Rightarrow x=1\)
c, x+1/10+x+2/9=x+3/8+x+4/7
=> (x+1/10 +1) +(x+2/9 +1)= ( x+3/8 +1) +(x+4/7 +1)
=> x+11/10 + x+11/9 = x+11/8 + x+11/7
...............
1 tháng 4 2020
a) \(3\left(x+1\right)+4x=10\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow3x+4x=10-3\)
\(\Rightarrow7x=7\)
\(\Rightarrow x=7\)
NN
0
A
1
A
0
a) \(x+\left|x-2\right|=7\)
\(\Leftrightarrow\left|x-2\right|=7-x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=7-x\\x-2=-7+x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=9\\0x=-5\left(loại\right)\end{matrix}\right.\) \(\Leftrightarrow x=\dfrac{9}{2}\)
b) \(\left|x-3\right|+\left|x-5\right|=9\left(1\right)\)
Ta thấy :
\(\left|x-3\right|+\left|x-5\right|\ge\left|x-3+x-5\right|=\left|2x-8\right|\)
\(pt\left(1\right)\Leftrightarrow\left|2x-8\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-8=9\\2x-8=-9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=17\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
c) \(\left|x-1\right|+\left|x+1\right|=10\left(1\right)\)
Ta thấy :
\(\left|x-1\right|+\left|x+1\right|\ge\left|x-1+x+1\right|=\left|2x\right|\)
\(pt\left(1\right)\Leftrightarrow\left|2x\right|=10\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
a) \(x+\left|x-2\right|=7\)
\(\Rightarrow\left\{{}\begin{matrix}x+\left(x-2\right)=7\left(x\ge2\right)\\x-\left(x-2\right)=7\left(x< 2\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+x-2=7\\x-x+2=7\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x-2=7\\2=7\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=9\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow2x=-9\)
\(\Rightarrow x=-\dfrac{9}{2}\)