y x 2,5 + y x 7,5 = 967,8
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 0,09 x m - 5,8 = 34,52
=> 0,09 x m = 34,52 + 5,8
=> 0,09 x m = 40,32
=> m = 40,32 : 0,09
=> m = 448
b,m x 2,5 + m x 7,5 = 967,8
=> m x ( 2,5 + 7,5 ) = 967,8
=> m x 10 = 967,8
=> m = 967,8 : 10
=> m = 96,78
a,0.9*m=34.52+5.8
0.9*m= 40.32
40.32:0.9
m=44.8
b,(2.5+7.5)*m=967.8
m=967.8:10
=96.78
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y:2,5+9,6\cdot y=7,5-3\)
\(y:\frac{25}{10}+9,6\cdot y=4,5\)
\(y\cdot\frac{10}{25}+9,6\cdot y=4,5\)
\(y\cdot0,4+9,6\cdot y=4,5\)
\(y\left(0,4+9,6\right)=4,5\)
\(y\cdot10=4,5\)
\(\Rightarrow y=0,45\)
P/s: Hoq chắc :<
y : 2,5 + 9,6 x y = 7,5 - 3
y : \(\frac{25}{10}\) + \(\frac{96}{10}\) x y = 3,5
y x \(\frac{10}{25}\)+ \(\frac{96}{10}\) x y = 3,5
y x (\(\frac{2}{5}+\frac{48}{5}\)) = \(\frac{35}{10}\)
y x \(\frac{50}{5}\)= \(\frac{7}{2}\)
y x 10 = \(\frac{7}{2}\)
y = \(\frac{7}{2}\) : 10
y = \(\frac{7}{2}\) x \(\frac{1}{10}\)
y = \(\frac{7}{20}\)
Ko bik đúng hay ko nữa, sai thì thui đừng k sai nha
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(y:10+y.3,9=4,8\)
\(\Rightarrow y.0,1+y.3,9=4,8\)
\(\Rightarrow y.\left(0,1+3,9\right)=4,8\)
\(\Rightarrow y.4=4,8\)
\(\Rightarrow y=1,2\)
Vậy y = 1,2
b) \(y.2,5+y.7,5=12,8\)
\(\Rightarrow y.\left(2,5+7,5\right)=12,8\)
\(\Rightarrow y.10=12,8\)
\(\Rightarrow y=1,28\)
Vậy y = 1,28
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\left|2x\right|-\left|-2,5\right|=\left|-7,5\right|\)
\(\Rightarrow\left|2x\right|-2,5=7,5\)
\(\Rightarrow\left|2x\right|=10\)
\(\Rightarrow\left[{}\begin{matrix}2x=10\Rightarrow x=5\\2x=-10\Rightarrow x=-5\end{matrix}\right.\)
b) \(\left|2x-3\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=1\Rightarrow2x=4\Rightarrow x=2\\2x-3=-1\Rightarrow2x=2\Rightarrow x=1\end{matrix}\right.\)
c) \(\left|x-3,5\right|+\left|y-1,3\right|=0\)
Ta có: \(\left|x-3,5\right|\ge0\forall x\)
\(\left|y-1,3\right|\ge0\forall y\)
\(\Rightarrow\left|x-3,5\right|+\left|y-1,3\right|\ge0\forall x,y\)
Dấu "=" xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\Rightarrow x=3,5\\y-1,3=0\Rightarrow y=1,3\end{matrix}\right.\)
\(a)\left|2x\right|-\left|-2,5\right|=\left|-7,5\right|\)
\(\Rightarrow\left|2x\right|-2,5=7,5\)
\(\Rightarrow\left|2x\right|=10\)
\(\Rightarrow\left[{}\begin{matrix}2x=10\Rightarrow x=5\\2x=-10\Rightarrow x=-5\end{matrix}\right.\)
Vậy ...............
\(b)\left|2x-3\right|=1\)
\(\Rightarrow\left|2x\right|-3=1\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=1\Rightarrow2x=4\Rightarrow x=2\\2x-3=-1\Rightarrow2x=2\Rightarrow x=1\end{matrix}\right.\)
Vậy .........
\(c)\left|x-3,5\right|+\left|y-1,3\right|=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3,5=0\Rightarrow x=3,5\\y-1,3=0\Rightarrow y=1,3\end{matrix}\right.\)
Vậy ..............
Chúc bạn học tốt!
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