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y*2+y*1/2=10
y*(2+1/2)=10
y*(4/2+1/2)=10
y*5/2=10
y=10:5/2=4
Vậy y=4
Ta có: y x 2 + y : 2 = 10
=> y x 2 + y x 1/2 = 10
=> y x (2 + 1/2) = 10
=> y x 5/2 = 10
=> y = 20 : 5/2
=> y = 4
![](https://rs.olm.vn/images/avt/0.png?1311)
1/2-2y=9/20
=>2y=1/2-9/20=1/20
=>y=1/20:2=1/40
b,3/5:4/3:y=2+7/10=9/20:y=27/10
=>y=9/20:27/10=1/6
c,y+y*3/2-y*1/2=1/10
=>y(1+3/2-1/2)=1/10
=>2y=1/10
=>y=1/10:2=1/20
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
y x 2 + y : 2 = 10
y + y + y: 2 = 10
y : ( 1 + 1 + 2 ) = 10
y : 4 = 10
y = 10: 4
y= \(\frac{10}{4}\)
y = \(\frac{5}{2}\)
Vậy y = \(\frac{5}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y\times4,9+y\div10=1,2\)
\(y\times4,9+y\times\dfrac{1}{10}=1,2\)
\(y\times4,9+y\times0,1=1,2\)
\(y\times\left(4,9+0,1\right)=1,2\)
\(y\times5=1,2\)
\(y=1,2\div5\)
\(y=0,24\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(\hept{\begin{cases}\left(2x-3\right)^{10}\ge0\forall x\\\left[100\left(x+2y\right)\right]^{100}\ge0\forall x;y\end{cases}}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-3=0\\x+2y=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1,5\\y=-0,75\end{cases}}\)
Vậy x = 1,5 ; y = -0,75
\(\left(2x-3\right)^{10}+\left(x+2y\right)^{100}\le0\)
Ta có: \(\left(2x-3\right)^{10}\)và \(\left(x+2y\right)^{100}\) là số chính phương. => \(\left(2x-3\right)^{10}\ge0;\left(x+2y\right)^{100}\ge0\)
Mà \(\left(2x-3\right)^{10}+\left(x+2y\right)^{100}\le0\)
=> \(\left(2x-3\right)^{10}=0;\left(x+2y\right)^{100}=0\)
=> 2x - 3 = 0; x + 2y = 0. => x = 3/2; y = -3/4.
y * 2+ y/2 = 10
y(2+1/2) = 10
y* 5/2 = 10
y= 4
y x 2 + y/2 = 10
y x 2 + y x 1/2 = 10
y x ( 2 + 1/2 ) = 10
y x 5/2 = 10
y = 10 : 5/2
y = 10 x 2/5
y = 4