y : 25% + 50% + y : 0,2 - y = 2023
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y:\frac{1}{4}+y:\frac{1}{2}+y:\frac{1}{5}-y\)\(=2017\)
\(y.4+y.2+y.5-y=2017\)
\(y.\left(4+2+5-1\right)=2017\)
\(y.10=2017\)
\(y=2017:10\)
\(y=201,7\)
CHÚC BN HOK TỐT!
đÚNG THÌ K CHO MK NHA!
này ! đề bài có nhầm k ! 2 con ở trên thfi có % 2 con ở cuối lại hông cóa !
![](https://rs.olm.vn/images/avt/0.png?1311)
y :25% + y : 50% + y : 0,2 - y = 2017 .Làm xong giải thích hộ mình nhaaaaaaaaa!!!!!!!!!!!!!!!!!!! :)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y:25\%+y:50\%+y:0,2-y=2017\)
=> \(y:\frac{1}{4}+y:\frac{1}{2}+y:\frac{1}{5}-y=2017\)
=> \(y\cdot4+y\cdot2+y\cdot5-y=2017\)
=> \(y\left(4+2+5\right)-y=2017\)
=> \(y\cdot11-y=2017\)
=> \(10y=2017\)
=> \(y=201,7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y:50\%+y:25\%+y:0,2-y=1590\)
\(\Leftrightarrow y.2+y.4+y.5-y=1590\)
\(\Leftrightarrow y\left(2+4+5-1\right)=1590\)
\(\Leftrightarrow y.10=1590\)
\(\Leftrightarrow y=159\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(2x-y+7\right)^{2022}>=0\forall x,y\)
\(\left|x-1\right|^{2023}>=0\forall x\)
=>\(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}>=0\forall x,y\)
mà \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}< =0\forall x,y\)
nên \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}=0\)
=>\(\left\{{}\begin{matrix}2x-y+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2x+7=9\end{matrix}\right.\)
\(P=x^{2023}+\left(y-10\right)^{2023}\)
\(=1^{2023}+\left(9-10\right)^{2023}\)
=1-1
=0
c: \(\left|x-3\right|>=0\forall x\)
=>\(\left|x-3\right|+2>=2\forall x\)
=>\(\left(\left|x-3\right|+2\right)^2>=4\forall x\)
mà \(\left|y+3\right|>=0\forall y\)
nên \(\left(\left|x-3\right|+2\right)^2+\left|y+3\right|>=4\forall x,y\)
=>\(P=\left(\left|x-3\right|+2\right)^2+\left|y-3\right|+2019>=4+2019=2023\forall x,y\)
Dấu '=' xảy ra khi x-3=0 và y-3=0
=>x=3 và y=3
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\left|x+5\right|\ge0\forall x\)
\(\Rightarrow\left|x+5\right|+2023\ge2023\forall x\)
\(\Rightarrow A\ge2023\forall x\)
Dấu \("="\) xảy ra khi: \(x+5=0\Leftrightarrow x=-5\)
Vậy \(Min_A=2023\) khi \(x=-5\).
b) Ta có: \(\left\{{}\begin{matrix}\left|2x+6\right|\ge0\forall x\\\left|y+3x\right|\ge0\forall x,y\end{matrix}\right.\)
\(\Rightarrow\left|2x+6\right|+\left|y+3x\right|\ge0\forall x,y\)
\(\Rightarrow\left|2x+6\right|+\left|y+3x\right|+25\ge25\forall x,y\)
\(\Rightarrow B\ge25\forall x,y\)
Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}2x+6=0\\y+3x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-6\\y=-3x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-6:2=-3\\y=-3\cdot\left(-3\right)=9\end{matrix}\right.\)
Vậy \(Min_B=25\) khi \(x=-3;y=9\).
c) Ta có: \(\left\{{}\begin{matrix}\left|12-3x\right|\ge0\forall x\\\left|-y-4x\right|\ge0\forall x,y\end{matrix}\right.\)
\(\Rightarrow\left|12-3x\right|+\left|-y-4x\right|\ge0\forall x,y\)
\(\Rightarrow\left|12-3x\right|+\left|-y-4x\right|-12\ge-12\forall x,y\)
\(\Rightarrow C\ge-12\forall x,y\)
Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}12-3x=0\\-y-4x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=12\\y=-4x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=12:3=4\\y=-4\cdot4=-16\end{matrix}\right.\)
Vậy \(Min_C=-12\) khi \(x=4;y=-16\).
\(\mathit{Toru}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
y:5%+y:25%+y:0.2-y*10=100
y:5/100+y:25/100+y*5-y*10=100
y*20+y*4+y*5-y*10=100
y*(20+4+5-10)=100
y*19=100
y=100:19
y=100/19
tick mình đấy nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y\times2023-y=2023\times2021+2023\)
\(y\times\left(2023-1\right)=2023\times\left(2021+1\right)\)
\(y\times2022=2023\times2022\)
\(y=2023\times2022\div2022\)
\(y=2023\)
\(y:25\%+50\%+y:0,2-y=2023\)
=>\(y\cdot4+0,2+y\cdot5-y=2023\)
=>9y=2023-0,2=2022,8
=>\(y=\dfrac{2022.8}{9}=\dfrac{10114}{45}\)
y : 25% + 50% + y: 0,2 - y = 2023
y x 4 + 0,5 + y x 5 - y = 2023
y x (4 + 5 - 1) + 0,5 = 2023
y x 8 + 0,5 = 2023
y x 8 = 2023 - 0,5
y x 8 = 2022,5
y = 2022,5 : 8
y = 252,8125