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M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2
(2x-5)^2020+(3y+4)^2022<=0
=>x=5/2 và y=-4/3
M=25/4+11*5/2*(-4/3)-16/9=-1159/36
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa: \(\left(\dfrac{1}{3}-2x\right)^{2020}+\left(3y-x\right)^{2022}\le0\)
Mà \(\left(\dfrac{1}{3}-2x\right)^{2020}+\left(3y-x\right)^{2022}\ge0\) với mọi x,y
Do đó \(\left\{{}\begin{matrix}\dfrac{1}{3}-2x=0\\3y-x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=\dfrac{1}{18}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}=6+18=24\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-1\right)^{2020}+\left(y-\frac{2}{5}\right)^{2022}+\left|x+y-z\right|=0\)
Ta có : \(\left(2x-1\right)^{2020}\ge0\forall x;\left(y-\frac{2}{5}\right)^{2022}\ge0\forall x;\left|x+y-z\right|\ge0\forall x;y;z\)
Dấu bằng xảy ra <=> \(x=\frac{1}{2};y=\frac{2}{5};z=x+y=\frac{1}{2}+\frac{2}{5}=\frac{9}{10}\)
Vậy \(x=\frac{1}{2};y=\frac{2}{5};z=\frac{9}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(2x-8\right)^{2000}+\left(3y+4\right)^{2022}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-8=0\\3y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=8\\3y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-\dfrac{4}{3}\end{matrix}\right.\)
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Lời giải:
Ta thấy:
$(7x-5y)^{2018}\geq 0, \forall x,y$
$(3x-2z)^{2020}\geq 0, \forall x,z$
$(xy+yz+xz-4500)^{2022}\geq 0, \forall x,y,z$
Do đó để tổng $(7x-5y)^{2018}+(3x-2z)^{2020}+(xy+yz+xz-4500)^{2022}=0$ thì:
$(7x-5y)^{2018}=(3x-2z)^{2020}=(xy+yz+xz-4500)^{2022}=0$
$\Leftrightarrow$ \(\left\{\begin{matrix} 7x=5y(1)\\ 3x=2z(2)\\ xy+yz+xz=4500(3)\end{matrix}\right.\)
Từ $(1);(2)\Rightarrow y=\frac{7}{5}x; z=\frac{3}{2}x$
Thay vào $(3)$:
$x.\frac{7}{5}x+\frac{7}{5}x.\frac{3}{2}x+x.\frac{3}{2}x=4500$
$\Leftrightarrow x^2=900\Rightarrow x=\pm 30$
Nếu $x=30\Rightarrow y=42; z=45$
Nếu $x=-30\Rightarrow y=-42; z=-45$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\\ \Leftrightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(2x-5\right)^{2018}=0\\\left(3y+4\right)^{2020}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{4}{3}\end{matrix}\right.\\ \Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2\\ \Leftrightarrow M=\dfrac{25}{4}-11\cdot\dfrac{4}{3}\cdot\dfrac{5}{2}-\dfrac{16}{9}=\dfrac{25}{4}-\dfrac{110}{3}-\dfrac{16}{9}=-\dfrac{1159}{36}\)
Vì \(\left(2x-5\right)^{2020}\ge0\forall x\); \(\left(5y+1\right)^{2022}\ge0\forall y\)
\(\Rightarrow\left(2x-5\right)^{2020}+\left(5y+1\right)^{2022}\ge0\forall x,y\)
mà \(\left(2x-5\right)^{2020}+\left(5y+1\right)^{2022}\le0\)( giả thuyết )
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}2x-5=0\\5y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=5\\5y=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=\frac{-1}{5}\end{cases}}\)
Vậy \(x=\frac{5}{2}\)và \(y=\frac{-1}{5}\)
( 2x - 5 )2020 + ( 5y + 1 )2022 ≤ 0
Ta có : ( 2x - 5 )2020 ≥ 0 ∀ x
( 5y + 1 )2022 ≥ 0 ∀ y
=> ( 2x - 5 )2 + ( 5y + 1 )2022 ≥ 0 ∀ x, y
Kết hợp với đề bài => Chỉ xảy ra trường hợp ( 2x - 5 )2020 + ( 5y + 1 )2022 = 0
Khi đó \(\hept{\begin{cases}2x-5=0\\5y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{1}{5}\end{cases}}\)