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Ta có: \(\left|x+5\right|\ge0\forall x\)
\(\left(3y-4\right)^{2020}\ge0\forall y\)
Do đó: \(\left|x+5\right|+\left(3y-4\right)^{2020}\ge0\forall x,y\)
mà \(\left|x+5\right|+\left(3y-4\right)^{2020}=0\)
nên \(\left\{{}\begin{matrix}\left|x+5\right|=0\\\left(3y-4\right)^{2020}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\3y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=\frac{4}{3}\end{matrix}\right.\)
Vậy: x=-5 và \(y=\frac{4}{3}\)
\(\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\le0\left(1\right)\)
Ta có: \(\hept{\begin{cases}\left(2x-5\right)^{2020}\ge0;\forall x,y\\\left(3y+4\right)^{2018}\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\ge0;\forall x,y\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(2x-5\right)^{2020}=0\\\left(3y+4\right)^{2018}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=\frac{-4}{3}\end{cases}}\)
Vậy...
\(\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\le0\)
Ta có:
\(\left\{{}\begin{matrix}\left(2x-5\right)^{2020}\ge0\\\left(3y+4\right)^{2018}\ge0\end{matrix}\right.\forall xy.\)
\(\Rightarrow\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\ge0\) \(\forall xy.\)
Mà \(\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\le0.\)
\(\Rightarrow\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}=0\)
\(\Rightarrow\left(2x-5\right)+\left(3y+4\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=5\\3y=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5:2\\y=\left(-4\right):3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=-\frac{4}{3}\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\frac{5}{2};-\frac{4}{3}\right\}.\)
Chúc bạn học tốt!
\(\left(\dfrac{3x-5}{9}\right)^{2018}>=0\forall x\)
\(\left(\dfrac{3y+0,4}{3}\right)^{2020}>=0\forall y\)
Do đó: \(\left(\dfrac{3x-5}{9}\right)^{2018}+\left(\dfrac{3y+0,4}{3}\right)^{2020}>=0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}\dfrac{3x-5}{9}=0\\\dfrac{3y+0,4}{3}=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x-5=0\\3y+0,4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-\dfrac{0.4}{3}=-\dfrac{2}{15}\end{matrix}\right.\)
c) Ta có(x-1)2 >= 0 với mọi x
(y+3)2>=0 với mọi c
=> (x-1)2+(y+3)2 >= 0 với mọi x,y
Dấu bằng xảy ra khi và chỉ khi
(x-1)2=0 và (y+3)2=0
=> x=1 và y=-3
\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\\ \Leftrightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(2x-5\right)^{2018}=0\\\left(3y+4\right)^{2020}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{4}{3}\end{matrix}\right.\\ \Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2\\ \Leftrightarrow M=\dfrac{25}{4}-11\cdot\dfrac{4}{3}\cdot\dfrac{5}{2}-\dfrac{16}{9}=\dfrac{25}{4}-\dfrac{110}{3}-\dfrac{16}{9}=-\dfrac{1159}{36}\)
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
Ta thấy:|x+5| lớn hơn hoặc bằng 0
(3y-4)2020 cũng lớn hơn hoặc bằng 0
Mà |x+5| + (3y-4)2020 = 0
=>|x+5| = (3y-4)2020 = 0
=>x+5 = 3y-4 = 0
=>x=0-5=-5 ; 3y=0+4=4 =>y = 4:3=4/3