Em xem lại số mũ của 2x - 5y nhé

2023 hay 2024?

Ta có: \(\left(2x-8\right)^{2000}+\left(3y+4\right)^{2022}\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-8=0\\3y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=8\\3y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-\dfrac{4}{3}\end{matrix}\right.\)

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

\(\left(2x+4\right)^{2024}+\left(\left|3y-9\right|\right)^{2023}=0\) (*) 

Ta có: \(\left(2x+4\right)^{2024}\ge0\forall x\) (vì có số mũ chẵn) (1)

\(\left(\left|3y-9\right|\right)^{2023}\ge0\forall y\) (vì giá trị tuyệt đối luôn ≥0) (2) 

Từ (1) và (2) ta có: 

\(\Rightarrow\left\{{}\begin{matrix}2x+4=0\\3y-9=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=3\end{matrix}\right.\)

Vậy: ... 

`(2x-5)^2024 + (3y+4)^26 <= 0`

  Vì `(2x-5)^2024 >= 0 AA x`

       `(3y+4)^26 >= 0 AA x`

 `=>{(2x-5=0),(3y+4=0):}`

`<=>{(x=5/2),(x=-4/3):}`

Ta thấy: (2x - 5)²⁰²⁴≥ 0 ∀ x ∈ R

              (3y + 4)²⁶ ≥ 0 ∀ y ∈ R

=> (2x - 5)²⁰²⁴ + (3y + 4)²⁶ ≥ 0 

Mặt khác:  (2x - 5)²⁰²⁴ + (3y + 4)²⁶ ≤ 0

Suy ra:  (2x - 5)²⁰²⁴ + (3y + 4)²⁶ = 0

\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\)                          \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy...

\(\left(2x-1\right)^{2020}+\left(y-\frac{2}{5}\right)^{2022}+\left|x+y-z\right|=0\)

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 \(x=\frac{1}{2};y=\frac{2}{5};z=\frac{9}{10}\)

=>2x-1=0 và x+2y=0

=>x=1/2 và y=-x/2=-1/4

\(a,\left\{{}\begin{matrix}\left|x-3y\right|\ge0\\\left|y+4\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-3y=0\\y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3y=-12\\y=-4\end{matrix}\right.\)

\(b,Sửa:\left|x-y-5\right|+\left(y+3\right)^2=0\\ \left\{{}\begin{matrix}\left|x-y-5\right|\ge0\\\left(y+3\right)^2\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-y-5=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+5=2\\y=-3\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}\left|x+y-1\right|\ge0\\\left(y-2\right)^4\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-y=-1\\y=2\end{matrix}\right.\)

\(d,\left\{{}\begin{matrix}\left|x+3y-1\right|\ge0\\3\left|y+2\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+3y-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-3y=7\\y=-2\end{matrix}\right.\)

\(e,Sửa:\left|2021-x\right|+\left|2y-2022\right|=0\\ \left\{{}\begin{matrix}\left|2021-x\right|\ge0\\\left|2y-2022\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}2021-x=0\\2y-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\y=1011\end{matrix}\right.\)

\(\left(x-2022\right)^{2024}+\left|y-2023\right|\le0\left(1\right)\)

Nhận thấy : \(\left(x-2022\right)^{2024}\ge0\forall x\inℝ,\left|y-2023\right|\ge0\forall y\inℝ\)

\(=>\left(x-2022\right)^{2024}+\left|y-2023\right|\ge0\forall x,y\inℝ\)

Do đó (1) xảy ra khi :

\(\left(x-2022\right)^{2024}=0,\left|y-2023\right|=0\)

\(=>\left(x;y\right)=\left(2022;2023\right)\)