=>(x-2023)[(x-2023)^21-1]=0

=>x-2023=0 hoặc x-2023=1

=>x=2023 hoặc x=2024

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

a, 7\(x\).(2\(x\) + 10) =0

    \(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy \(x\in\) {-5; 0}

b, -9\(x\) : (2\(x\) - 10) = 0

    9\(x\)                   = 0 

     \(x\)                    = 0 

c, (4 - \(x\)).(\(x\) + 3)  = 0

    \(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

Vậy \(x\in\) {-3; 4}

d, (\(x\) + 2023).(\(x\) - 2024) = 0

     \(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)

Vậy \(x\in\) {-2023; 2024}

ĐKXĐ: y>=0

\(\left(x+1\right)^{2024}>=0\forall x\)

\(\left(\sqrt{y-1}\right)^{2023}>=0\forall y\) thỏa mãn ĐKXĐ

=>\(\left(x+1\right)^{2024}+\left(\sqrt{y-1}\right)^{2023}>=0\forall x,y\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+1=0\\y-1=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

\(\dfrac{x-2023}{6}+\dfrac{x-2023}{10}+\dfrac{x-2023}{15}+\dfrac{x-2023}{21}=\dfrac{8}{21}\)

\(\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)

\(\left(x-2023\right).\dfrac{8}{21}=\dfrac{8}{21}\)

\(x-2023=1\)

\(x=2024\)

Vậy..............

\(...\Rightarrow\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right)\left(\dfrac{35+21+14+1}{210}\right)=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}.\dfrac{210}{71}=\dfrac{80}{71}\)

\(\Rightarrow x-2023=\dfrac{80}{71}\Rightarrow x=\dfrac{80}{71}+2023=\dfrac{143713}{71}\)

\(\dfrac{x-2023}{6}+\dfrac{x-2023}{10}+\dfrac{x-2023}{15}+\dfrac{x-2023}{21}=\dfrac{8}{21}\)

\(\Leftrightarrow\left(x-2023\right).\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)

\(\Leftrightarrow\left(x-2023\right).\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)=\dfrac{4}{21}\)

\(\Leftrightarrow\left(x-2023\right).\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\right)=\dfrac{4}{21}\)

\(\Leftrightarrow\left(x-2023\right).\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)=\dfrac{4}{21}\)

\(\Leftrightarrow\left(x-2023\right).\left(\dfrac{1}{3}-\dfrac{1}{7}\right)=\dfrac{4}{21}\)

\(\Leftrightarrow x-2023=1\Leftrightarrow x=2024\)

Ta có:

\(x^2+5y^2-4x-4xy+6y+5=0\\\Rightarrow[(x^2-4xy+4y^2)-(4x-8y)+4]+(y^2-2y+1)=0\\\Rightarrow[(x-2y)^2-4(x-2y)+4]+(y-1)^2=0\\\Rightarrow(x-2y-2)^2+(y-1)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-2y-2\right)^2\ge0\forall x,y\\\left(y-1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-2y-2\right)^2+\left(y-1\right)^2\ge0\forall x,y\)

Mà: \(\left(x-2y-2\right)^2+\left(y-1\right)^2=0\)

nên: \(\left\{{}\begin{matrix}x-2y-2=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2y+2\\y=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot1+2=4\\y=1\end{matrix}\right.\)

Thay \(x=4;y=1\) vào \(P\), ta được:

\(P=\left(4-3\right)^{2023}+\left(1-2\right)^{2023}+\left(4+1-5\right)^{2023}\)

\(=1^{2023}+\left(-1\right)^{2023}+0^{2023}\)

\(=1-1=0\)

Vậy \(P=0\) khi \(x=4;y=1\).

- Với \(0< x;y< 1\)

\(x^2>x^{2003}\left(1\right)\)

\(y^2>y^{2003}\left(2\right)\)

\(z^2>z^{2003}\left(3\right)\)

\(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow M=x^2+y^2+z^2>x^{2003}+y^{2003}+z^{2003}=3\)

\(\Rightarrow\) Không có giá trị max của M.

- Với \(x;y\ge1\)

\(x^2\le x^{2003}\left(1\right)\)

\(y^2\le y^{2003}\left(2\right)\)

\(z^2\le z^{2003}\left(3\right)\)

\(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow x^2+y^2+z^2\le x^{2003}+y^{2003}+z^{2003}=3\)

\(\Rightarrow Max\left(M\right)=3\left(x=y=z=1\right)\)

Em xem lại số mũ của 2x - 5y nhé

2023 hay 2024?