
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


Ta có: x+y+z=0
=>\(\left(x+y+z\right)^2=0^2=0\)
=>\(x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
=>\(x^2+y^2+z^2=0\)
mà \(x^2\ge0\forall x;y^2\ge0\forall y;z^2\ge0\forall z\)
nên \(\begin{cases}x=0\\ y=0\\ z=0\end{cases}\)
\(\left(x-1\right)^{2023}+y^{2024}+\left(z+1\right)^{2025}\)
\(=\left(0-1\right)^{2023}+0^{2024}+\left(0+1\right)^{2025}\)
=-1+0+1
=0

Ta có: x+y+z=0
=>\(\left(x+y+z\right)^2=0^2=0\)
=>\(x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
=>\(x^2+y^2+z^2=0\)
mà \(x^2\ge0\forall x;y^2\ge0\forall y;z^2\ge0\forall z\)
nên \(\begin{cases}x=0\\ y=0\\ z=0\end{cases}\)
\(\left(x-1\right)^{2023}+y^{2024}+\left(z+1\right)^{2025}\)
\(=\left(0-1\right)^{2023}+0^{2024}+\left(0+1\right)^{2025}\)
=-1+0+1
=0

\(\sqrt{x^2+2024}=\sqrt{x^2+xy+yz+zx}=\sqrt{\left(x+y\right)\left(z+x\right)}\ge\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}=\sqrt{xy}+\sqrt{xz}\)
Tương tự: \(\sqrt{y^2+2024}\ge\sqrt{xy}+\sqrt{yz}\)
\(\sqrt{z^2+2024}\ge\sqrt{xz}+\sqrt{yz}\)
Cộng vế:
\(P\ge\dfrac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}=2\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{2024}{3}\)

Sửa đề: \(5x^2+5y^2+8xy-2x+2y+2=0\)
=>\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)
=>\(\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
=>\(\left\{{}\begin{matrix}2x+2y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(M=\left(x-y\right)^{2023}-\left(x-2\right)^{2024}+\left(y+1\right)^{2023}\)
\(=\left(1+1\right)^{2023}-\left(1-2\right)^{2024}+\left(-1+1\right)^{2023}\)
\(=2^{2023}-1\)

\(5x^2+5y^2+8xy-2x+2y+2=0\)
=>\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)
=>\(4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
=>x=1 và y=-1
\(M=\left(1-1\right)^{2023}+\left(1-2\right)^{2024}+\left(-1+1\right)^{2025}=1\)

=2024.2024-2024.4046+2023.2023
=2024.(2024-2023)+2023.(2023-2024
=1-1
=0
Đề không đầy đủ. Bạn coi lại.