Lời giải:
$4x^2+2y^2+2z^2-4xy-4xz+2yz-6y-10z+34=0$

$(4x^2+y^2+z^2-4xy-4xz+2yz)+y^2+z^2-6y-10z+34=0$

$(2x-y-z)^2+(y^2-6y+9)+(z^2-10z+25)=0$
$(2x-y-z)^2+(y-3)^2+(z-5)^2=0$

Vì $(2x-y-z)^2\geq 0; (y-3)^2\geq 0; (z-5)^2\geq 0$ với mọi $x,y,z$

Do đó để tổng của chúng bằng $0$ thì bản thân mỗi số đó bằng $0$

$\Rightarrow 2x-y-z=y-3=z-5=0$

$\Rightarrow y=3; z=5; x=4$

Khi đó:
$P=0^{2023}+(-1)^{2025}+(5-4)^{2027}=0$

Sửa đề: Cho x,y,z đôi một khác nhau và \(x^3+y^3+z^3=3xyz\)

Ta có: \(x^3+y^3+z^3=3xyz\)

=>\(\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)

=>\(\left(x+y+z\right)\left\lbrack\left(x+y\right)^2-z\left(x+y\right)+z^2\right\rbrack-3xy\left(x+y+z\right)=0\)

=>\(\left(x+y+z\right)\left\lbrack x^2+2xy+y^2-xz-zy+z^2\right\rbrack-3xy\left(x+y+z\right)=0\)

=>\(\left(x+y+z\right)\left\lbrack x^2+y^2+z^2-xy-xz-yz\right\rbrack=0\)

=>\(\left(x+y+z\right)\left\lbrack2x^2+2y^2+2z^2-2xy-2yz-2xz\right\rbrack=0\)

=>\(\left(x+y+z\right)\left\lbrack\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right\rbrack=0\)

mà \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2>0\) vì x,y,z đôi một khác nhau

nên x+y+z=0

=>y+z=-x

Sửa đề: \(A=2025+\left(y+z\right)^{2025}+x^{2025}\)

\(=2025+\left(-x\right)_{}^{2025}+x^{2025}\)

\(=2025-x^{2025}+x^{2025}=2025\)

\(A=x^2+y^2+1-2xy-2x+2y+2\left(x^2-4x+4\right)+2018\)

\(A=\left(x-y-1\right)^2+2\left(x-2\right)^2+2018\ge2018\)

\(A_{min}=2018\) khi \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Chứng minh BĐT phụ :

\(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

Thật vậy : \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

\(\Leftrightarrow x^2+y^2-2xy\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\) ( luôn đúng )

Áp dụng vào bài toán ta có : \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

\(\Leftrightarrow2025\ge\left(x+y\right)^2\)

\(\Leftrightarrow-45\le x+y\le45\)

Vậy : \(min\left(x+y\right)=-45,max\left(x+y\right)=45\)

Bài 1 

ta có a+3+b-3 =a +b chia hết cho 4

nên (b-a )(a+b) cũng chia hết cho 4

bài 2.

ta có: \(M=6x^2-5x-6-12xy+6y^2+6y-3x+2y+2027\)

\(=6\left(x-y\right)^2-8\left(x-y\right)+2021=24-16+2021=2029\)

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

Đăng câu hỏi 1 lần thôi em

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

a: \(=\left(328+172\right)\left(328^2+328\cdot172+172^2\right)\)

\(=5000\cdot4\left(26896+328\cdot43+7396\right)⋮20000\)

b: \(=69\left(69-5\right)=69\cdot64⋮32\)

\(x^2+2y^2+2xy+6x+2y+2027\)

\(=x^2+2x\left(y+3\right)+\left(y+3\right)^2+\left(y^2-4y+4\right)+2014\)

\(=\left(x+y+3\right)^2+\left(y-2\right)^2+2014\)

Ta có: \(\left\{{}\begin{matrix}\left(x+y+3\right)^2\ge0\forall x;y\\\left(y-2\right)^2\ge0\forall y\end{matrix}\right.\)\(\Leftrightarrow\)\(\Rightarrow\left(x+y+3\right)^2+\left(y-2\right)^2+2014\ge2014\)\(\forall x;y\)

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