ta có biểu thức:

x⁵ − 2023x⁴ − 2023x³ − 2023x² − 2023x − 2010

thay x = 2024.

cách tính hợp lí: đặt 2024 = 2023 + 1

gọi a = 2023

-> x = a + 1

thay vào:
p(x) = (a + 1)⁵ − a(a + 1)⁴ − a(a + 1)³ − a(a + 1)² − a(a + 1) − 2010

thay a = 2023, ta được kết quả là 14

vậy p(2024) = 14

Khi x=2024 nên x-1=2023

\(x^5-2023x^4-2023x^3-2023x^2-2023x-2010\)

\(=x^5-x^4\left(x-1\right)-x^3\left(x-1\right)-x^2\left(x-1\right)-x\left(x-1\right)-2010\)

\(=x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-2010\)

=x-2010

=2024-2010

=14

x=2022 

=>x+1=2023

A=x^50-x^49(x+1)+x^48(x+1)-...+x^2(x+1)-x(x+1)+x+2

=x^50-x^50-x^49+x^49+...+x^3+x^2-x^2-x+x+2

=2

\(1,\left(x+2022\right)\left(x-1\right)=x^2+2021x-2022\left(B\right)\\ 2,\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\left(A\right)\)

Đề không đầy đủ. Bạn coi lại.

Sửa đề: \(5x^2+5y^2+8xy-2x+2y+2=0\)

=>\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)

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

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

\(M=\left(x-y\right)^{2023}-\left(x-2\right)^{2024}+\left(y+1\right)^{2023}\)

\(=\left(1+1\right)^{2023}-\left(1-2\right)^{2024}+\left(-1+1\right)^{2023}\)

\(=2^{2023}-1\)

a, tách 80=79+1

sau đó phân phối ra

b, tách 10=9+1

phân phối tiếp

phần a ) là \(P\left(x\right)=x^7-80x^6-80x^5-80x^4\)\(+...+80x+5\)nha ình chép thiếu

Sửa đề: x=8

Khi x=8 thì x+1=9

\(x^{13}-9x^{12}+9x^{11}-9x^{10}+\cdots-9x^2+9x-2\)

\(=x^{13}-x^{12}\left(x+1\right)+x^{11}\left(x+1\right)-x^{10}\left(x+1\right)+\cdots-x^2\left(x+1\right)+x\left(x+1\right)-2\)

\(=x^{13}-x^{13}-x^{12}+x^{12}+x^{11}-x^{11}-x^{10}+\cdots-x^3-x^2+x^2+x-2\)

=x-2

=8-2

=6

Viết đề có sai không bạn:")?

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