=(1-2-3+4)+(5-6-7+8)+...+(2017-2018-2019+2020)+2021-2022-2023

=0+0+...+0-1-2023

=-2024

Bằng 2024

Áp dụng tính chất của dãy tỉ số bằng nhau,ta có:
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}=\dfrac{x+y+z}{y+z+x}=\dfrac{x+y+z}{x+y+z}=1\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\)
Do đó \(\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\)
Thay vào biểu thức \(P=\left(x-y\right)^{2022}+\left(y-z\right)^{2023}+\left(x-z-1\right)^{202}\),ta có:
\(P=0^{2022}+0^{2023}+\left(-1\right)^{202}\)
\(=0+0+1\)
\(=1\)

`a, A = 3020 xx 3110 - 5 = 3020 xx 3109 + 3020 - 5`

`= 3020 xx 3109 + 3015 = B`.

`b, B = (2022-2)(2022+2) = 2022^2-4 < 2022^2 = A.`

Ta có: \(\left\{{}\begin{matrix}x^2+2y+1=0\\y^2+2z+1=0\\z^2+2x+1=0\end{matrix}\right.\)

\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0\)

\(\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)

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

a) \(A=x^{2020}+y^{2020}+z^{2020}=\left(-1\right)^{2020}+\left(-1\right)^{2020}+\left(-1\right)^{2020}=1+1+1=3\)

b) \(B=\dfrac{1}{x^{2020}}+\dfrac{1}{y^{2020}}+\dfrac{1}{z^{2020}}=\dfrac{1}{\left(-1\right)^{2020}}+\dfrac{1}{\left(-1\right)^{2020}}+\dfrac{1}{\left(-1\right)^{2020}}=\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}=3\)

A.

$a^2+4b^2+9c^2=2ab+6bc+3ac$

$\Leftrightarrow a^2+4b^2+9c^2-2ab-6bc-3ac=0$

$\Leftrightarrow 2a^2+8b^2+18c^2-4ab-12bc-6ac=0$

$\Leftrightarrow (a^2+4b^2-4ab)+(a^2+9c^2-6ac)+(4b^2+9c^2-12bc)=0$

$\Leftrightarrow (a-2b)^2+(a-3c)^2+(2b-3c)^2=0$

$\Rightarrow a-2b=a-3c=2b-3c=0$

$\Rightarrow A=(0+1)^{2022}+(0-1)^{2023}+(0+1)^{2024}=1+(-1)+1=1$

B.

$x^2+2xy+6x+6y+2y^2+8=0$

$\Leftrightarrow (x^2+2xy+y^2)+y^2+6x+6y+8=0$

$\Leftrightarrow (x+y)^2+6(x+y)+9+y^2-1=0$

$\Leftrightarrow (x+y+3)^2=1-y^2\leq 1$ (do $y^2\geq 0$ với mọi $y$)

$\Rightarrow -1\leq x+y+3\leq 1$

$\Rightarrow -4\leq x+y\leq -2$

$\Rightarrow 2020\leq x+y+2024\leq 2022$

$\Rightarrow A_{\min}=2020; A_{\max}=2022$

A

Dẫu '' = '' xảy ra khi và chỉ khi ( x - y )² + (x – 1)² + (y + 2)² = 0