3x^2+3y^2+4xy-2x+2y+2=0

=>2x^2+4xy+2y^2+x^2-2x+1+y^2+2y+1=0

=>x=1 và y=-1

M=(1-1)^2017+(1-2)^2018+(-1+1)^2015=1

x^2+y^2-2x-4y+6=1-(x-y+1)^2

=>x^2-2x+1+y^2-4y+4=-(x-y+1)^2

=>(x-1)^2+(y-2)^2=-(x-y+1)^2

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

=>x=1;y=2

A=2022+2023*2

=2022+4046

=6068

2x² + 2y² + 3xy - x + y + 1 = 0

2x² + 2y² + 4xy - xy - x + y + 1 = 0

(2x² + 2y² + 4xy) + (-xy - x) + (y + 1) = 0

2(x + y)² - x(y + 1) + (y + 1) = 0

2(x + y)² + (y + 1)(1 - x) = 0

Do (x + y)² \(\ge0\)

\(\Rightarrow\) 2(x + y)² \(\ge0\)

\(\Rightarrow\) 2(x + y)² + (y + 1)(1 - x) = 0 \(\Leftrightarrow\) (y + 1)(1 - x) = 0

\(\Rightarrow y+1=0;1-x=0\)

*) y + 1 = 0

y = -1

*) 1 - x = 0

x = 1

Với x = 1; y = -1, ta có:

B = [1 + (-1)]²⁰¹⁸ + (1 - 2)²⁰¹⁸ + (-1 - 1)²⁰¹⁸

= 1 + 2²⁰¹⁸

(x-1)^2+|2y-3|=0

=>x-1=0 và 2y-3=0

=>x=1 và y=1,5

B=4*1^20+5*1^2*1,5-6*1,5+2

=4+7,5-9+2

=4,5

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 $y$)

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

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

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

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

Ko thèm tick cho người ta mà đòi hỏi câu khác ✅

Ta có

ĐÁP ÁN C