a) \(M=2022-\left|x-9\right|\le2022\)

\(maxM=2022\Leftrightarrow x=9\)

b) \(N=\left|x-2021\right|+2022\ge2022\)

\(minN=2022\Leftrightarrow x=2021\)

đợi tý

Đã trả lời rồi còn độ tí đồ ngull

A = (x+5)²⁰²² + | y - 2021| + 2022

vì ( x+5)²⁰²² \(\ge\) 0; 

    |y-2021|   \(\ge\) 0

    2022      = 2022

Cộng vế với vế ta được : A = (x+5)²⁰²²+|y-2021|+2022\(\ge\) 2022

Vậy A(min) = 2022 dấu bằng xảy ra khi : \(\left\{{}\begin{matrix}x+5=0\\y-2021=0\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=-5\\y=2021\end{matrix}\right.\)

a: \(a^2+4b^2+9c^2=2ab+6bc+3ac\)

=>\(2a^2+8b^2+18c^2-4ab-12bc-6ac=0\)

=>\(a^2-4ab+4b^2+4b^2-12bc+9c_{}^2+a^2-6ac+9c^2=0\)

=>\(\left(a-2b\right)^2+\left(2b-3c\right)^2+\left(a-3c\right)^2=0\)

=>\(\begin{cases}a-2b=0\\ 2b-3c=0\\ 3c-a=0\end{cases}\Rightarrow a=2b=3c\)

\(A=\left(a-2b+1\right)^{2022}+\left(2b-3c-1\right)^{2023}+\left(3c-a+1\right)^{2024}\)

\(=\left(a-a+1\right)^{2022}+\left(2b-2b-1\right)^{2023}+\left(a-a+1\right)^{2024}\)

=1-1+1

=1

b: \(x^2+2xy+6x+6y+2y^2+8=0\)

=>\(x^2+2xy+y^2+6\left(x+y\right)+9+y^2-1=0\)

=>\(\left(x+y+3\right)^2-1=-y^2\)

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

=>\(-y^2=\left(x+y+2024-2022\right)\left(x+y+2024-2020\right)\)

=>\(-y^2=\left(A-2022\right)\left(A-2020\right)\)

mà \(-y^2\le0\forall y\)

nên (A-2022)(A-2020)<=0

=>2020<=A<=2022

\(A_{\min}=2020\) khi x+y+2=0 và y=0

=>y=0 và x=-2-y=-2-0=-2

\(A\max=2022\) khi x+y+4=0 và y=0

=>y=0 và x=-y-4=-4

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 $y$)

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

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

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

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

\(M=\left|x-2021\right|+\left|2022-x\right|\ge\left|x-2021+2022-x\right|=1\\ M_{min}=1\Leftrightarrow\left(x-2021\right)\left(2022-x\right)\ge0\Leftrightarrow2021\le x\le2022\)

M=|x-2|+|2022-x|>=|x-2+2022-x|=2020

Dấu = xảy ra khi 2<=x<=2022

Sửa đề: \(\frac{x^2-42x+2022}{x^2}\)

ĐKXĐ: x<>0

Đặt \(A=\frac{x^2-42x+2022}{x^2}\)

\(=1-\frac{42}{x}+\frac{2022}{x^2}\)

\(=2022\left(\frac{1}{x^2}-\frac{21}{1011}\cdot\frac{1}{x}+\frac{1}{2022}\right)\)

\(=2022\left(\frac{1}{x^2}-2\cdot\frac{1}{x}\cdot\frac{21}{2022}+\frac{441}{2022^2}+\frac{1581}{2022^2}\right)=2022\left(\frac{1}{x}-\frac{21}{2022}\right)^2+\frac{1581}{2022^2}\cdot2022\)

\(=2022\left(\frac{1}{x}-\frac{21}{2022}\right)^2+\frac{1581}{2022}\ge\frac{1581}{2022}\forall x\)

Dấu '=' xảy ra khi \(\frac{1}{x}-\frac{21}{2022}=0\)

=>\(\frac{1}{x}=\frac{21}{2022}\)

=>\(x=\frac{2022}{21}=\frac{674}{7}\)

Sửa đề: \(\frac{x^2-42x+2022}{x^2}\)

ĐKXĐ: x<>0

Đặt \(A=\frac{x^2-42x+2022}{x^2}\)

\(=1-\frac{42}{x}+\frac{2022}{x^2}\)

\(=2022\left(\frac{1}{x^2}-\frac{21}{1011}\cdot\frac{1}{x}+\frac{1}{2022}\right)\)

\(=2022\left(\frac{1}{x^2}-2\cdot\frac{1}{x}\cdot\frac{21}{2022}+\frac{441}{2022^2}+\frac{1581}{2022^2}\right)=2022\left(\frac{1}{x}-\frac{21}{2022}\right)^2+\frac{1581}{2022^2}\cdot2022\)

\(=2022\left(\frac{1}{x}-\frac{21}{2022}\right)^2+\frac{1581}{2022}\ge\frac{1581}{2022}\forall x\)

Dấu '=' xảy ra khi \(\frac{1}{x}-\frac{21}{2022}=0\)

=>\(\frac{1}{x}=\frac{21}{2022}\)

=>\(x=\frac{2022}{21}=\frac{674}{7}\)

a: |x|+2003>=2003

=>A<=2022/2003

Dấu = xảy ra khi x=0

b: |x|+1>=1

=>(|x|+1)^10>=1

=>B>=2010

Dấu = xảy ra khi x=0