Vì x² ≥ 0 ∀ x 

=> -5x² ≤ 0

=> -5x² + 9 ≤ 9

Để A = -5x² + 9 nhận giá trị lớn nhất thì -5x² + 9 = 9 

=> A = 9

Vì ( 3x - 2 )² ≥ 0

=> 5 - ( 3x - 2 )² ≤ 5

Để B = 5 - ( 3x - 2 )² nhận giá trị lớn nhất thì 5 - ( 3x - 2 )² = 5 

=> B = 5

Để D = \(\frac{\text{2022}}{\left(\text{2 - x}\right)^2+\text{1}}\)nhận giá trị lớn nhất thì ( 2 - x )2 + 1 nhận giá trị nhỏ nhất

Mà ( 2 - x )2 + 1 ≠ 0

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

=> D = \(\frac{\text{2022}}{\left(\text{2 - x}\right)^2+\text{1}}=\frac{\text{2022}}{\text{1}}\)= 2022 

Ta có \(-5x^2\le0\Leftrightarrow-5x^2+9\le9\)  

=> Max A = 9 

Dấu "=" xảy ra <=> x² = 0 => x = 0

Vậy Max A = 9 <=> x = 0

b) Ta có \(-\left(3x-2\right)^2\le0\forall x\Rightarrow5-\left(3x-2\right)^2\le5\)

=> Max B = 5 

Dấu "=" xảy ra <=> 3x - 2 = 0 <=> x = 2/3

Vậy Max = 5 <=> x = 2/3

c) Ta có \(2x^2+3\ge3\forall x\Rightarrow\frac{1}{2x^2+3}\le\frac{1}{3}\)

=> Max C = 1/3 

Dấu "=" xảy  ra <=> x²  = 0 => x = 0

Vậy Max C = 1/3 <=> x = 0

d) Ta có \(\left(2-x\right)^2+1\ge1\forall x\Leftrightarrow\frac{2022}{\left(2-x\right)^2+1}\le2022\)

=> Max D = 2022

 Dấu "=" xảy ra <=> 2 - x = 0 => x = 2

Vậy Max D = 2022 <=> x = 2

a) Ta có \(\left(x-2\right)^2\ge0\forall x\)

=> Min A = 0

Dấu "=" xảy ra <=> x - 2 = 0 <=> x = 2

Vậy Min A = 0 <=> x = 2

b) Ta có \(\left(2x+1\right)^4\ge0\forall x\Rightarrow\left(2x+1\right)^4-98\ge-98\)

=> Min B = -98

Dấu "=" xảy ra <=> 2x + 1= 0 <=> x = -0,5

Vậy Min B = -98 <=> x = -0,5

c) Ta có  C = |x - 10| + |x - 11| 

= |x - 10| + |11 - x| \(\ge\left|x-10+11-x\right|=\left|1\right|=1\)

=> Min C = 1

Dấu "=" xảy ra <=> \(\left(x-10\right)\left(11-x\right)\ge0\)

TH1 : \(\hept{\begin{cases}x-10\ge0\\11-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge10\\x\le11\end{cases}}\Leftrightarrow10\le x\le11\)

TH2 : \(\hept{\begin{cases}x-10\le0\\11-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le10\\x\ge11\end{cases}}\Leftrightarrow x\in\varnothing\)

Vậy Min C = 1 <=> \(10\le x\le11\)

Ta có:B = \(\frac{1}{2}+\frac{3}{2^2}+\frac{7}{2^3}+...+\frac{2^{100}-1}{2^{100}}=\frac{2-1}{2}+\frac{2^2-1}{2^2}+\frac{2^3-1}{2^3}+...+1-\frac{1}{2^{100}}\)

\(=1-\frac{1}{2}+1-\frac{1}{2^2}+1-\frac{1}{2^3}+...+1-\frac{1}{2^{100}}=100-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

=> \(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(A=1-\frac{1}{2^{100}}\)

=> \(B=100-\left(1-\frac{1}{2^{100}}\right)=100-1+\frac{1}{2^{100}}=99+\frac{1}{2^{100}}>99\) (Đpcm)