Với x ≥ 0 thì \(\sqrt{x}\ge0\) nên \(\sqrt{x}+1\ge1\)

Khi đó \(B=\left(\sqrt{x}+1\right)^{99}+2022\ge1^{99}+2022\)

Hay \(B=\left(\sqrt{x}+1\right)^{99}+2022\ge2023\)

Dấu "=" xảy ra khi \(\sqrt{x}=0\) hay x = 0

Vậy GTNN của \(B=\left(\sqrt{x}+1\right)^{99}+2022\) là 2023 khi x = 0

\(B=\left(\sqrt{x}+1\right)^{99}+2022\left(x\ge0\right)\)

Vì: \(x\ge0\)

Nên => \(\left(\sqrt{x}+1\right)^{99}\ge0\)

=> \(\left(\sqrt{x}+1\right)^{99}+2022\ge2022\)

=> \(B\ge2022\)

Dấu " = " xảy ra khi: \(\Leftrightarrow\sqrt{x}+1=0\Leftrightarrow\sqrt{x}=-1\left(voli\right)\)

Vậy: B không có giá trị nhỏ nhất

Ta có tính chất: \(\dfrac{a}{b}>\dfrac{a-m}{b-m}\)

\(A=\dfrac{2022^{99}-1}{2022^{100}-1}>\dfrac{2022^{99}-1-2021}{2022^{100}-1-2021}\)

\(A>\dfrac{2022^{99}-2022}{2022^{100}-2022}\)

\(A>\dfrac{2022\left(2022^{98}-1\right)}{2022\left(2022^{99}-1\right)}\)

\(A>\dfrac{2022^{98}-1}{2022^{99}-1}\)

\(A>B\)

Ta có \(x+1=2022\)

\(P\left(x\right)=x^{101}-\left(x+1\right)x^{100}+...+\left(x+1\right)x-1\)

\(=x^{101}-x^{101}-x^{100}+...+x^2+x-1=x-1\)

-> P(x) = 2020 

\(\dfrac{1}{2022}\cdot A=\dfrac{2022^{100}+1}{2022^{100}+100}=1-\dfrac{99}{2022^{100}+100}\)

\(\dfrac{1}{2022}B=\dfrac{2022^{101}+1}{2022^{101}+100}=1-\dfrac{9}{2022^{101}+100}\)

2022^100+100<2022^101+100

=>-99/2022^100+100<-99/2022^101+100

=>A<B

=> A/2022 = 2022^100+1/2022^100+2022 = 1- 2021/2022^100+2022

=> B/2022 = 2022^101+1/2022^101+2022 = 1- 2021/2022^101+2022

Nhận thấy 2022^101 + 2022 > 2022^100 + 2022

=> 2021/2022^101 + 2022 < 2021/2022^100 + 2022

=> B/2022 > A/2022 => B>A

Vậy A<B

Tìm GTNN chứ nhỉ e

\(D=\left|2022-x\right|+\left|x-1\right|\ge\left|2022-x+x-1\right|=2021\)

Dấu "=" xảy ra \(\Leftrightarrow\left(2022-x\right)\left(x-1\right)\ge0\)

\(\Leftrightarrow1\le x\le2022\)

Vậy Min D=2021 \(\Leftrightarrow1\le x\le2022\)

a: \(98^{10}\cdot A=\dfrac{98^{98}+98^{10}}{98^{98}+1}=1+\dfrac{98^{10}-1}{98^{98}+1}\)

\(98^{10}\cdot B=\dfrac{98^{99}+98^{10}}{98^{99}+1}=1+\dfrac{98^{10}-1}{98^{99}+1}\)

98^88+1>98^99+1

=>A<B

b: \(\dfrac{1}{2022^2}\cdot C=\dfrac{2022^{2023}+1}{2022^{2023}+2022^2}=1+\dfrac{1-2022^2}{2022^{2023}+2022^2}\)

\(\dfrac{1}{2022^2}\cdot D=\dfrac{2022^{2021}+1}{2022^{2021}+2022^2}=1+\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

2022^2023>2022^2021

=>2022^2023+2022^2>2022^2021+2022^2

=>\(\dfrac{2022^2-1}{2022^{2023}+2022^2}< \dfrac{2022^2-1}{2022^{2021}+2022^2}\)

=>\(\dfrac{1-2022^2}{2022^{2023}+2022^2}>\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

=>C>D

đợi tý

a) Để \(A=\dfrac{2022}{\left|x\right|+2023}\) đạt Max thì |x| + 2023 phải đạt Min

Ta có \(\left|x\right|\ge0\forall x\Rightarrow\left|x\right|+2023\ge2023\forall x\)

\(\Rightarrow\dfrac{2022}{\left|x\right|+2023}\le\dfrac{2022}{2023}\forall x\)

Dấu "=" xảy ra khi \(\left|x\right|=0\Rightarrow x=0\)

Vậy Max \(A=\dfrac{2022}{\left|x\right|+2023}=\dfrac{2022}{2023}\) đạt được khi x = 0

b) Để \(B=\left(\sqrt{x}+1\right)^{99}+2022\) đạt Min với \(x\ge0\) thì \(\sqrt{x}+1\) phải đạt Min

Ta có \(\sqrt{x}\ge0\forall x\ge0\Rightarrow\sqrt{x}+1\ge1\forall x\ge0\)

\(\Rightarrow\left(\sqrt{x}+1\right)^{99}+2022\ge1+2022\ge2023\forall x\ge0\)

Dấu "=" xảy ra khi \(\sqrt{x}=0\Rightarrow x=0\)

Vậy Max \(B=\left(\sqrt{x}+1\right)^{99}+2022=2023\) đạt được khi x = 0

Câu c) và d) thì tự làm, ko có rảnh =))))

Đã trả lời rồi còn độ tí đồ ngull

a/Thay a = 1; b = 0 vào biểu thức C, ta có:
\(C=\left(2022\times1+2022\times0\right)-2021\times0\)
\(=\left(2022+0\right)-0\)
\(=2022\)
b/Thay a = 1; b = 0 vào biểu thức D, ta có:
\(D=\left(999\times1-99\times0\right)+201\times\left(1-0\right)\)
\(=\left(999-0\right)+201\times1\)
\(=999+201\)
\(=1200\)
#deathnote