\(a=2022.\left|x^2+1\right|+2023\)

\(\Rightarrow a=2022.\left(x^2+1\right)+2023\left(\left|x^2+1\right|>0,\forall x\right)\)

mà \(\left(x^2+1\right)\ge1,\forall x\)

\(\Rightarrow a=2022.\left(x^2+1\right)+2023\ge2022.1+2023=4045\)

\(\Rightarrow GTNN\left(a\right)=4045\left(x=0\right)\)

GTNN(a) = 4045 khi x = 0

Ta có :A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\) -\(\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}\) 

=\(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)-2

=\(\dfrac{-\sqrt{x}}{\sqrt{x}+1}\)

thay vào A=\(\dfrac{-2}{3}\)

b)

A=-1+\(\dfrac{1}{\sqrt{x}+1}\) \(\ge\) -1+\(\dfrac{1}{1}\)=1(vì \(\sqrt{x}\)\(\ge\) 0)

Dấu bằng xẩy ra\(\Leftrightarrow\) x=0

chỗ đó cho thêm x-1 nha

đấu >= thay thành <= rùi nhân thêm x-1>=-1 nữa là lớn nhất bằng 0

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

a) \(M=2022-\left|x-9\right|\le2022\)

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

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

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

A = \(\dfrac{1}{\left|x+1\right|+\left|x-2022\right|}\)

Đặt B = \(\left|x+1\right|+\left|x-2022\right|\)

\(\left|x-2022\right|\) = \(\left|2022-x\right|\) ⇒ B = \(\left|x+1\right|+\left|2022-x\right|\)

B =\(\left|x+1\right|+\left|2022-x\right|\) ≥ \(\left|x+1+2022-x\right|\) = 2023

B(min) = 2023 ⇔ (\(x+1\))(2022-\(x\)) \(\ge\) 0

Lập bảng ta có: 

Theo bảng trên ta có: B(min) = 2023 ⇔ -1 ≤ \(x\) ≤ 2022

A = \(\dfrac{1}{\left|x+1\right|+\left|x-2022\right|}\) 

Vì A dương nên A(max) ⇔ B(min) ⇔ B = 2023

A(max) = \(\dfrac{1}{2023}\) ⇔ -1 ≤ \(x\) ≤ 2022

a) Ta có: 

\(Q=\sqrt{\left(1-3x\right)\left(x+\dfrac{1}{2}\right)}\) Q có nghĩa khi:

\(\left(1-3x\right)\left(x+\dfrac{1}{2}\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1-3x\ge0\\x+\dfrac{1}{2}\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}1-3x\le0\\x+\dfrac{1}{2}\le\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x\le1\\x\ge-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}3x\ge1\\x\le-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\x\ge-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\x\le-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{1}{2}\le x\le\dfrac{1}{3}\\x\in\varnothing\end{matrix}\right.\)

\(\Leftrightarrow-\dfrac{1}{2}\le x\le\dfrac{1}{3}\)

b) Ta có: \(Q=\sqrt{\left(1-3x\right)\left(x+\dfrac{1}{2}\right)}\)

\(Q=\sqrt{x+\dfrac{1}{2}-3x^2-\dfrac{3}{2}x}\)

\(Q=\sqrt{-\left(3x^2+\dfrac{1}{2}x-\dfrac{1}{2}\right)}\)

\(Q=\sqrt{-3\left(x^2+\dfrac{1}{6}x-\dfrac{1}{6}\right)}\)

\(Q=\sqrt{-3\left(x^2+2\cdot\dfrac{1}{12}\cdot x+\dfrac{1}{144}-\dfrac{25}{144}\right)}\)

\(Q=\sqrt{-3\left(x+\dfrac{1}{12}\right)^2+\dfrac{25}{144}}\)

Mà: \(Q=\sqrt{-3\left(x+\dfrac{1}{12}\right)^2+\dfrac{25}{144}}\le\sqrt{\dfrac{25}{144}}=\dfrac{5}{12}\)

Dấu "=" xảy ra khi:

\(\Leftrightarrow-3\left(x+\dfrac{1}{12}\right)^2=0\)

\(\Leftrightarrow x+\dfrac{1}{12}=0\)

\(\Leftrightarrow x=-\dfrac{1}{12}\)

Vậy: \(Q_{max}=\dfrac{5}{12}.khi.x=-\dfrac{1}{12}\)

a: Ta có: \(x^2=3-2\sqrt{2}\)

nên \(x=\sqrt{2}-1\)

Thay \(x=\sqrt{2}-1\) vào A, ta được:

\(A=\dfrac{\left(\sqrt{2}+1\right)^2}{\sqrt{2}-1}=\dfrac{3+2\sqrt{2}}{\sqrt{2}-1}=7+5\sqrt{2}\)

`C=(sqrtx+3)/(sqrtx-2)=(sqrtx-2+5)/(sqrtx-2)=1+5/(sqrtx-2)`

Ta cần tìm `max(5/(sqrtx-2))`

Nếu `0<=x<4` thì `5/(sqrtx-2)<0`

Nếu `x>4` thì `5/(sqrtx-2)>0`

Do đó ta chỉ xét `x>4` hay `x>=5(` Do `x` nguyên `)`

`=>sqrtx-2>=sqrt5-2`

`=>5/(sqrtx-2)<=5/(sqrt5-2)`

`=>C<=1+5/(sqrt5-2)=11+sqrt5`

Vậy `C_(max)=11+sqrt5<=>x=5`