\(A\le\sqrt{2\left(3x-5+7-3x\right)}=2\)

\(A_{max}=2\) khi \(3x-5=7-3x\Leftrightarrow x=2\)

ĐKXĐ x x > hoặc bằng 0
Do x > hoặc bằng 0 nên (x^5 + 3x^3 + 2 căn x)  > hoặc bằng 0
=> x^5 + 3x^3 + 2 căn x + 4 > hoặc bằng 4
=> A < hoặc bằng 3
Vậy max A bằng 3 khi và chỉ khi x = 0
Ko liên quan nhưng tick cho mình bạn nhé ^^

a) \(P=\dfrac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

\(=\dfrac{3x+3\sqrt{x}-9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b) \(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+2}{\sqrt{x}+2}=3+\dfrac{2}{\sqrt{x}+2}\)

Để \(P\in Z\Rightarrow2⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2=2\left(\sqrt{x}+2\ge2\right)\)

\(\Rightarrow x=0\)

c) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\Rightarrow3+\dfrac{2}{\sqrt{x}+2}\le4\)

\(\Rightarrow P_{max}=4\) khi \(x=0\)

1) \(P=\dfrac{5-3\sqrt{x}}{\sqrt{x}-1}\left(đk:x\ge0,x\ne1\right)\)

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

\(\Rightarrow\sqrt{x}-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Do \(x\ge0,x\ne1\) và x là số chính phương

\(\Rightarrow x\in\left\{0;4;9\right\}\)

2) \(3x^2-5x+1=3\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)-\dfrac{13}{12}=3\left(x-\dfrac{5}{6}\right)^2-\dfrac{13}{12}\ge-\dfrac{13}{12}\)

\(\Rightarrow C=\dfrac{2022}{3x^2-5x+1}\le2022:\left(-\dfrac{13}{12}\right)=-\dfrac{24264}{13}\)

\(minC=-\dfrac{24624}{13}\Leftrightarrow x=\dfrac{5}{6}\)

Bài 1:

Ta thấy:\(2x^2\ge0\Rightarrow-2x^2\le0\)

\(\Rightarrow-2x^2-1\le-1\Rightarrow C\le-1\)

Dấu "=" khi \(-2x^2=0\Leftrightarrow x=0\)

Vậy \(Max_C=-1\) khi x=0

Ta thấy: \(3\sqrt{x-5}\ge0\)

\(\Rightarrow-3\sqrt{x-5}\le0\)

\(\Rightarrow-3\sqrt{x-5}+2\le2\)

\(\Rightarrow D\le2\)

Dấu "=" khi \(-3\sqrt{x-5}=0\Leftrightarrow\sqrt{x-5}=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)

Vậy \(Max_D=2\) khi \(x=5\)

Bài 2:

Ta thấy: \(3x^2\ge0\Rightarrow3x^2-5\ge-5\)

\(\Rightarrow A\ge-5\)

Dấu "=" khi \(3x^2=0\Leftrightarrow x=0\)

Vậy \(Min_A=-5\) khi x=0

Ta thấy: \(2\left(x-3\right)^2\ge0\)

\(\Rightarrow B\ge0\)

Dấu "=" khi \(2\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy \(Min_B=0\) khi x=3

\(A=\sqrt{3x-5}+\sqrt{7-3x}\)

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

\(=2+2\sqrt{\left(3x-5\right)\left(7-3x\right)}\)

\(\le2+\left(3x-5\right)+\left(7-3x\right)\)(Bđt Cô-si)

\(=2+2=4\)

\(\Rightarrow A^2\le4\Rightarrow A\le2\)

Dấu = khi \(\sqrt{3x-5}=\sqrt{7-3x}\Leftrightarrow x=2\)

Vậy....

tks bạn nha