Sửa đề :

Tìm max \(\sqrt{3x-9}+\sqrt{7-x}\)

Áp dụng BĐT Cô-si:

\(\sqrt{3x-9}=\frac{3\cdot\sqrt{3x-9}}{3}=\frac{\sqrt{9\cdot\left(3x-9\right)}}{3}\le\frac{\frac{9+3x-9}{2}}{3}=\frac{x}{2}\)

\(\sqrt{7-x}=\sqrt{1\cdot\left(7-x\right)}\le\frac{1+7-x}{2}=\frac{8-x}{2}\)

Cộng theo vế :

\(\sqrt{3x-9}+\sqrt{7-x}\le\frac{x+8-x}{2}=4\)

Dấu "=" xảy ra \(\Leftrightarrow x=6\)

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 \(x=0\)

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

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

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)

\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

\(=\dfrac{3}{\sqrt{x}+3}\)

b: Thay x=16 vào A, ta được:

\(A=\dfrac{3}{4+3}=\dfrac{3}{7}\)

c)\(A=\dfrac{3}{\sqrt{x}+3}=\dfrac{1}{3}\)

\(\Rightarrow\sqrt{x}+3=9\\ \Rightarrow\sqrt{x}=6\\ \Rightarrow x=36\)

d) \(A=\dfrac{3}{\sqrt{x}+3}\)

Vì \(3>0;\sqrt{x}+3>0\Rightarrow\dfrac{3}{\sqrt{x}+3}>0\)

e) \(2A\in Z\Rightarrow\dfrac{6}{\sqrt{x}+3}\in Z \Rightarrow6⋮x+3\\\Rightarrow\sqrt{x}+3\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\Rightarrow x=\left\{0;9\right\}\)

Áp dụng BĐT bu - nhi -a cốp - xki 

ta có \(B^2=\left(1.\sqrt{2x-3}+1.\sqrt{x-1}+1.\sqrt{7-3x}\right)^2\le\left(1^2+1^2+1^2\right)\left(2x-3+x-1+7-3x\right)\)

       <=> \(b^2\le3.3=9\Rightarrow B\le3\)

Dấu '=' xảy ra khi x = 2 

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

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

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

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