\(A\le\sqrt{\left(3^2+4^2\right)\left(x-1\right)\left(5-x\right)}=10\)

\(A_{max}=10\) khi \(\dfrac{\sqrt{x-1}}{3}=\dfrac{\sqrt{5-x}}{4}\Rightarrow x=\dfrac{61}{25}\)

\(A=3\left(\sqrt{x-1}+\sqrt{5-x}\right)+\sqrt{5-x}\ge3\left(\sqrt{x-1}+\sqrt{5-x}\right)\ge3\sqrt{x-1+5-x}=6\)

\(A_{min}=6\) khi \(x=5\)

Áp dụng BĐT Bunhiacopxki , ta có :

\(\left(x-3+5-x\right)\left(1^2+1^2\right)\) ≥ \(\left(\sqrt{x-3}+\sqrt{5-x}\right)^2\)

⇔ \(B^2\) ≤ \(2.2=4\)

⇔ \(B\) ≤ \(2\)

⇒ B_Min = 2 ⇔ \(\sqrt{x-3}=\sqrt{5-x}\) ⇔ x = 4

1:

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

căn x+1>=1

=>2/căn x+1<=2

=>-2/căn x+1>=-2

=>A>=-2+1=-1

Dấu = xảy ra khi x=0

b: 

ĐKXĐ: x>=0; x<>1/4

Ta có: \(A=\frac{\sqrt{x}+1}{2\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{x+6\sqrt{x}+2}{2x+5\sqrt{x}-3}\)

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

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+\sqrt{x}\left(2\sqrt{x}-1\right)-x-6\sqrt{x}-2}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+4\sqrt{x}+3+2x-\sqrt{x}-x-6\sqrt{x}-2}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{2x-3\sqrt{x}+1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

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

Ta có: P=A*B

\(=\frac{\sqrt{x}-1}{\sqrt{x}+3}\cdot\frac{\sqrt{x}+3}{x+8}=\frac{\sqrt{x}-1}{x+8}\)

=>\(\frac{1}{P}=\frac{x+8}{\sqrt{x}-1}=\frac{x-1+9}{\sqrt{x}-1}=\sqrt{x}+1+\frac{9}{\sqrt{x}-1}=\sqrt{x}-1+\frac{9}{\sqrt{x}-1}+2\ge2\cdot\sqrt{\left(\sqrt{x}-1\right)\cdot\frac{9}{\sqrt{x}-1}}+2=2\cdot3+2=8\forall x\) thỏa mãn ĐKXĐ

=>\(P\le\frac18\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\left(\sqrt{x}-1\right)^2=9;\sqrt{x}-1>0\)

=>\(\sqrt{x}-1=3\)

=>\(\sqrt{x}=4\)

=>x=16(nhận)