Bài 1: 

a: \(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)

b: \(=\sqrt{11}+\sqrt{2}-\sqrt{11}-\sqrt{7}-\sqrt{2}=-\sqrt{7}\)

\(P=\sqrt[]{x}+\dfrac{3}{\sqrt[]{x}-1}\left(x>1\right)\)

\(P=\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}+1\)

Áp dụng bất đẳng thức Cauchy cho 2 số \(\sqrt[]{x}-1;\dfrac{3}{\sqrt[]{x}-1}\) ta được :

\(\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}\ge2\sqrt[]{\sqrt[]{x}-1.\dfrac{3}{\sqrt[]{x}-1}}\)

\(\Rightarrow\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}\ge2\sqrt[]{3}\)

\(\Rightarrow P=\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}+1\ge2\sqrt[]{3}+1\)

\(\Rightarrow Min\left(P\right)=2\sqrt[]{3}+1\)

sorry mn cho e sửa lại đề ạ

tìm gtln của p ạ

a: AC=4cm

AH=2,4cm

BH=1,8cm

CH=3,2cm

\(C=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}+2\right)+6\sqrt{x}}{x-4}.\left(x-4\right)=2\sqrt{x}\)

Đề bài đâu rồi em?

a: \(Q=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{x-4}:\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)

\(=\dfrac{x+3\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{x-4}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)

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

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

b: Khi x=4-2căn 3 thì \(Q=\dfrac{\sqrt{3}-1+2}{\sqrt{3}-1-3}=\dfrac{\sqrt{3}+1}{\sqrt{3}-4}=\dfrac{-7-5\sqrt{3}}{13}\)

c: Q>1/6

=>Q-1/6>0

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{1}{6}>0\)

=>\(\dfrac{6\sqrt{x}+12-\sqrt{x}+3}{6\left(\sqrt{x}-3\right)}>0\)

=>\(\dfrac{5\sqrt{x}+9}{6\left(\sqrt{x}-3\right)}>0\)

=>căn x-3>0

=>x>9

thay \(x=3-2\sqrt{2}\) vào P ta có:

\(\dfrac{x+8}{\sqrt{x}+1}=\dfrac{3-2\sqrt{2}+8}{\sqrt{3-2\sqrt{2}}+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}-1+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}}\)

\(b,x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\)

Thay vào P, ta được:

\(P=\dfrac{3-2\sqrt{2}+8}{\sqrt{\left(\sqrt{2}-1\right)^2}+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}}=\dfrac{11\sqrt{2}-4}{2}\)