\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x+1}{x-1}\)   \(\left(\text{Đ}K:x\ge0;x\ne1\right)\)

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

    \(=\dfrac{x+1}{x+1}=1\)

A =\(\sqrt{9}\) + \(\sqrt{16}\) + 2\(\sqrt{2}\) - \(\sqrt{8}\)

A = \(\sqrt{3^2}\) + \(\sqrt{4^2}\) + 2\(\sqrt{2}\) - 2\(\sqrt{2^3}\)

A = 3 + 4 + 2\(\sqrt{2}\) - 2\(\sqrt{2}\)

A = 7 + 0

A = 7

a: góc CAO+góc CBO=180độ

=>CAOB nội tiếp

b: Xét ΔCEB và ΔCBD có

góc CEB=góc CBD

góc ECB chung

=>ΔCEB đồng dạng vơi ΔCBD

=>CE/CB=CB/CD

=>CB^2=CE*CD

BĐT cần chứng minh tương đương:

\(\dfrac{a}{a+\sqrt{3a+bc}}+\dfrac{b}{b+\sqrt{3b+ca}}+\dfrac{c}{c+\sqrt{3c+ab}}\le1\)

Ta có:

\(\dfrac{a}{a+\sqrt{3a+bc}}=\dfrac{a}{a+\sqrt{a\left(a+b+c\right)+bc}}=\dfrac{a}{a+\sqrt{\left(a+b\right)\left(c+a\right)}}\le\dfrac{a}{a+\sqrt{\left(\sqrt{ab}+\sqrt{ac}\right)^2}}\)

\(=\dfrac{a}{a+\sqrt{ab}+\sqrt{ac}}=\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Tương tự:

\(\dfrac{b}{b+\sqrt{3b+ca}}\le\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

\(\dfrac{c}{c+\sqrt{3c+ab}}\le\dfrac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Cộng vế:

\(\dfrac{a}{a+\sqrt{3a+bc}}+\dfrac{b}{b+\sqrt{3b+ca}}+\dfrac{c}{c+\sqrt{3c+ab}}\le\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)