\(D=\sqrt{2+1-2\sqrt{2}}-\sqrt{2+1+2\sqrt{2}}\)

\(D=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)

\(D=\sqrt{2}-1-\left(\sqrt{2}+1\right)\)

\(D=\sqrt{2}-1-\sqrt{2}-1\)

\(D=-2\)

CÂU THỨ 2 NHA !!!!!!

XÉT:        \(2VT=2a\sqrt{b-1}+2b\sqrt{a-1}\)

=>    \(2VT=a.2.\sqrt{1}.\sqrt{b-1}+b.2.\sqrt{1}.\sqrt{a-1}\)

TA ÁP DỤNG BĐT CAUCHY 2 SỐ SẼ ĐƯỢC: 

=>    \(2VT\le a\left(1+b-1\right)+b\left(1+a-1\right)\)

=>   \(2VT\le ab+ab\)

=>   \(2VT\le2ab\)

=>   \(VT\le ab\)

=> TA CÓ ĐIỀU PHẢI CHỨNG MINH.

\(D=\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}-\sqrt{2+2\sqrt{2}+1}\)

\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}=\left|\sqrt{2}-1\right|-\left|\sqrt{2}+1\right|=\sqrt{2}-1-1-\sqrt{2}=-2\)

Bài làm:

Ta có: \(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}\)

\(=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

=> đpcm

Ta có: \(VT=\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\frac{4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{a+2\sqrt{ab}+b-\left(a-2\sqrt{ab}+b\right)+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{4b+4\sqrt{ab}}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{4\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}{2\left(\sqrt{b}+\sqrt{a}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}=VP\)(đpcm)

\(D=\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)

\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)

\(=\sqrt{2}-1-\sqrt{2}-1=-2\)

___

Ta có: \(\left(\sqrt{a-1}-1\right)^2\ge0\forall a\ge1\)

\(\Leftrightarrow a-2\sqrt{a-1}\ge0\)

\(\Leftrightarrow\frac{\sqrt{a-1}}{a}\le\frac{1}{2}\)

Tương tự: \(\frac{\sqrt{b-1}}{b}\le\frac{1}{2}\)

\(\Rightarrow\frac{\sqrt{a-1}}{a}+\frac{\sqrt{b-1}}{b}\le1\)

\(\Leftrightarrow b\sqrt{a-1}+a\sqrt{b-1}\le ab\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=2\)

Nhận thấy \(a+b< a+b+2\sqrt{ab}\)<=>\(a+b< \left(\sqrt{a}+\sqrt{b}\right)^2\)

Do a,b đều dương, lấy căn 2 vế ta được: 

\(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)(đpcm)

Chúc bạn học tốt!

\(0<2\sqrt{ab}\) cộng 2 vế với a+ b

a+b< a+b+ 2.căn(ab)

\(a+b<\left(\sqrt{a}+\sqrt{b}\right)^2\)

lấy căn 2 vế  là xong