Dạng tổng quát:   \(\sqrt{a-b}\ge\sqrt{a}-\sqrt{b}\)         với   \(a\ge b\ge0\)

Chứng minh:   

Ta có:   \(\sqrt{a-b}\ge\sqrt{a}-\sqrt{b}\)

\(\Rightarrow\)\(\left(\sqrt{a-b}\right)^2\ge\left(\sqrt{a}-\sqrt{b}\right)^2\)

\(\Rightarrow\)\(a-b\ge a+b-2\sqrt{ab}\)

\(\Rightarrow\)\(-2b\ge-2\sqrt{ab}\)

\(\Rightarrow\)\(b\le\sqrt{ab}\)

\(\Rightarrow\)\(b^2\le ab\)  luôn đúng do  \(a\ge b\ge0\)

Vậy   \(\sqrt{a-b}\ge\sqrt{a}-\sqrt{b}\)

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

\(\sqrt{35}+\sqrt{99}< \sqrt{36}+\sqrt{100}=6+10=16\)

Vậy \(\sqrt{35}+\sqrt{99}< 16\)

\(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2017}+\sqrt{2016}}\)

\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)

2017>2015

=>căn 2017>căn 2015

=>\(\sqrt{2017}+\sqrt{2016}>\sqrt{2016}+\sqrt{2015}\)

=>\(\dfrac{1}{\sqrt{2017}+\sqrt{2016}}< \dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)

=>\(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}-\sqrt{2015}\)

1) \(A=\left(\sqrt{7-\sqrt{21}+4\sqrt{5}}\right)^2=7-\sqrt{21}+4\sqrt{5}\)

\(B=\left(\sqrt{5}-1\right)^2=6-2\sqrt{5}\)

\(\Rightarrow A-B=1-\sqrt{21}+6\sqrt{5}=\left(1+\sqrt{180}\right)-\sqrt{21}>0\)

\(\Rightarrow A>B\Rightarrow\sqrt{7-\sqrt{21}+4\sqrt{5}}>\sqrt{5}-1\)

2) \(C=\left(\sqrt{5}+\sqrt{10}+1\right)^2=5+10+1+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}\)

\(=26+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}>26+10>35=\left(\sqrt{35}\right)^2\)

Vậy \(\sqrt{5}+\sqrt{10}+1>\sqrt{35}\)

3) \(\left(\frac{15-2\sqrt{10}}{3}\right)^2=\frac{225-60\sqrt{10}+40}{9}=\frac{265-60\sqrt{10}}{9}=\frac{265}{9}-\frac{20\sqrt{10}}{3}< 15\)

Vậy nên \(\frac{15-2\sqrt{10}}{3}< \sqrt{15}\)

b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)

mà 80>75

nên \(4\sqrt{5}>5\sqrt{3}\)

tại sao ko sử dụng máy tính

Vì \(\sqrt{2,5+6,5}\ge0\Rightarrow\sqrt{2,5+6,5}< \sqrt{2,5+6,5}+1\)

\(\sqrt{7}+\sqrt{11}\)\(+\sqrt{32}+\sqrt{40}\) < 18

k mk nha

tại sao v bạn ???