Ta có : \(\sqrt{61-35}=\sqrt{26}>\sqrt{25}=5\)(1)

           \(\sqrt{61}-\sqrt{35}< \sqrt{64}-\sqrt{36}=8-6=2\)(2)

Từ (1) và (2) ta được :  \(\sqrt{61-35}>5>2>\sqrt{61}-\sqrt{35}\)

\(\Rightarrow\sqrt{61-35}>\sqrt{61}-\sqrt{35}\)

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  \(\Leftrightarrow\)\(a=b\)

\(\sqrt{10}+\sqrt{17}+1>\sqrt{9}+\sqrt{16}+1=3+4+1=8\)
\(\sqrt{61}< \sqrt{64}=8\)
Vậy \(\sqrt{10}+\sqrt{17}+1>\sqrt{61}\)

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 \(4\sqrt{5}>5\sqrt{3}\)

Có:\(\sqrt{48}< \sqrt{49}=7\)

\(13-\sqrt{35}>13-\sqrt{36}=7\)

\(\Rightarrow\sqrt{48}< 13-\sqrt{35}\)

\(\sqrt{48}+\sqrt{35}< \sqrt{49}+\sqrt{36}=7+6=13\)

\(\rightarrow\sqrt{48}< 13-\sqrt{35}\)

a) <

b) <

c) >

d) <

      a <

            b <

                           c >

                   d <

\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)

\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)

\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)

Áp dụng t/c dtsbn:

\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)

\(\sqrt{27}\) +  \(\sqrt{6}\)>  \(\sqrt{35}\)

căn 27 + căn 6 = 7,196156423

căn 35 = 5,916079783

=>căn 27 + căn 6 > căn 35