\(\sqrt{4}=2\)

7=2+5

5=\(\sqrt{25}\)

\(\sqrt{25}>\sqrt{5}\)

=>\(\sqrt{4}+\sqrt{5}>7\)

\(7=2+5=\sqrt{4}+\sqrt{25}.\)

Ta có : \(25>5\Rightarrow\sqrt{25}>\sqrt{5}\Rightarrow\sqrt{4}+\sqrt{25}>\sqrt{4}+\sqrt{5}\)

Vậy : \(\sqrt{4}+\sqrt{5}< 7\)

\(\left(5-2\sqrt{7}\right)^2=53-20\sqrt{7}=19+34-20\sqrt{7}\)

\(\left(3-\sqrt{10}\right)^2=19-6\sqrt{10}\)

mà \(34-20\sqrt{7}>-6\sqrt{10}\)

nên \(5-2\sqrt{7}>3-\sqrt{10}\)

\(5\sqrt{3}-4< 3\sqrt{5}\)

1/ bình phương hai vế được (căn11)^2+(căn5)^2=11+5   4^2=16 vậy căn 11+căn 5=4

2/ tương tự (3 căn3 )^2=27   (căn19)^2-(căn 2)^2=19-2=17  vậy 3 căn 3 >căn 19-căn2

\(\left(\sqrt{5}+\sqrt{7}\right)^2=12+2\sqrt{35}>12=\left(\sqrt{12}\right)^2\\ \Rightarrow\sqrt{5}+\sqrt{7}>\sqrt{12}\)

\(\sqrt{5}+\sqrt{7}\) và \(\sqrt{12}\)

Giả sử: \(\sqrt{5}+\sqrt{7}>\sqrt{12}\)

=> \(\left(\sqrt{5}+\sqrt{7}\right)^2>\left(\sqrt{12}\right)^2\)

<=> \(5+2\sqrt{35}+7>12\)

<=> \(12+2\sqrt{35}>12\) (thỏa mãn giả sử)

Vậy \(\sqrt{5}+\sqrt{7}>\sqrt{12}\)

b: \(\dfrac{3}{\sqrt{7}-2}-\dfrac{4}{\sqrt{7}+\sqrt{3}}\)

\(=\sqrt{7}+2-\sqrt{7}+\sqrt{3}=2+\sqrt{3}\)