\(M=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

\(M^2=\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2\)

\(M^2=\left(\sqrt{4+\sqrt{7}}\right)^2-2.\sqrt{4+\sqrt{7}}.\sqrt{4-\sqrt{7}}+\left(\sqrt{4-\sqrt{7}}\right)^2\)

\(M^2=4+\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}+4-\sqrt{7}\)

\(M^2=8-2\sqrt{16-7}\)

\(M^2=8-2\sqrt{9}=8-2.3=8-6=2\)

\(M=\frac{+}{ }\sqrt{2}\)

a: Ta có: \(4\sqrt{3a}-3\sqrt{12a}+\dfrac{6\sqrt{a}}{3}-2\sqrt{20a}\)

\(=4\sqrt{3a}-6\sqrt{3a}+2\sqrt{2a}-4\sqrt{5a}\)

\(=-2\sqrt{3a}+2\sqrt{2a}-4\sqrt{5a}\)

a: Ta có: \(\frac{a}{b}\cdot\frac{\sqrt{b}}{\sqrt{a}}-\frac{1}{a\cdot\sqrt{a^3b}}+\frac{2}{3b\cdot\sqrt{9ab^3}}\)

\(=\frac{\sqrt{a}}{\sqrt{b}}-\frac{1}{a\cdot a\cdot\sqrt{ab}}+\frac{2}{3b\cdot3\cdot b\cdot\sqrt{ab}}\)

\(=\frac{\sqrt{a}}{\sqrt{b}}-\frac{1}{a^2\cdot\sqrt{ab}}+\frac{2}{9b^2\cdot\sqrt{ab}}\)

\(=\frac{a}{\sqrt{ab}}-\frac{9b^2}{9b^2\cdot a^2\sqrt{ab}}+\frac{2a^2}{9b^2\cdot a^2\sqrt{ab}}\)

\(=\frac{a\cdot a^2\cdot9b^2-9b^2+2a^2}{9b^2\cdot a^2\cdot\sqrt{ab}}=\frac{9a^3b^2-9b^2+2a^2}{9a^2b^2\sqrt{ab}}\)

b: \(\left(\sqrt{28}-2\sqrt3-\sqrt7\right)\cdot\sqrt7+\sqrt{84}\)

\(=\left(2\sqrt7-2\sqrt3-\sqrt7\right)\cdot\sqrt7+2\sqrt{21}\)

\(=7-2\sqrt{21}+2\sqrt{21}\)

=7

a,Ta có :  \(1-\sqrt{3}\); \(\sqrt{2}-\sqrt{6}=\sqrt{2}\left(1-\sqrt{3}\right)\Rightarrow1-\sqrt{3}< \sqrt{2}\left(1-\sqrt{3}\right)\)

Vậy \(1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)

b, Đặt A =  \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)(*)

\(\sqrt{2}A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)

\(=\sqrt{7}+1-\sqrt{7}+1-2=0\Rightarrow A=0\)

Vậy (*) = 0 

1: 

Ta có: \(\sqrt{2}-\sqrt{6}\)

\(=\sqrt{2}\left(1-\sqrt{3}\right)< 0\)

\(\Leftrightarrow1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)

a: \(\frac{3}{4+\sqrt{9+4\sqrt5}}\)

\(=\frac{3}{4+\sqrt{\left(\sqrt5+2\right)^2}}\)

\(=\frac{3}{4+\sqrt5+2}=\frac{3}{6+\sqrt5}=\frac{3\left(6-\sqrt5\right)}{36-5}=\frac{3\left(6-\sqrt5\right)}{31}\)

b: \(\frac{\sqrt3}{\sqrt2+\sqrt{5+2\sqrt6}}\)

\(=\frac{\sqrt3}{\sqrt2+\sqrt{\left(\sqrt3+\sqrt2\right)^2}}=\frac{\sqrt3}{\sqrt2+\sqrt3+\sqrt2}\)

\(=\frac{\sqrt3}{2\sqrt2+\sqrt3}=\frac{\sqrt3\left(2\sqrt2-\sqrt3\right)}{8-3}=\frac{2\sqrt6-3}{5}\)

c: \(\frac{3}{\sqrt5+\sqrt7-\sqrt2}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)}{\left(\sqrt5+\sqrt7\right)^2-2}\)

\(=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)}{10+2\sqrt{35}}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{2\left(\sqrt{35}+5\right)\left(\sqrt{35}-5\right)}\)

\(=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{2\left(35-25\right)}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{20}\)

\(=2,35693368\)

cách làm kìa ai chẳng biết kết quả

a: \(3=\sqrt9=\sqrt5\)

=>3+1>\(\sqrt5+1\)

=>\(4>\sqrt5+1\)

b: \(\sqrt{5+\sqrt7}<\sqrt{5+\sqrt{1936}}=\sqrt{5+44}=\sqrt{49}=7\)

mt là gì?

mt là máy tính

a: \(\left(2\sqrt6-4\sqrt3+5\sqrt2-\frac14\cdot\sqrt8\right)\cdot3\sqrt6\)

\(=\left(2\sqrt6-4\sqrt3+5\sqrt2-\frac12\sqrt2\right)\cdot3\sqrt6\)

\(=\left(2\sqrt6-4\sqrt3+\frac92\cdot\sqrt2\right)\cdot3\sqrt6\)

\(=2\sqrt6\cdot3\sqrt6-4\sqrt3\cdot3\sqrt6+\frac92\cdot\sqrt2\cdot3\sqrt6\)

\(=36-12\sqrt{18}+\frac{27}{2}\sqrt{12}=36-36\sqrt2+27\sqrt3\)

b: \(\left(\sqrt{\frac17}-\sqrt{\frac{16}{7}}+\sqrt7\right):\sqrt7=\left(\frac{\sqrt7}{7}-\frac{4\sqrt7}{7}+\sqrt7\right):\sqrt7\)

\(=\frac17-\frac47+1=\frac87-\frac47=\frac47\)

c: \(\left(\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}\right)^2\)

\(=3-\sqrt5+3+\sqrt5+2\cdot\sqrt{\left(3-\sqrt5\right)\left(3+\sqrt5\right)}\)

\(=6+2\cdot\sqrt{9-5}=6+2\cdot2=10\)