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

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

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

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

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

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

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

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: \(\sqrt{2}-\sqrt{6}\)

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

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

\(=2,35693368\)

cách làm kìa ai chẳng biết kết quả

mt là gì?

mt là máy tính