Lời giải:

Nếu .... là vô hạn thì:

$M=\sqrt{15-2M}$

$\Rightarrow M^2=15-2M$

$\Leftrightarrow M^2+2M-15=0$

$\Leftrightarrow (M-3)(M+5)=0$

$\Leftrightarrow M=3$ (do $M>0$)

C=√15+√15+....

C^2=15+√15+√15....=15+C.Ta có phương trình :

C^2-C-15=0.Sau đó giải ra C=(1+√61)/2(do C>0)

a) \(\sqrt{6-\sqrt{11}}\cdot\sqrt{6+\sqrt{11}}\)

\(=\sqrt{\left(6-\sqrt{11}\right)\left(6+\sqrt{11}\right)}\)

\(=\sqrt{6^2-\left(\sqrt{11}\right)^2}\)

\(=\sqrt{36-11}\)

\(=\sqrt{25}\)

\(=\sqrt{5^2}\)

\(=5\)

b) \(\sqrt{8+\sqrt{15}}\cdot\sqrt{8-\sqrt{15}}\)

\(=\sqrt{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}\)

\(=\sqrt{8^2-\left(\sqrt{15}\right)^2}\)

\(=\sqrt{64-15}\)

\(=\sqrt{49}\)

\(=\sqrt{7^2}\)

\(=7\)

a: \(=\sqrt{6^2-11}=\sqrt{25}=5\)

b: \(=\sqrt{8^2-15}=\sqrt{49}=7\)

\(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}=\sqrt{\left(\sqrt{\dfrac{5}{2}}-\sqrt{\dfrac{3}{2}}\right)^2}-\sqrt{\left(\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{3}{2}}\right)^2}=\sqrt{\dfrac{5}{2}}-\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{5}{2}}-\sqrt{\dfrac{3}{2}}=-2\sqrt{\dfrac{3}{2}}=-\sqrt{6}\)

\(B=\sqrt{\dfrac{8-2\sqrt{15}}{2}}-\sqrt{\dfrac{8+2\sqrt{15}}{2}}\\ B=\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^2}{2}}\\ B=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\\ B=\dfrac{-2\sqrt{3}}{\sqrt{2}}=\dfrac{-2\sqrt{6}}{2}=-\sqrt{6}\)

$(4+\sqrt{15})(\sqrt{10}-\sqrt6)\sqrt{4-\sqrt{15}}$

$=\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}.(\sqrt{10}-\sqrt6)\sqrt{4-\sqrt{15}}$

$=(\sqrt{10}-\sqrt6)\sqrt{4+\sqrt{15}}\sqrt{16-15}$

$=\sqrt2(\sqrt5-\sqrt3)\sqrt{4+\sqrt{15}}$

$=(\sqrt5-\sqrt3)\sqrt{8+2\sqrt{15}}$

$=(\sqrt5-\sqrt3)\sqrt{5+2\sqrt{5}.\sqrt3+3}$

$=(\sqrt5-\sqrt3)\sqrt{(\sqrt5+\sqrt3)^2}$

$=(\sqrt5-\sqrt3)(\sqrt5+\sqrt3)=5-3=2$

\(B=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\\ B=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\\ B=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\\ B=2\)

\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\sqrt{4+\sqrt{15}}.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{16-15}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}.\sqrt{5}+\left(\sqrt{5}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}.\left(\sqrt{5}-\sqrt{3}\right)=\left|\sqrt{5}+\sqrt{3}\right|\left(\sqrt{5}-\sqrt{3}\right)=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)

Ta có: \(B=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

=2

\(=\sqrt{4+\sqrt{15}}\left(\sqrt{4+\sqrt{15}}\cdot\sqrt{4-\sqrt{15}}\right)\left(\sqrt{10}-\sqrt{6}\right)\\ =\sqrt{4+\sqrt{15}}\left(16-15\right)\left(\sqrt{10}-\sqrt{6}\right)\\ =\sqrt{2\left(4+\sqrt{15}\right)}\left(\sqrt{5}-\sqrt{3}\right)\\ =\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\\ =\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)

(\(\sqrt{8-2\sqrt{15}}\)+ \(\sqrt{8+2\sqrt{15}}\)-  \(2\sqrt{6-2\sqrt{5}}\))/2

= (\(\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)+ \(\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)-  \(2\sqrt{\left(\sqrt{5}-1\right)^2}\))/2

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

= 2/2 = 1

bài của   TuanMinhAms  sai nha

\(A=\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)

\(\Rightarrow\)\(\sqrt{2}A=\sqrt{8-2\sqrt{15}}+\sqrt{8+2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)

                       \(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}\)

                       \(=\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}-2\left(\sqrt{5}-1\right)=2\)

\(\Rightarrow\)\(A=\sqrt{2}\)