a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)

\(=4-\sqrt{15}+\sqrt{15}=4\)

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

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

\(=2\sqrt{3}\)

c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

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

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

\(A=\sqrt{8-2\sqrt{15}}=\sqrt{5-2\sqrt{15}+3}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\)

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

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

\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}-\left|4-\sqrt{7}\right|\)

\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=\sqrt{16-7}-4+\sqrt{7}\)

\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=3-4+\sqrt{7}=-1+\sqrt{7}\)

\(\Leftrightarrow B=\frac{-1+\sqrt{7}}{\sqrt{4-\sqrt{7}}}\)

tíck mình nha bn thanks !!!!!!!!!!

cảm ơn b nhìu nha mik k giùm b rr đó

mik k cho b rr đó

\(M=\dfrac{8\left(\sqrt{5}+\sqrt{3}\right)}{2}-\dfrac{7\left(2+\sqrt{3}\right)}{4-3}+\dfrac{4\left(\sqrt{2}+1\right)}{2-1}+\dfrac{\sqrt{15}\left(\sqrt{3}-1\right)}{\sqrt{15}}\)

\(=4\left(\sqrt{5}+\sqrt{3}\right)-14-7\sqrt{3}+4\sqrt{2}+4+\sqrt{3}-1\)

\(=4\sqrt{5}+4\sqrt{3}-6\sqrt{3}+4\sqrt{2}-11\)

\(=4\sqrt{5}-2\sqrt{3}+4\sqrt{2}-11\)

\(M=\dfrac{8\left(\sqrt{5}+\sqrt{3}\right)}{5-3}+\dfrac{7\left(\sqrt{3}+2\right)}{3-4}+\dfrac{4\left(\sqrt{2}+1\right)}{2-1}+\dfrac{\sqrt{15}\left(\sqrt{3}-1\right)}{\sqrt{15}}\)

\(=4\sqrt{5}+4\sqrt{3}-7\sqrt{3}-14+4\sqrt{2}+4+\sqrt{3}-1\)

\(=4\sqrt{5}-2\sqrt{3}+4\sqrt{2}-11\)

\(D=\left(\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}-\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right).\frac{\sqrt{3}-1}{3-\sqrt{3}}=\left(\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{-1}-\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{-1}\right).\frac{\sqrt{3}-1}{\sqrt{3}\left(\sqrt{3}-1\right)}\)

\(=\left(-5-2\sqrt{6}+5-2\sqrt{6}\right).\frac{1}{\sqrt{3}}=\frac{-4\sqrt{6}}{\sqrt{3}}=-4\sqrt{2}\)

\(E=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}=\frac{\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)

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

\(=\sqrt{2}\)

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

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

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

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

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

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

Trả lời:

\(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(A^2=\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)

\(A^2=4+\sqrt{10+2\sqrt{5}}+2.\sqrt{4+\sqrt{10+2\sqrt{5}}}.\sqrt{4-\sqrt{10+2\sqrt{5}}}+4-\sqrt{10+2\sqrt{5}}\)

\(A^2=8+2\sqrt{16-10-2\sqrt{5}}\)

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

\(A^2=8+2\sqrt{5-2\sqrt{5}+1}\)

\(A^2=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(A^2=8+2.\left(\sqrt{5}+1\right)\)

\(A^2=8+2\sqrt{5}-2\)

\(A^2=6+2\sqrt{5}\)

\(A^2=5+2\sqrt{5}+1\)

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

\(A=\sqrt{5}+1\)

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

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

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

\(\sqrt{2}B=\sqrt{5+2\sqrt{15}+3}+\sqrt{5-2\sqrt{15}+3}-2\sqrt{5-2\sqrt{5}+1}\)

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

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

\(\sqrt{2}B=2\)

\(B=\sqrt{2}\)

Cảm ơn bạn nhiều nha UvU 

`2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{6-2\sqrt5}}`

`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{(\sqrt5-1)^2}}`

`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt5-1}`

`=`=2(\sqrt{10}-\sqrt2)\sqrt{3+\sqrt5)`

`=2\sqrt2(\sqrt5-1)\sqrt{3+\sqrt5}`

`=2(\sqrt5-1)sqrt{6+2\sqrt5}`

`=2(\sqrt5-1)(\sqrt5+1)`

`=2(5-1)`

`=8`

`2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{6-2\sqrt5}}`

`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{(\sqrt5-1)^2}}`

`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt5-1}`

`=2(\sqrt{10}-\sqrt2)\sqrt{3+\sqrt5)`

`=2\sqrt2(\sqrt5-1)\sqrt{3+\sqrt5}`

`=2(\sqrt5-1)sqrt{6+2\sqrt5}`

`=2(\sqrt5-1)(\sqrt5+1)`

`=2(5-1)`

`=8`

`(4\sqrt2+\sqrt{30})(\sqrt5-\sqrt3)\sqrt{4-\sqrt{15}}`

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

`=(4+\sqrt{15})(\sqrt5-\sqrt3)\sqrt{8-2\sqrt{15}}`

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

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

`=2(4+\sqrt{15})(4-\sqrt{15})`

`=2(16-15)`

`=2`

a) \(2\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{6-2\sqrt{5}}}\)

\(=2\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{5}-1}\)

\(=\dfrac{2\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{5}+1\right)}{\sqrt{2}}\)

\(=2\cdot4=8\)

\(=2\sqrt{80\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{5\sqrt{3}}-4\sqrt{45\sqrt{3}}\)

\(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-12\sqrt{5\sqrt{3}}\)

=0

`c)1/(2sqrt2)-3/2sqrt{4,5}+2/5sqrt{50}`

`=1/(2sqrt2)-3/2sqrt{9/2}+2/5sqrt{25.2}`

`=1/(2sqrt2)-9/(2sqrt2)+2sqrt2`

`=2sqrt2-8/(2sqrt2)`

`=2sqrt2-sqrt2=sqrt2`

`d)4/(3+sqrt5)-8/(1+sqrt5)+15/sqrt5`

`=(4(3-sqrt5))/(9-5)-(8(sqrt5-1))/(5-1)+3sqrt5`

`=3-sqrt5-2(sqrt5-1)+3sqrt5`

`=3+3sqrt5-3sqrt5+2=5`