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Bài 1
a) Đặt VT = A
<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)
<=> \(2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
<=> \(2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2\)
<=> 2A = \(\left(5-3\right)^2=4\)
<=> A = 2
b) Đặt VT = B
<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)
<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)
<=> B = 8
Bài 2
Đặt VT = A
<=> A2 = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)
<=> A2 = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)
<=> \(A=\sqrt{\sqrt{5}+1}\)
\(f,\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\\ =\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}{4-3}}\\ =\sqrt{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}\\ =\sqrt{\dfrac{\left(6-2\sqrt{5}\right)\left(4+2\sqrt{3}\right)}{4}}\\ =\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{3}+1\right)}{2}\)
\(a,\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\\ =\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\\ =2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\\ =2.4\\ =8\)
`1)A=sqrt{4+sqrt{10+2sqrt5}}+sqrt{4-sqrt{10+2sqrt5}}`
`<=>A^2=4+sqrt{10+2sqrt5}+4-sqrt{10+2sqrt5}+2sqrt{16-10-2sqrt5}`
`<=>A^2=8+2sqrt{6-2sqrt5}`
`<=>A^2=8+2sqrt{(sqrt5-1)^2}`
`<=>A^2=8+2(sqrt5-1)`
`<=>A^2=6+2sqrt5=(sqrt5+1)^2`
`<=>A=sqrt5+1(do \ A>0)`
`b)B=sqrt{35+12sqrt6}-sqrt{35-12sqrt6}`
Vì `35+12sqrt6>35-12sqrt6`
`=>B>0`
`B^2=35+12sqrt6+35-12sqrt6-2sqrt{35^2-(12sqrt6)^2}`
`<=>B^2=70-2sqrt{361}`
`<=>B^2=70-2sqrt{19^2}=70-38=32`
`<=>B=sqrt{32}=4sqrt2(do \ B>0)`
`3)(4+sqrt{15})(sqrt{10}-sqrt6)sqrt{4-sqrt{15}}`
`=sqrt{4+sqrt{15}}.sqrt{4-sqrt{15}}.sqrt{4+sqrt{15}}(sqrt{10}-sqrt6)`
`=sqrt{16-15}.sqrt2(sqrt5-sqrt3).sqrt{4+sqrt{15}}`
`=sqrt{8+2sqrt{15}}(sqrt5-sqrt3)`
`=sqrt{5+2sqrt{5.3}+3}(sqrt5-sqrt3)`
`=sqrt{(sqrt5+sqrt3)^2}(sqrt5-sqrt3)`
`=(sqrt5+sqrt3)(sqrt5-sqrt3)`
`=5-3=2`
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(A=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(A=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\)
\(B=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(B=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(B=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)\)
\(B=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)=2\)
Lời giải:
a)
\(\frac{4}{\sqrt{10}}(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}})=\frac{4}{\sqrt{20}}(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}})\)
\(=\frac{4}{2\sqrt{5}}(\sqrt{5+1+2\sqrt{5}}+\sqrt{5+1-2\sqrt{5}})=\frac{2}{\sqrt{5}}[\sqrt{(\sqrt{5}+1)^2}+\sqrt{(\sqrt{5}-1)^2}]\)
\(=\frac{2}{\sqrt{5}}(\sqrt{5}+1+\sqrt{5}-1)=\frac{2}{\sqrt{5}}.2\sqrt{5}=4\)
b)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{3+5-2\sqrt{3.5}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})\)
\(=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)
c)
\(=\sqrt{4\sqrt{2}(\sqrt{3}+1)+8\sqrt{3}+18}=\sqrt{4\sqrt{2}(\sqrt{3}+1)+4(3+1+2\sqrt{3})+2}\)
\(=\sqrt{4\sqrt{2}(\sqrt{3}+1)+4(\sqrt{3}+1)^2+2}\)
\(=\sqrt{(2\sqrt{3}+2)^2+(\sqrt{2})^2+2.(2\sqrt{3}+2).\sqrt{2}}\)
\(=\sqrt{(2\sqrt{3}+2+\sqrt{2})^2}=2\sqrt{3}+2+\sqrt{2}\)
a, \(VT=\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{20}-2\right)}{2}\)
\(=\frac{\sqrt{5-2\sqrt{5}+1}\left(3+\sqrt{5}\right)\left(2\sqrt{5}-2\right)}{2}\)
\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)2\left(\sqrt{5}-1\right)}{2}\)
\(=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=18-6\sqrt{5}+6\sqrt{5}-10=8=VP\)
b, \(VT=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2\sqrt{5}\sqrt{3}+3}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)
\(=2\left(16-15\right)=2=VP\)