Cho a =\(\sqrt{3+\sqrt{5}}\), b = \(\sqrt{3-\sqrt{5}}\).
a, Tính A = a.b
b, Tính B = a2 + b2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
H
Hoaa
5 tháng 7 2020
Đúng(0)
Những câu hỏi liên quan
LL
1
6 tháng 10 2020
Ta có:
1) \(A=a\cdot b=\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\sqrt{9-5}=\sqrt{4}=2\)
2) \(B=a^2+b^2=\left(\sqrt{3+\sqrt{5}}\right)^2+\left(\sqrt{3-\sqrt{5}}\right)^2\)
\(=3+\sqrt{5}+3-\sqrt{5}=6\)
3) Xét: \(\left(a+b\right)^2=a^2+2ab+b^2=10\)
\(\Rightarrow a+b=\sqrt{10}\)
\(C=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=\sqrt{10}\cdot\left(6-2\right)\)
\(=4\sqrt{10}\)
4) \(a^5+b^5=\left(a+b\right)^5-\left(5a^4b+10a^3b^2+10a^2b^3+5ab^4\right)\)
\(=\left(\sqrt{10}\right)^5-5ab\left(a^3+b^3\right)-10a^2b^2\left(a+b\right)\)
\(=100\sqrt{10}-5\cdot2\cdot4\sqrt{10}-10\cdot2^2\cdot\sqrt{10}\)
\(=100\sqrt{10}-40\sqrt{10}-40\sqrt{10}\)
\(=20\sqrt{10}\)
17 tháng 8 2016
Ta có : \(\frac{23\sqrt{2}}{\sqrt{2}+\sqrt{14+5\sqrt{3}}}=\frac{46}{2+\sqrt{28+10\sqrt{3}}}=\frac{46}{2+\sqrt{\left(\sqrt{3}+5\right)^2}}=\frac{46}{7+\sqrt{3}}\)
\(=\frac{46\left(7-\sqrt{3}\right)}{\left(7+\sqrt{3}\right)\left(7-\sqrt{3}\right)}=\frac{46\left(7-\sqrt{3}\right)}{46}=7-\sqrt{3}\)
Suy ra a = 7 , b = -1
=> a x b = -7
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 8 2021
\(\dfrac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}=\dfrac{9\sqrt{5}+9\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\dfrac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)
b.
\(=\sqrt{3-\sqrt{5}}.\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}+\sqrt{3+\sqrt{5}}.\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(=\sqrt{3-\sqrt{5}}.\sqrt{9-5}+\sqrt{3+\sqrt{5}}.\sqrt{9-5}\)
\(=\sqrt{12-4\sqrt{5}}+\sqrt{12+4\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{10}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{10}+\sqrt{2}\right)^2}\)
\(=\sqrt{10}-\sqrt{2}+\sqrt{10}+\sqrt{2}=2\sqrt{10}\)
c.
\(\dfrac{a-\sqrt{b}}{\sqrt{b}}:\dfrac{\sqrt{b}}{a+\sqrt{b}}=\dfrac{\left(a-\sqrt{b}\right)\left(a+\sqrt{b}\right)}{\sqrt{b}.\sqrt{b}}=\dfrac{a^2-b}{b}\)
5 tháng 4 2022
1: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-8}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-5\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-4}{\sqrt{x}}\)
2: \(P=A\cdot B=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
\(\Leftrightarrow P-2=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}>0\)
=>P>2
25 tháng 6 2021
a) Ta có: \(A=\sqrt{\sqrt{3}+\sqrt{2}}\cdot\sqrt{\sqrt{3}-\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\sqrt{3-2}=1\)
b) Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{3}\)
25 tháng 6 2021
`A=sqrt{sqrt3+sqrt2}.sqrt{sqrt3-sqrt2}`
`=sqrt{(sqrt3+sqrt2)(sqrt3-sqrt2)}`
`=sqrt{3-2}=1`
`b)B=sqrt{5-2sqrt6}+sqrt{5+2sqrt6}`
`=sqrt{3-2sqrt6+2}+sqrt{3+2sqrt6+2}`
`=sqrt{(sqrt3-sqrt2)^2}+sqrt{(sqrt3+sqrt2)^2}`
`=sqrt3-sqrt2+sqrt3+sqrt2=2sqrt3`
`c)C=3-sqrt{3-sqrt5}`
`=3-sqrt{(6-2sqrt5)/2}`
`=3-sqrt{(sqrt5-1)^2/2}`
`=3-(sqrt5-1)/sqrt2`
`=3-(sqrt{10}-sqrt2)/2`
`=(6-sqrt{10}+sqrt2)/2`