Cho \(a=\sqrt{3+\sqrt{5}}b=\sqrt{3-\sqrt{5}}\)
Tính\(A=a.b\)
\(B=a^2+b^2\)
\(C=a^3+b^3\)
\(D=a^5+b^5\)
K
Khách
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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`
25 tháng 9 2021
a)A=\(2\sqrt{3}-8\sqrt{3}+7\sqrt{3}=\sqrt{3}\)
b)B\(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}=3-\sqrt{5}+\sqrt{5}-2=1\)
d)\(=\dfrac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)}{1}+1-\sqrt{5}-\dfrac{11\left(2\sqrt{5}-3\right)}{11}=5\sqrt{5}+5-10-2\sqrt{5}+1-\sqrt{5}-2\sqrt{5}+3=-1\)
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.
\(\dfrac{a-\sqrt{b}}{\sqrt{b}}:\dfrac{\sqrt{b}}{a+\sqrt{b}}=\dfrac{\left(a-\sqrt{b}\right)\left(a+\sqrt{b}\right)}{\sqrt{b}.\sqrt{b}}=\dfrac{a^2-b}{b}\)
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
7 tháng 7 2023
a: Sửa đề: \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}=4\)
b: \(A=2-\sqrt{3}+\sqrt{3}-1=1\)
c: \(C=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
d: Sửa đề: \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3\)
=6
H9
HT.Phong (9A5)
CTVHS
7 tháng 7 2023
a) \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(A=\left|4-\sqrt{15}\right|+\sqrt{15}\)
\(A=4-\sqrt{15}+\sqrt{15}\)
\(A=4\)
b) \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)}\)
\(B=\left|2-\sqrt{3}\right|+\left|1-\sqrt{3}\right|\)
\(B=2-\sqrt{3}-1+\sqrt{3}\)
\(B=1\)
c) \(C=\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(C=\sqrt{\left(3\sqrt{5}\right)^2-2\cdot3\sqrt{15}\cdot2+2^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\sqrt{5}\cdot2+2^2}\)
\(C=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(C=\left|3\sqrt{5}-2\right|-\left|3\sqrt{5}+2\right|\)
\(C=3\sqrt{5}-2-3\sqrt{5}-2\)
\(C=-4\)
d) \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(D=\sqrt{\left(2\sqrt{5}\right)^2+2\cdot2\sqrt{5}\cdot3+3^2}-\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot3+3^3}\)
\(D=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(D=\left|2\sqrt{5}+3\right|-\left|2\sqrt{5}-3\right|\)
\(D=2\sqrt{5}+3-2\sqrt{5}+3\)
\(D=6\)
NK
24 tháng 3 2019
Đề này nằm trong đề ôn ấy mà,nó ghi sao mình viết lại vậy thôi.:) Đừng hiểu nhầm nhé!
7 tháng 7 2021
a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)
\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3\cdot A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 8 2023
Bài 1:
a) \(\sqrt{1,44\cdot1,21-1,44\cdot0,4}\)
\(=\sqrt{1,44\cdot\left(1,21-0,4\right)}\)
\(=\sqrt{1,44\cdot0,81}\)
\(=\sqrt{1,44}\cdot\sqrt{0,81}\)
\(=1,2\cdot0,9\)
\(=1,08\)
b) \(\dfrac{\sqrt{5}-2}{\sqrt{5}+2}+\sqrt{80}\)
\(=\dfrac{\left(\sqrt{5}-2\right)^2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+4\sqrt{5}\)
\(=\dfrac{5-4\sqrt{5}+4}{1}+4\sqrt{5}\)
\(=9-4\sqrt{5}+4\sqrt{5}\)
\(=9\)
c) \(\sqrt[3]{16}+\sqrt[3]{2}\left(\sqrt[3]{4}-\sqrt[3]{2}\right)\)
\(=\sqrt[3]{2^3\cdot2}+\sqrt[3]{2\cdot4}-\sqrt[3]{2\cdot2}\)
\(=2\sqrt[3]{2}+\sqrt[3]{8}-\sqrt[3]{4}\)
\(=2\sqrt[3]{2}+2-\sqrt[3]{4}\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 8 2023
Bài 2: Ta có:
\(VT=\dfrac{1}{\sqrt{a}-\sqrt{b}}:\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}:\dfrac{\sqrt{ab}\cdot\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\cdot\dfrac{1}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{\left(a-b\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{1}{a-b}=VP\left(dpcm\right)\)
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) \(a^5+b^5=\left(a+b\right)^5-\left(5a^4b+10a^3b^2+10a^2b^3+5ab^4\right)\)
\(=\left(\sqrt{10}\right)^5-5ab\left(a^3+b^3\right)-10a^2b^2\left(a+b\right)\)
\(=100\sqrt{10}-5\cdot2\cdot4\sqrt{10}-10\cdot2^2\cdot\sqrt{10}\)
\(=100\sqrt{10}-40\sqrt{10}-40\sqrt{10}\)
\(=20\sqrt{10}\)