Tính:
A= \(\sqrt{5+\sqrt{3}}+\sqrt{-\sqrt{3}+5}\)
Giúp em với, em đang cần gấp!!!
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Những câu hỏi liên quan
27 tháng 7 2021
Đặt \(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
\(\Leftrightarrow A^3=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\cdot\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(\Leftrightarrow A^3=4+3\cdot\left(-1\right)\cdot A\)
\(\Leftrightarrow A^3=4-3A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A^2+A^2-A+4A-4=0\)
\(\Leftrightarrow A^2\left(A-1\right)+A\left(A-1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
24 tháng 8 2021
`a)sqrt{4+sqrt7}-sqrt{4-sqrt7}`
`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`
`=sqrt{(7+2sqrt7+1)/2}-sqrt{(7-2sqrt7+1)/2}`
`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7-1)^2/2}`
`=(sqrt7+1)/sqrt2-(sqrt7-1)/sqrt2`
`=2/sqrt2=sqrt2`
`b)sqrt{4--sqrt15}-sqrt{4+sqrt15}`
`=sqrt{(8-2sqrt15)/2}-sqrt{(8+2sqrt15)/2}`
`=sqrt{(5-2sqrt{5.3}+3)/2}-sqrt{(5+2sqrt{5.3}+3)/2}`
`=sqrt{(sqrt5-sqrt3)^2/2}-sqrt{(sqrt5+sqrt3)^2/2}`
`=(sqrt5-sqrt3)/sqrt2-(sqrt5+sqrt3)/sqrt2`
`=(-2sqrt3)/sqrt2=-sqrt6`
`c)sqrt{2+sqrt3}+sqrt{2-sqrt3}`
`=sqrt{(4+2sqrt3)/2}+sqrt{(4-2sqrt3)/2}`
`=sqrt{(3+2sqrt3+1)/2}+sqrt{(3-2sqrt3+1)/2}`
`=sqrt{(sqrt3+1)^2/2}+sqrt{(sqrt3-1)^2/2}`
`=(sqrt3+1)/sqrt2+(sqrt3-1)/sqrt2`
`=(2sqrt3)/sqrt2=sqrt6`
`d)sqrt{9+sqrt17}-sqrt{9-sqrt17}`
`=sqrt{(18+2sqrt17)/2}-sqrt{(18-2sqrt17)/2}`
`=sqrt{(17+2sqrt17+1)/2}-sqrt{(17-2sqrt17+1)/2}`
`=sqrt{(sqrt17+1)^2/2}-sqrt{(sqrt17-1)^2/2}`
`=(sqrt17+1)/sqrt2-(sqrt17-1)/sqrt2`
`=2/sqrt2=sqrt2`
25 tháng 8 2021
a: Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\sqrt{2}\)
b: Ta có: \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
24 tháng 8 2021
a: Ta có: \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)
=3
26 tháng 6 2023
Giải
Ta có:
\(x=\sqrt{2+\sqrt{2+\sqrt{3}}-\sqrt{6-3\sqrt{2+\sqrt{3}}}}\)
Khi đó:
\(x^2=\left(\sqrt{2+\sqrt{2+\sqrt{3}}-\sqrt{6-3\sqrt{2+\sqrt{3}}}}\right)^2\\ =2+\sqrt{2+\sqrt{3}}+6-3\sqrt{2+\sqrt{3}}-2\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(6-3\sqrt{2+\sqrt{3}}\right)}\\ =8-2\sqrt{2+\sqrt{3}}-2\sqrt{12-3\left(2+\sqrt{3}\right)}\\ =8-\sqrt{2}.\sqrt{4+2\sqrt{3}}-2\sqrt{6-3\sqrt{3}}\\ =8-\sqrt{2}.\sqrt{4+2\sqrt{3}}-\sqrt{2}.\sqrt{12-6\sqrt{3}}\\ =8-\sqrt{2}.\left(\sqrt{4+2\sqrt{3}}+\sqrt{12-6\sqrt{3}}\right)\\ =8-\sqrt{2}.\left(\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}+1}+\sqrt{9-2.3\sqrt{3}+\left(\sqrt{3}\right)^2}\right)\\ 8-\sqrt{2}.\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(3-\sqrt{3}\right)^2}\right)\\ =8-\sqrt{2}.\left(\sqrt{3}+1+3-\sqrt{3}\right)\\ =8-4\sqrt{2}\\ \Rightarrow x^4-16x^2=\left(8-4\sqrt{2}\right)^2-16.\left(8-4\sqrt{2}\right)\\ =96-64\sqrt{2}-128+64\sqrt{2}=-32\)
Vậy \(S=-32\)
20 tháng 6 2019
\(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}.\left(\sqrt{5}-1\right).\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=\sqrt{9-5}\left(\sqrt{5}-1\right)\sqrt{6+2\sqrt{5}}\)
\(=2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\)
\(=2\left(5-1\right)\)
\(=8\)
16 tháng 2 2022
\(a,A=\left(\dfrac{x+14\sqrt{x}-5}{x-25}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{x+14\sqrt{x}-5+x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{2x+10\sqrt{x}-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}+5\right)-\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}\)
TG
30 tháng 9 2021
\(A=\dfrac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}=\dfrac{x-\sqrt{2}}{x+\sqrt{2}}\)
\(B=\dfrac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\dfrac{1}{x+\sqrt{5}}\)
B
0
8 tháng 5 2019
d) \(\frac{1}{\sqrt{3}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{5}}=\frac{\sqrt{3}+\sqrt{5}}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}-\frac{\sqrt{3}-\sqrt{5}}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}=\frac{\sqrt{3}+\sqrt{5}-\sqrt{3}+\sqrt{5}}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}=\frac{2\sqrt{5}}{3-5}=\frac{2\sqrt{5}}{-2}=-\sqrt{5}\)c) \(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}+\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
b) \(\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}=\sqrt{5+2.\sqrt{5}.2+4}+\sqrt{5-2.\sqrt{5}.2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5}\)a) \(\sqrt{27}+\sqrt{243}-6\sqrt{12}=\sqrt{9.3}+\sqrt{81.3}-6\sqrt{4.3}=3\sqrt{3}+9\sqrt{3}-12\sqrt{3}=0\)
\(A=\sqrt{5+\sqrt{3}}+\sqrt{5-\sqrt{3}}\)
=> \(A^2=\left(\sqrt{5+\sqrt{3}}+\sqrt{5-\sqrt{3}}\right)^2\)
=> \(A^2=5+\sqrt{3}+2\left(5^2-\left(\sqrt{3}\right)^2\right)+5-\sqrt{3}\)
=> \(A^2=10+2.22\)
=> \(A^2=54\)
=> \(A=\sqrt{54}=\sqrt{9.6}=3\sqrt{6}\)