\(\left(3+\sqrt{5}\right).\sqrt{4-6\sqrt{5}}\)
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20 tháng 6 2021
`a)A=(3-sqrt5)sqrt{3+sqrt5}+(3+sqrt5)sqrt{3-sqrt5}`
`=sqrt{3-sqrt5}sqrt{3+sqrt5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`
`=sqrt{9-5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`
`=2(sqrt{3+sqrt5}+sqrt{3-sqrt5})`
`=sqrt2(sqrt{6+2sqrt5}+sqrt{6-2sqrt5})`
`=sqrt2(sqrt{(sqrt5+1)^2}+sqrt{(sqrt5+1)^2})`
`=sqrt2(sqrt5+1+sqrt5-1)`
`=sqrt{2}.2sqrt5`
`=2sqrt{10}`
20 tháng 6 2021
`b)B=(5+sqrt{21})(sqrt{14}-sqrt6)sqrt{5-sqrt{21}}`
`=sqrt{5+sqrt{21}}sqrt{5-sqrt{21}}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`
`=sqrt{25-21}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`
`=2sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`
`=2sqrt2sqrt{5+sqrt{21}}(sqrt{7}-sqrt3)`
`=2sqrt{10+2sqrt{21}}(sqrt{7}-sqrt3)`
`=2sqrt{(sqrt3+sqrt7)^2}(sqrt{7}-sqrt3)`
`=2(sqrt3+sqrt7)(sqrt{7}-sqrt3)`
`=2(7-3)`
`=8`
`c)C=sqrt{4+sqrt7}-sqrt{4-sqrt7}`
`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`
`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7+1)^2/2}`
`=(sqrt7+1)/sqrt2-(sqrt7-1)/2`
`=2/sqrt2=sqrt2`
H9
HT.Phong (9A5)
CTVHS
1 tháng 10 2023
\(A=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)
\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)
\(A=2^2-\left(\sqrt{5}\right)^2\)
\(A=4-5\)
\(A=-1\)
____
\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(B=6-121\)
\(B=-115\)
23 tháng 7 2021
e) Ta có: \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2}+1\)
=2
24 tháng 9 2021
\(a,=2\sqrt{6}-4+\sqrt{\left(3-\sqrt{6}\right)^2}=2\sqrt{6}-4+3-\sqrt{6}=\sqrt{6}-1\\ b,=3-2\sqrt{2}+\sqrt{\left(3\sqrt{2}+1\right)^2}=3-2\sqrt{2}+3\sqrt{2}+1=4+\sqrt{2}\\ c,=\sqrt{\left(\sqrt{5}+2\right)^2}-\left(\sqrt{5}-1\right)=\sqrt{5}+2-\sqrt{5}+1=3\)
24 tháng 9 2021
a) \(=2\sqrt{6}-4+\sqrt{\left(3-\sqrt{6}\right)^2}=2\sqrt{6}-4+3-\sqrt{6}=-1+\sqrt{6}\)
b) \(=\left|3-2\sqrt{2}\right|+\sqrt{\left(3\sqrt{2}+1\right)^2}=3-2\sqrt{2}+3\sqrt{2}+1=4+\sqrt{2}\)
c) \(=\sqrt{\left(\sqrt{5}+2\right)^2}-\left|1-\sqrt{5}\right|=\sqrt{5}+2+1-\sqrt{5}=3\)
BL
12 tháng 11 2017
\(f,\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\\ =\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}{4-3}}\\ =\sqrt{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}\\ =\sqrt{\dfrac{\left(6-2\sqrt{5}\right)\left(4+2\sqrt{3}\right)}{4}}\\ =\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{3}+1\right)}{2}\)
BL
12 tháng 11 2017
\(a,\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\\ =\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\\ =2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\\ =2.4\\ =8\)
21 tháng 12 2023
Bài 1:
a: \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
\(=5\cdot2\sqrt{2}-4\cdot3\sqrt{3}-2\cdot5\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-16\sqrt{3}\)
b: \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|1-\sqrt{6}\right|\)
\(=3-\sqrt{6}+\sqrt{6}-1\)
=3-1=2
c: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\dfrac{1\left(4-\sqrt{15}\right)}{16-15}\)
\(=\sqrt{15}+4-\sqrt{15}=4\)
d: \(\dfrac{2\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
\(=\dfrac{\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\dfrac{\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=3+\sqrt{5}-\dfrac{\sqrt{5}}{2}=3+\dfrac{\sqrt{5}}{2}\)
Bài 2:
Vẽ đồ thị:
Phương trình hoành độ giao điểm là:
\(\dfrac{1}{2}x-4=-3x+3\)
=>\(\dfrac{1}{2}x+3x=3+4\)
=>\(\dfrac{7}{2}x=7\)
=>x=2
Thay x=2 vào y=-3x+3, ta được:
\(y=-3\cdot2+3=-3\)
Vậy: (d1) cắt (d2) tại A(2;-3)