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AT
29 tháng 7 2021
câu đầu có \(3-12\sqrt{6}< 0\) nên không căn được nên đề bạn sai
\(\sqrt{31-8\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{4^2-2.4.\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2.\sqrt{15}.3+3^2}\)
\(=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}=\left|4-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)
\(=4-\sqrt{15}+\sqrt{15}-3=1\)
\(\sqrt{49-5\sqrt{96}}-\sqrt{49+5\sqrt{96}}=\sqrt{49-20\sqrt{6}}-\sqrt{49+20\sqrt{6}}\)
\(=\sqrt{5^2-2.5.2\sqrt{6}+\left(2\sqrt{6}\right)^2}-\sqrt{5^2+2.5.4\sqrt{6}+\left(2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(5+2\sqrt{6}\right)^2}=\left|5-2\sqrt{6}\right|-\left|5+2\sqrt{6}\right|\)
\(=5-2\sqrt{6}-5-2\sqrt{6}=-4\sqrt{6}\)
29 tháng 7 2021
\(\sqrt{31-8\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=4-\sqrt{15}+\sqrt{15}-3\)
=1
CD
0
NT
1
DH
23 tháng 6 2017
\(\dfrac{\sqrt{\sqrt{15}-\sqrt{6}}+\sqrt{\sqrt{15}+\sqrt{6}}}{\sqrt{\sqrt{15}+3}}\)
\(=\dfrac{\sqrt{\sqrt{3}.\left(\sqrt{5}-\sqrt{3}\right)}+\sqrt{\sqrt{3}.\left(\sqrt{5}+\sqrt{3}\right)}}{\sqrt{\sqrt{3}.\left(\sqrt{5}+\sqrt{3}\right)}}\)
\(=\dfrac{\sqrt{\sqrt{3}}.\left[\left(\sqrt{5}-\sqrt{3}\right)+\left(\sqrt{5}+\sqrt{3}\right)\right]}{\sqrt{\sqrt{3}.\left(\sqrt{5}+\sqrt{3}\right)}}\)
\(=\dfrac{\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\dfrac{2\sqrt{5}}{\sqrt{5}+\sqrt{3}}\)
Chúc bạn học tốt!!!
H9
HT.Phong (9A5)
CTVHS
24 tháng 8 2023
a) \(\sqrt{6-\sqrt{11}}\cdot\sqrt{6+\sqrt{11}}\)
\(=\sqrt{\left(6-\sqrt{11}\right)\left(6+\sqrt{11}\right)}\)
\(=\sqrt{6^2-\left(\sqrt{11}\right)^2}\)
\(=\sqrt{36-11}\)
\(=\sqrt{25}\)
\(=\sqrt{5^2}\)
\(=5\)
b) \(\sqrt{8+\sqrt{15}}\cdot\sqrt{8-\sqrt{15}}\)
\(=\sqrt{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}\)
\(=\sqrt{8^2-\left(\sqrt{15}\right)^2}\)
\(=\sqrt{64-15}\)
\(=\sqrt{49}\)
\(=\sqrt{7^2}\)
\(=7\)
24 tháng 8 2023
a: \(=\sqrt{6^2-11}=\sqrt{25}=5\)
b: \(=\sqrt{8^2-15}=\sqrt{49}=7\)
Y
26 tháng 6 2023
\(a,\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\\ =\sqrt{\sqrt{5^2}+2\sqrt{5}.\sqrt{3}+\sqrt{3^2}}-\sqrt{\sqrt{5^2}-2\sqrt{5}.\sqrt{3}+\sqrt{3^2}}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\\ =\left|\sqrt{5}+\sqrt{3}\right|-\left|\sqrt{5}-\sqrt{3}\right|\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}\\ =2\sqrt{3}\)
\(b,\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}\\ =\sqrt{\sqrt{2^2}+2.\sqrt{3}.\sqrt{2}+\sqrt{3^2}}+\sqrt{\sqrt{2^2}-2.\sqrt{3}.\sqrt{2}+\sqrt{3^2}}\\ =\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\\ =\left|\sqrt{2}+\sqrt{3}\right|+\left|\sqrt{2}-\sqrt{3}\right|\\ =\sqrt{2}+\sqrt{3}-\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
H9
HT.Phong (9A5)
CTVHS
26 tháng 6 2023
a) \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
\(=\sqrt{5-2\cdot\sqrt{5\cdot3}+3}-\sqrt{5+2\cdot\sqrt{5\cdot3}+1}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
\(=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}\)
\(=-2\sqrt{3}\)
b. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}-\sqrt{3-2\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{2}\)
26 tháng 6 2021
`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`
HT
1
6 tháng 11 2021
\(=\sqrt{\dfrac{\left(5-2\sqrt{6}\right)^2}{25-24}+\sqrt{\left(3-\sqrt{6}\right)^2}}=\sqrt{25+24-20\sqrt{6}+3-\sqrt{6}}=\sqrt{52-21\sqrt{6}}\)
=\(\sqrt{9-2.3\sqrt{6}+6}+\sqrt{9+2.3\sqrt{6}+6}\)
=\(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3+\sqrt{6}\right)^2}\)
=\(\left|3-\sqrt{6}\right|+\left|3+\sqrt{6}\right|\)
=\(3-\sqrt{6}+3+\sqrt{6}\)
=6