\(\sqrt{12}\)+\(\sqrt{4}\)=
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
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`
AH
Akai Haruma
Giáo viên
20 tháng 7 2020
7.
\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+3+2\sqrt{4.3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{(\sqrt{4}+\sqrt{3})^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10(2+\sqrt{3})}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25+3-2.5\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{(5-\sqrt{3})^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5(5-\sqrt{3})}}=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=3\)
AH
Akai Haruma
Giáo viên
20 tháng 7 2020
5.
\(\sqrt{6+2\sqrt{5}-\sqrt{29+12\sqrt{5}}}=\sqrt{6+2\sqrt{5}-\sqrt{20+9+2\sqrt{20.9}}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{(\sqrt{20}+3)^2}}=\sqrt{6+2\sqrt{5}-(\sqrt{20}+3)}=\sqrt{3}\)
6.
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{\sqrt{49}+\sqrt{40}}\)
\(=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{(2+5+2\sqrt{2.5})+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{2+5+2\sqrt{2.5}}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5})^2+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{(\sqrt{2}+\sqrt{5})^2}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5}+1)^2}-\sqrt{(\sqrt{2}+\sqrt{5})^2}=|\sqrt{2}+\sqrt{5}+1|-|\sqrt{2}+\sqrt{5}|=1\)
12 tháng 10 2023
a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}=4\)
b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3=6\)
25 tháng 9 2021
1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)
H9
HT.Phong (9A5)
CTVHS
17 tháng 7 2023
1) \(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{2^2-2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|\)
\(=2+\sqrt{5}+2-\sqrt{5}\)
\(=4\)
2) \(\sqrt{12-6\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)
\(=\sqrt{3^2-2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{3^2+2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(=\left|3-\sqrt{3}\right|+\left|3+\sqrt{3}\right|\)
\(=3-\sqrt{3}+3+\sqrt{3}\)
\(=6\)
12 tháng 10 2021
c: Ta có: \(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)
27 tháng 7 2021
a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
\(=2\sqrt{5}+2+\sqrt{5}-2\)
\(=3\sqrt{5}\)
b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
\(=3-2\sqrt{2}+2\sqrt{2}-1\)
=2
c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
\(=2-\sqrt{2}+3\sqrt{2}-2\)
\(=2\sqrt{2}\)
27 tháng 7 2021
1: \(\sqrt{3+\sqrt{5}}\cdot\sqrt{2}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
3) \(\left(\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\dfrac{4}{3}}\right)\cdot\sqrt{12}\)
\(=\left(\dfrac{\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{2}+5\cdot\dfrac{2}{\sqrt{3}}\right)\cdot\sqrt{12}\)
\(=\dfrac{17\sqrt{3}}{6}\cdot2\sqrt{3}\)
\(=\dfrac{34\cdot3}{6}=\dfrac{102}{6}=17\)
17 tháng 12 2021
\(a,=4\sqrt{6}-15\sqrt{6}+\sqrt{\left(2+\sqrt{6}\right)^2}=-11\sqrt{6}+2+\sqrt{6}=2-10\sqrt{6}\\ b,=\dfrac{\sqrt{3}\left(\sqrt{6}-2\right)}{\sqrt{6}-2}+\dfrac{4\left(\sqrt{3}-1\right)}{2}+\left|3\sqrt{3}-12\right|=\sqrt{3}+2\sqrt{3}-2+12-3\sqrt{3}=10\)
mình biết nội quy rồi nên đưng đăng nội quy
ai chơi bang bang 2 kết bạn với mình
mình có nick có 54k vàng đang góp mua pika
ai kết bạn mình cho
\(\sqrt{12}+\sqrt{4}=2+2\sqrt{3}\)