Rút gọn
a, √21+3√48 - √21-3√48
b, √-2√10 - √7 + 2√10
c.ơn các bạn trước :3
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NP
3
27 tháng 6 2021
`a)sqrt{8-2sqrt7}+sqrt{16-6sqrt7}`
`=sqrt{(sqrt7-1)^2}+sqrt{(3-sqrt7)^2}`
`=sqrt7-1+3-sqrt7=2`
`b)sqrt{(sqrt7-1)^2}-sqrt{11+4sqrt7}`
`=sqrt7-1-sqrt{(2+sqrt7)^2}`
`=sqrt7-1-2-sqrt7=-3`
27 tháng 6 2021
a, \(=\sqrt{7-2\sqrt{7}+1}+\sqrt{7-2.3\sqrt{7}+9}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(3-\sqrt{7}\right)^2}=\left|\sqrt{7}-1\right|+\left|3-\sqrt{7}\right|\)
\(=\sqrt{7}-1+3-\sqrt{7}=2\)
\(b,=\left|\sqrt{7}-1\right|-\sqrt{7+2.2\sqrt{7}+4}\)
\(=\left|\sqrt{7}-1\right|-\sqrt{\left(\sqrt{7}+2\right)^2}=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+2\right|\)
\(=\sqrt{7}-1-\sqrt{7}-2=-3\)
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`
NA
9 tháng 2 2022
1:
a: Vì \(\dfrac{-4}{3}=\dfrac{-4\cdot3}{3\cdot3}=\dfrac{-12}{9}=\dfrac{12}{9}\\ \Rightarrow\dfrac{-4}{3}=\dfrac{12}{9}\)
b: Vì : \(-2\cdot3=-6\\ -6\cdot8=-48\)
nên 2 p/s ko bằng nhau
HC
2
Y
20 tháng 6 2023
\(1,\)
\(a,\sqrt{6-2\sqrt{5}}=\sqrt{\sqrt{5^2}-2.\sqrt{5}.1+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)
\(b,\sqrt{8+2\sqrt{7}}=\sqrt{\sqrt{7^2}+2.\sqrt{7}.1+1}=\sqrt{\left(\sqrt{7}+1\right)^2}=\left|\sqrt{7}+1\right|=\sqrt{7}+1\)
\(2,\)
\(a,\sqrt{\left(\sqrt{10}-3\right)^2}-\sqrt{10}\)
\(=\left|\sqrt{10}-3\right|-\sqrt{10}\)
\(=\sqrt{10}-\sqrt{10}-3\)
\(=-3\)
\(b,\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}\)
\(=\left|5+\sqrt{7}\right|-\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(=5+\sqrt{7}-\left|\sqrt{7}-1\right|\)
\(=5+\sqrt{7}-\sqrt{7}+1\)
\(=6\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 10
a; \(\dfrac{9}{27}\) + \(\dfrac{7}{-49}\)
= \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\)
= \(\dfrac{7}{21}\) - \(\dfrac{3}{21}\)
= \(\dfrac{4}{21}\)
b; - \(\dfrac{12}{10}\) + \(\dfrac{-25}{30}\)
= - \(\dfrac{6}{5}\) - \(\dfrac{5}{6}\)
= -\(\dfrac{36}{30}\) - \(\dfrac{25}{30}\)
= \(\dfrac{-61}{30}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 10
c; \(\dfrac{-20}{35}\) + \(\dfrac{-16}{-24}\)
= - \(\dfrac{4}{7}\) + \(\dfrac{2}{3}\)
= - \(\dfrac{12}{21}\) + \(\dfrac{14}{21}\)
= \(\dfrac{2}{21}\)
d; - \(\dfrac{21}{77}\) + \(\dfrac{10}{-35}\)
= - \(\dfrac{3}{11}\) - \(\dfrac{2}{7}\)
= - \(\dfrac{21}{77}\) - \(\dfrac{22}{77}\)
= - \(\dfrac{43}{77}\)
9 tháng 7 2023
b: \(=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}\)
\(=\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)\)
\(=50-10\sqrt{21}+10\sqrt{21}-42=8\)
a: \(A=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}\)
=>\(A^2=\sqrt{2}-1+\sqrt{2}+1+2\sqrt{2-1}=2\sqrt{2}+2\)
=>\(A=\sqrt{2\sqrt{2}+2}\)
Đặt \(B=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}-\sqrt{2+\sqrt{2}}\)
=>\(B=\sqrt{2\sqrt{2}+2}-\sqrt{2+\sqrt{2}}\)
=>\(B^2=2\sqrt{2}+2+2+\sqrt{2}-2\sqrt{\sqrt{2}\left(2+\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)
=>\(B^2=4+3\sqrt{2}-2\sqrt[4]{2}\left(2+\sqrt{2}\right)\)
=>\(B\simeq0,35\)
`a)Đặt \, A=sqrt{21+3sqrt{48}}-sqrt{21-3sqrt{48}}`
Vì `21+3sqrt{48}>21-3sqrt{48}`
`=>sqrt{21+3sqrt{48}}-sqrt{21-3sqrt{48}}>0`
Hay `A>0`
`<=>A^2=21+3sqrt{48}+21-3sqrt{48}-2sqrt{21^2-9.48}`
`<=>A^2=42-2sqrt{9}=32-2.3=26`
`<=>A=sqrt{26}(do \ A>0)`
b)Chắc đề là như này:
`sqrt{7-2sqrt{10}}-sqrt{7+2sqrt{10}}`
`=sqrt{5-2sqrt{5}.sqrt2+2}-sqrt{5+2sqrt{5}.sqrt2+2}`
`=sqrt{(sqrt5-sqrt2)^2}-sqrt{(sqrt5+sqrt2)^2}`
`=sqrt5-sqrt2-sqrt5-sqrt2=-2sqrt2`