1. Rút Gọn
a)√6-2√5
b)√8+2√7
2 Tính
a) √(√10-3)2 -√10
b)√(5+√7)2 - √8-2√7
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Những câu hỏi liên quan
1 tháng 6 2023
i: =-12*căn 3/2căn 3=-6
h: =72căn 2/12căn 2=6
g: =25căn 12/5căn 6=5căn 2
f: =(15:5)*căn 6:3=3căn 2
d: =-1/2*6*căn 10=-3căn 10
28 tháng 6 2021
`a)((sqrt(14)-sqrt7)/(1-sqrt2)+(sqrt{15}-sqrt5)/(1-sqrt3)):1/(sqrt7-sqrt5)`
`=((sqrt7(sqrt2-1))/(1-sqrt2)+(sqrt5(sqrt3-1))/(1-sqrt3)).(sqrt7-sqrt5)`
`=(-sqrt7-sqrt5)*(sqrt7-sqrt5)`
`=-(sqrt7+sqrt5)(sqrt7+sqrt5)`
`=-(7-5)=-2`
`b)sqrt2+1/sqrt{5+2sqrt6}+2/sqrt{8+2sqrt{15}}`
`=sqrt2+1/sqrt{3+2sqrt{3}.sqrt2+2}+2/sqrt{5+2sqrt{5}.sqrt3+3}`
`=sqrt2+1/sqrt{(sqrt3+sqrt2)^2}+2/sqrt{(sqrt5+sqrt3)^2}`
`=sqrt2+1/(sqrt3+sqrt2)+2/(sqrt5+sqrt3)`
`=sqrt2+((sqrt3+sqrt2)(sqrt3-sqrt2))/(sqrt3+sqrt2)+((sqrt5+sqrt3)(sqrt5-sqrt3))/(sqrt5+sqrt3)`
`=sqrt2+sqrt3-sqrt2+sqrt5-sqrt3=sqrt5`
28 tháng 6 2021
a) Ta có: \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(-\dfrac{\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\dfrac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=-2\)
b) Ta có: \(\sqrt{2}+\dfrac{1}{\sqrt{5+2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)
\(=\sqrt{2}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}\)
\(=\sqrt{5}\)
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, \(=\sqrt{7-2\sqrt{7}+1}+\sqrt{7-2.3\sqrt{7}+9}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(3-\sqrt{7}\right)^2}=\left|\sqrt{7}-1\right|+\left|3-\sqrt{7}\right|\)
\(=\sqrt{7}-1+3-\sqrt{7}=2\)
\(b,=\left|\sqrt{7}-1\right|-\sqrt{7+2.2\sqrt{7}+4}\)
\(=\left|\sqrt{7}-1\right|-\sqrt{\left(\sqrt{7}+2\right)^2}=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+2\right|\)
\(=\sqrt{7}-1-\sqrt{7}-2=-3\)
NT
1
13 tháng 12 2022
a: \(=\left(\dfrac{5}{8}-\dfrac{28}{8}\right)\cdot\dfrac{5}{6}=\dfrac{-23}{8}\cdot\dfrac{5}{6}=\dfrac{-115}{48}\)
b: \(=\dfrac{4}{7}\cdot\dfrac{24+25}{15}=\dfrac{4}{7}\cdot\dfrac{49}{15}=\dfrac{28}{15}\)
c: \(=\dfrac{7}{2}:\dfrac{24-10}{15}=\dfrac{7}{2}\cdot\dfrac{15}{14}=\dfrac{15}{4}\)
TH
9
27 tháng 2 2022
Bài 2:
a: 2/6x5/3=10/18=5/9
b: 11/9x5/10=55/90=11/18
c: 3/9x6/8=1/3x3/4=1/4
d: 4/9x12/16=48/144=1/3
e: 25/15x6/7=5/3x6/7=30/21=10/7
f: 6/10x15/20=90/200=9/20
LM
27 tháng 2 2022
Bài 1
4/5 x 6/7= 24/35
2/9 x 1/2= 2/18= 1/9
1/2 x 8/3= 8/6= 4/3
7/9 x 6/5= 42/45= 14/15
8/7 x 5/9= 40/63
10/11 x 22/15= 220/165= 4/3
Bài 2
2/6 x 5/3= 1/3 x 5/3=5/9
11/9 x 5/10= 11/9 x 1/2= 11/18
3/9 x 6/8= 1/3 x 3/4 =3/12= 1/4
4/9 x 12/16= 4/9 x 3/4= 12/36= 1/3
25/15 x 6/7= 5/3 x 6/7= 30/21= 10/7
6/10 x 15/20= 3/5 x 3/4= 9/20
\(1,\)
\(a,\sqrt{6-2\sqrt{5}}=\sqrt{\sqrt{5^2}-2.\sqrt{5}.1+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)
\(b,\sqrt{8+2\sqrt{7}}=\sqrt{\sqrt{7^2}+2.\sqrt{7}.1+1}=\sqrt{\left(\sqrt{7}+1\right)^2}=\left|\sqrt{7}+1\right|=\sqrt{7}+1\)
\(2,\)
\(a,\sqrt{\left(\sqrt{10}-3\right)^2}-\sqrt{10}\)
\(=\left|\sqrt{10}-3\right|-\sqrt{10}\)
\(=\sqrt{10}-\sqrt{10}-3\)
\(=-3\)
\(b,\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}\)
\(=\left|5+\sqrt{7}\right|-\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(=5+\sqrt{7}-\left|\sqrt{7}-1\right|\)
\(=5+\sqrt{7}-\sqrt{7}+1\)
\(=6\)