Tính
a.\(\sqrt{6,4.361}\)
b.\(\sqrt{9,9.1,1}\)
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28 tháng 10 2023
1:
a: \(\sqrt{36}-\sqrt{100}=6-10=-4\)
b: Để \(\sqrt{\dfrac{2}{2x-1}}\) có nghĩa thì \(\dfrac{2}{2x-1}>=0\)
=>2x-1>0
=>x>1/2
2:
a: \(A=\dfrac{\left(15\sqrt{180}-5\sqrt{200}-3\sqrt{450}\right)}{\sqrt{10}}\)
\(=15\sqrt{\dfrac{180}{10}}-5\sqrt{\dfrac{200}{10}}-3\sqrt{\dfrac{450}{10}}\)
\(=15\sqrt{18}-5\sqrt{20}-3\sqrt{45}\)
\(=45\sqrt{2}-10\sqrt{5}-9\sqrt{5}\)
\(=45\sqrt{2}-19\sqrt{5}\)
b: \(B=\sqrt{32}-\sqrt{50}-16\sqrt{\dfrac{1}{8}}\)
\(=4\sqrt{2}-5\sqrt{2}-\dfrac{16}{\sqrt{8}}\)
\(=-\sqrt{2}-2\sqrt{8}=-\sqrt{2}-4\sqrt{2}=-5\sqrt{2}\)
29 tháng 8 2021
Bài 1:
a: \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)
\(=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}\)
\(=8\sqrt{7}\)
30 tháng 8 2021
Bài 3:
a: \(\sqrt{27^2-23^2}=10\sqrt{2}\)
b: \(\sqrt{37^2-35^2}=12\)
c: \(\sqrt{65^2-63^2}=16\)
d: \(\sqrt{117^2-108^2}=45\)
TN
1
19 tháng 8 2021
a: Ta có: \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)
b: Ta có: \(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\left(\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)
13 tháng 12 2023
a: \(\sqrt{169}-\sqrt{225}\)
\(=\sqrt{13^2}-\sqrt{15^2}\)
=13-15
=-2
b: \(\dfrac{\sqrt{144}}{9}\)
\(=\dfrac{\sqrt{12^2}}{9}\)
\(=\dfrac{12}{9}=\dfrac{4}{3}\)
c: \(\sqrt{18}:\sqrt{2}=\sqrt{\dfrac{18}{2}}=\sqrt{9}=3\)
LN
1
25 tháng 6 2023
`a,\sqrt(3+2sqrt2)=\sqrt((sqrt2)^2+2.sqrt2 .1+1^2)=\sqrt((sqrt2+1)^2)=|sqrt2+1|=sqrt2+1`
`b,\sqrt(7+4sqrt3)=\sqrt((sqrt3)^2+2.\sqrt3 .2 +2^2)=\sqrt((sqrt3+2)^2)=|sqrt3+2|=sqrt3+2`
`c,sqrt(14-6sqrt5)=\sqrt((sqrt5)^2-2.\sqrt5 .3+3^2)=sqrt((sqrt5-3)^2)=|sqrt5-3|+3-sqrt5`
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. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}-\sqrt{3-2\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{2}\)
8 tháng 2 2021
a, \(A=2\sqrt{3}-\sqrt{12}-\sqrt{9}\)
\(=2\sqrt{3}-2\sqrt{3}-3=-3\)
b, \(B=\sqrt{3}\left(\sqrt{12}+\sqrt{27}\right)\)
\(=\sqrt{3}\left(2\sqrt{3}+3\sqrt{3}\right)\)
\(=\sqrt{3}.5\sqrt{3}=5.3=15\)
NT
1
13 tháng 12 2023
a: \(\sqrt{\left(3-\sqrt{5}\right)^2}-\sqrt{5}\)
\(=\left|3-\sqrt{5}\right|-\sqrt{5}\)
\(=3-\sqrt{5}-\sqrt{5}=3-2\sqrt{5}\)
b: \(\sqrt{\left(4-2\sqrt{3}\right)^2}\)
\(=\left|4-2\sqrt{3}\right|\)
\(=4-2\sqrt{3}\)
22 tháng 9 2021
a) \(=\sqrt{\left(3\sqrt{5}-2\right)^2}+\sqrt{\left(3\sqrt{5}+2\right)^2}=3\sqrt{5}-2+3\sqrt{5}+2=6\sqrt{5}\)
b) \(=\sqrt{\left(2\sqrt{5}+3\right)^2}+\sqrt{\left(2\sqrt{5}-3\right)^2}=2\sqrt{5}+3+2\sqrt{5}-3=4\sqrt{5}\)
a) \(\sqrt{6,4.361}=\sqrt{6,4}.\sqrt{361}=\sqrt{16.0,4}.19\)
\(=\sqrt{16}.\sqrt{0,4}.19=4.\sqrt{0,4}.19=76.\sqrt{0,4}\)
b) \(\sqrt{9,9.1,1}=\sqrt{9.1,1.1,1}=\sqrt{9.1,1^2}=\sqrt{9}.\sqrt{1,1^2}=3.1,1=3,3\)
\(a,\sqrt{6,4.361}=\sqrt{2310,4}=\frac{76\sqrt{10}}{5}\)
\(b,\sqrt{9,9.1,1}=\sqrt{10,89}=3,3\)