Tính
a,8+2\(\sqrt{15}\)
b,12+2\(\sqrt{35}\)
c,8+\(\sqrt{60}\)
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NL
Nguyễn Lê Phước Thịnh
CTVHS
28 tháng 6 2021
c) \(\sqrt{5+\sqrt{24}}=\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)
d) \(\sqrt{12-\sqrt{140}}=\sqrt{12-2\sqrt{35}}=\sqrt{7}-\sqrt{5}\)
f) \(\sqrt{8-\sqrt{28}}=\sqrt{8-2\sqrt{7}}=\sqrt{7}-1\)
g) \(\sqrt{23-4\sqrt{15}}=\sqrt{23-2\cdot\sqrt{60}}=2\sqrt{5}-\sqrt{3}\)
h) \(\sqrt{9+4\sqrt{2}}=\sqrt{\left(2\sqrt{2}+1\right)^2}=2\sqrt{2}+1\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
20 tháng 8 2021
a: Ta có: \(\sqrt{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)
b: Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{2}}\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
=9-2
=7
c: Ta có: \(\left(\sqrt{7}+\sqrt{5}\right)\cdot\sqrt{12-2\sqrt{35}}\)
\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=2
NL
Nguyễn Lê Phước Thịnh
CTVHS
30 tháng 6
Bài 2:
a: \(\frac{21+8\sqrt5}{4+\sqrt5}\cdot\sqrt{9-4\sqrt5}\)
\(=\frac{\left(4+\sqrt5\right)^2}{4+\sqrt5}\cdot\sqrt{\left(\sqrt5-2\right)^2}\)
\(=\left(4+\sqrt5\right)\left(\sqrt5-2\right)=4\sqrt5-8+5-2\sqrt5=2\sqrt5-3\)
b: \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
\(=\sqrt{\left(\sqrt5-\sqrt3\right)^2}-\sqrt{\left(\sqrt5+\sqrt3\right)^2}\)
\(=\sqrt5-\sqrt3-\sqrt5-\sqrt3=-2\sqrt3\)
Bài 1:
a: \(\frac{\sqrt6-\sqrt{15}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt3\left(\sqrt2-\sqrt5\right)}{-\sqrt7\left(\sqrt2-\sqrt5\right)}=-\sqrt{\frac37}=-\frac{\sqrt{21}}{7}\)
b: \(\frac{10+2\sqrt{10}}{\sqrt5+\sqrt2}+\frac{8}{1-\sqrt5}\)
\(=\frac{2\sqrt5\left(\sqrt5+\sqrt2\right)}{\sqrt5+\sqrt2}-\frac{8\left(\sqrt5+1\right)}{\left(\sqrt5-1\right)\left(\sqrt5+1\right)}\)
\(=2\sqrt5-2\left(\sqrt5+1\right)=-2\)
c: \(\frac{\sqrt{3-\sqrt5}\left(3+\sqrt5\right)}{\sqrt{10}+\sqrt2}=\frac{\sqrt{6-2\sqrt5}\left(3+\sqrt5\right)}{\sqrt{20}+2}\)
\(=\frac{\sqrt{\left(\sqrt5-1\right)^2}\cdot\left(3+\sqrt5\right)}{2\left(\sqrt5+1\right)}=\frac{\left(\sqrt5-1\right)\left(3+\sqrt5\right)}{2\left(\sqrt5+1\right)}\)
\(=\frac{3\sqrt5+5-3-\sqrt5}{2\left(\sqrt5+1\right)}=\frac{2\sqrt5+2}{2\sqrt5+2}\)
=1
d: \(\frac{1}{\sqrt2+\sqrt{2+\sqrt3}}+\frac{1}{\sqrt2-\sqrt{2+\sqrt3}}\)
\(=\frac{\sqrt2}{2+\sqrt{4+2\sqrt3}}+\frac{\sqrt2}{2-\sqrt{4+2\sqrt3}}\)
\(=\frac{\sqrt2}{2+\sqrt{\left(\sqrt3+1\right)^2}}+\frac{\sqrt2}{2-\sqrt{\left(\sqrt3-1\right)^2}}=\frac{\sqrt2}{3+\sqrt3}+\frac{\sqrt2}{3-\sqrt3}\)
\(=\frac{\sqrt2\left(3-\sqrt3\right)+\sqrt2\left(3+\sqrt3\right)}{\left(3-\sqrt3\right)\left(3+\sqrt3\right)}=\frac{3\sqrt2-\sqrt6+3\sqrt2+\sqrt6}{9-3}=\frac{6\sqrt2}{6}=\sqrt2\)
TN
Y
26 tháng 6 2023
\(a,\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\\ =\sqrt{\sqrt{5^2}+2\sqrt{5}.\sqrt{3}+\sqrt{3^2}}-\sqrt{\sqrt{5^2}-2\sqrt{5}.\sqrt{3}+\sqrt{3^2}}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\\ =\left|\sqrt{5}+\sqrt{3}\right|-\left|\sqrt{5}-\sqrt{3}\right|\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}\\ =2\sqrt{3}\)
\(b,\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}\\ =\sqrt{\sqrt{2^2}+2.\sqrt{3}.\sqrt{2}+\sqrt{3^2}}+\sqrt{\sqrt{2^2}-2.\sqrt{3}.\sqrt{2}+\sqrt{3^2}}\\ =\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\\ =\left|\sqrt{2}+\sqrt{3}\right|+\left|\sqrt{2}-\sqrt{3}\right|\\ =\sqrt{2}+\sqrt{3}-\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
P
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. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}-\sqrt{3-2\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{2}\)
Y
25 tháng 6 2023
\(a,\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}\\ =\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{\sqrt{2}.\sqrt{4}-\sqrt{3}.\sqrt{4}}\\ =\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2^2}}\\ =\dfrac{\sqrt{5}}{2}\)
\(b,\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\\ =\dfrac{\sqrt{5}.\sqrt{3}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\\ =\dfrac{\sqrt{3}}{\sqrt{7}}\)
\(c,\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}=\dfrac{\sqrt{5}}{\sqrt{2}}\)
H
25 tháng 6 2023
\(a,=\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{2\sqrt{2}-2\sqrt{3}}\\ =\dfrac{\sqrt{5}.\left(\sqrt{2}-\sqrt{3}\right)}{2\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{2}\)
\(b,=\dfrac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\\ =\dfrac{\sqrt{3}}{\sqrt{7}}=\dfrac{\sqrt{21}}{7}\)
\(c,=\dfrac{\sqrt{5}.\sqrt{5}+\sqrt{5}}{\sqrt{2}.\sqrt{5}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{10}}{2}\)
8 tháng 11 2021
a) \(\sqrt{13^2-12^2}=\sqrt{\left(13-12\right)\left(13+12\right)}=\sqrt{25}=5\)
b) \(\sqrt{17^2-8^2}=\sqrt{\left(17-8\right)\left(17+8\right)}=\sqrt{9.25}=\sqrt{9}.\sqrt{25}=3.5=15\)
c) \(\sqrt{117^2-108^2}=\sqrt{\left(117-108\right)\left(117+108\right)}=\sqrt{9.225}=\sqrt{9}.\sqrt{225}=3.15=45\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
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: \(B=\sqrt{32}-\sqrt{50}-16\sqrt{\dfrac{1}{8}}\)
\(=4\sqrt{2}-5\sqrt{2}-\dfrac{16}{\sqrt{8}}\)
\(=-\sqrt{2}-2\sqrt{8}=-\sqrt{2}-4\sqrt{2}=-5\sqrt{2}\)
\(a,8+2\sqrt{15}=\left(\sqrt{3}+\sqrt{5}\right)^2\\ b,12+2\sqrt{35}=\left(\sqrt{7}+\sqrt{5}\right)^2\\ c,8+\sqrt{60}=8+2\sqrt{15}=\left(\sqrt{3}+\sqrt{5}\right)^2\)