\(\dfrac{\sqrt{5+\sqrt{3}}+\sqrt{5-\sqrt{3}}}{\sqrt{5+\sqrt{22}}}+\sqrt{11-6\sqrt{2}}\)
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30 tháng 9 2021
a: Ta có: \(A=\left(\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right):\left(2+\sqrt{2}\right)\)
\(=\left(2+\sqrt{2}\right):\left(2+\sqrt{2}\right)\)
=1
b: Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+1\)
=1
H9
HT.Phong (9A5)
CTVHS
1 tháng 10 2023
\(A=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)
\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)
\(A=2^2-\left(\sqrt{5}\right)^2\)
\(A=4-5\)
\(A=-1\)
____
\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(B=6-121\)
\(B=-115\)
RK
23 tháng 6 2021
a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\) = \(\sqrt{\dfrac{1}{8}\cdot2}.\sqrt{125\cdot\dfrac{1}{5}}=\sqrt{\dfrac{1}{4}}.\sqrt{25}=\dfrac{1}{2}\cdot5=2,5\)
RK
23 tháng 6 2021
b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2-1}=1\)
3 tháng 7 2023
a: \(=\dfrac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)
b: \(=\dfrac{\sqrt{10}\left(\sqrt{11}+\sqrt{7}\right)}{\sqrt{2}\left(\sqrt{11}+\sqrt{7}\right)}=\sqrt{\dfrac{10}{2}}=\sqrt{5}\)
c: \(=\dfrac{\sqrt{6}\left(\sqrt{7}-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{7}-\sqrt{6}\right)}=\sqrt{\dfrac{6}{3}}=\sqrt{2}\)
H9
HT.Phong (9A5)
CTVHS
4 tháng 7 2023
1) \(\dfrac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}\)
\(=\dfrac{9\sqrt{5}+3\sqrt{9\cdot3}}{\sqrt{5}+\sqrt{3}}\)
\(=\dfrac{9\sqrt{5}+3\cdot3\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
\(=\dfrac{9\cdot\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}\)
\(=\dfrac{9}{1}=9\)
2) \(\dfrac{\sqrt{110}+\sqrt{70}}{\sqrt{22}+\sqrt{14}}\)
\(=\dfrac{\sqrt{10}\cdot\sqrt{11}+\sqrt{10}\cdot\sqrt{7}}{\sqrt{2}\cdot\sqrt{11}+\sqrt{2}\cdot\sqrt{7}}\)
\(=\dfrac{\sqrt{10}\cdot\left(\sqrt{11}+\sqrt{7}\right)}{\sqrt{2}\cdot\left(\sqrt{11}+\sqrt{7}\right)}\)
\(=\dfrac{\sqrt{10}}{\sqrt{2}}=\sqrt{\dfrac{10}{2}}\)
\(=\sqrt{5}\)
3) \(\dfrac{\sqrt{42}-6}{\sqrt{21}-\sqrt{18}}\)
\(=\dfrac{\sqrt{6}\cdot\sqrt{7}-\sqrt{6}\cdot\sqrt{6}}{\sqrt{3}\cdot\sqrt{7}-\sqrt{3}\cdot\sqrt{6}}\)
\(=\dfrac{\sqrt{6}\cdot\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{3}\cdot\left(\sqrt{7}-\sqrt{3}\right)}\)
\(=\dfrac{\sqrt{6}}{\sqrt{3}}=\sqrt{\dfrac{6}{3}}\)
\(=\sqrt{2}\)
19 tháng 5 2022
b: \(=\sqrt{5}-1-\sqrt{5}-1=-2\)
c: \(=\dfrac{\left(2\sqrt{2}+\sqrt{3}-2\sqrt{2}+\sqrt{3}\right)}{2\sqrt{3}}=1\)
d: \(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\sqrt{2}\)
N
25 tháng 10 2021
\(\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\dfrac{11\left(4-\sqrt{5}\right)}{16-5}=\sqrt{5}-4+\sqrt{5}=2\sqrt{5}-4\)
H9
HT.Phong (9A5)
CTVHS
21 tháng 7 2023
a) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
\(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}}\)
\(=\dfrac{\sqrt{\left(3-\sqrt{5}\right)^2}}{\sqrt{3^2-\left(\sqrt{5}\right)^2}}\)
\(=\dfrac{\left|3-\sqrt{5}\right|}{\sqrt{9-5}}\)
\(=\dfrac{3-\sqrt{5}}{2}\)
b) \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{2^2-\left(\sqrt{3}\right)^2}}\)
\(=\dfrac{\left|2-\sqrt{3}\right|}{\sqrt{4-3}}\)
\(=\dfrac{2-\sqrt{3}}{1}\)
\(=2-\sqrt{3}\)
21 tháng 7 2023
a: \(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}}=\dfrac{3-\sqrt{5}}{2}\)
b: \(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{1}}=2-\sqrt{3}\)
d: \(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
=(căn 6-11)(căn 6+11)
=6-121=-115
Y
8 tháng 7 2023
\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)
Xét \(A=\sqrt{5+\sqrt{3}}+\sqrt{5-\sqrt{3}}\)
\(\Rightarrow A^2=10+2\sqrt{22}\Rightarrow A=\sqrt{2}\sqrt{5+\sqrt{22}}\)
\(\dfrac{\sqrt{5+\sqrt{3}}+\sqrt{5-\sqrt{3}}}{\sqrt{5+\sqrt{22}}}+\sqrt{11-6\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\sqrt{5+\sqrt{22}}}{\sqrt{5+\sqrt{22}}}+\sqrt{\left(\sqrt{2}-3\right)^2}\)
\(=\sqrt{2}-\sqrt{2}+3=3\)