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a)\(\sqrt{45.80}=\sqrt{9.400}=\sqrt{9}.\sqrt{400}=3.20=60\)
b) \(\sqrt{75.48}=\sqrt{25.3.16.3}=\sqrt{5^2.3^2.4^2}=5.4.3=60\)
c)\(\sqrt{90.6,4}=\sqrt{10.9.4.1,6}=\sqrt{4^2.3^2.2^2}=4.3.2=24\)
d) \(\sqrt{2,5.14,4}=\sqrt{\dfrac{25}{10}.\dfrac{144}{10}}=\sqrt{\dfrac{25.144}{100}}=\sqrt{\left(\dfrac{5.12}{10}\right)^2}=\dfrac{5.12}{10}=6\)
a) \(\sqrt{45.80}=\sqrt{9.400}=\sqrt{9}.\sqrt{400}=3.20=60\)
b)\(\sqrt{75.48}=\sqrt{25.3.3.16}=5.3.4=60\)
c)\(\sqrt{90.6,4}=\sqrt{9.64}=3.8=24\)
d)\(\sqrt{2,5.14,4}=\sqrt{\dfrac{25}{10}.\dfrac{144}{10}}=\sqrt{\dfrac{25.144}{100}=\dfrac{5.12}{10}=\dfrac{60}{10}=6}\)
a)\(\sqrt{75\cdot48}=\sqrt{25\cdot3\cdot48}=\sqrt{25\cdot144}=\sqrt{25}\cdot\sqrt{144}=5\cdot12=60\)
b) \(\sqrt{2,5\cdot14,4}=\sqrt{25\cdot144\cdot\frac{1}{100}}=\sqrt{25}\cdot\sqrt{144}\cdot\sqrt{\frac{1}{100}}=5\cdot12\cdot\frac{1}{10}=6\)
\(\frac{3\sqrt{128}}{\sqrt{2}}=\frac{\sqrt{9.128}}{\sqrt{2}}=\sqrt{\frac{1152}{2}}=\sqrt{576}=24\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=|2+\sqrt{3}|-|2-\sqrt{3}|\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
\(=3+2\sqrt{2}+3-2\sqrt{2}\)
\(=6\)
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)
\(=2+\sqrt{5}-\sqrt{5}+2\)
\(=4\)
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)
\(=1+\sqrt{5}-\sqrt{5}+1\)
\(=2\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(A=\sqrt{3}+2+2-\sqrt{3}\)
A = 2 + 2
A = 4
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(B=\sqrt{2}+3+3-\sqrt{2}\)
B = 3 + 3
B = 6
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(C=3+2\sqrt{2}+3-2\sqrt{2}\)
C = 3 + 3
C = 6
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(D=\sqrt{5}+2-\sqrt{5}+2\)
D = 2 + 2
D = 4
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(E=\sqrt{5}+1-\sqrt{5}+1\)
E = 1 + 1
E = 2
a: \(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}=12\sqrt{2}\)
b: \(=5\sqrt{7}-4\sqrt{7}+3\sqrt{7}=4\sqrt{7}\)
c: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}=\dfrac{1}{6}\sqrt{6}\)
d: \(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
e: \(=\sqrt{5}+\dfrac{2}{5}\sqrt{5}+\sqrt{5}=2.4\sqrt{5}\)
f: \(=\dfrac{1}{5}\sqrt{5}+\dfrac{3}{2}\sqrt{2}+\dfrac{5}{2}\sqrt{2}=\dfrac{1}{5}\sqrt{5}+4\sqrt{2}\)
Câu 1 nè
\(\sqrt{11}+6\sqrt{2}-3+\sqrt{2}=\sqrt{11}+7\sqrt{2}-3.\)
Câu 2 nè :
a) đề không rõ.
b) \(\sqrt{75.48}=\sqrt{25.16.3^2}=5.4.3=60\)