Áp dụng quy tắc chia hai căn bậc hai, hãy tính :
a) \(\dfrac{\sqrt{2300}}{\sqrt{23}}\)
b) \(\dfrac{\sqrt{12,5}}{\sqrt{0,5}}\)
c) \(\dfrac{\sqrt{192}}{\sqrt{12}}\)
d) \(\dfrac{\sqrt{6}}{\sqrt{150}}\)
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Y
22 tháng 6 2023
\(e,\sqrt{\dfrac{9}{169}}=\dfrac{\sqrt{9}}{\sqrt{169}}=\dfrac{\sqrt{3^2}}{\sqrt{13^2}}=\dfrac{3}{13}\)
\(f,\sqrt{1\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{\sqrt{25}}{\sqrt{16}}=\dfrac{\sqrt{5^2}}{\sqrt{4^2}}=\dfrac{5}{4}\)
\(g,\dfrac{\sqrt{2300}}{\sqrt{23}}=\sqrt{\dfrac{2300}{23}}=\sqrt{100}=\sqrt{10^2}=10\)
\(h,\dfrac{\sqrt{12,5}}{\sqrt{0,5}}=\sqrt{\dfrac{12,5}{0,5}}=\sqrt{25}=\sqrt{5^2}=5\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
e, \(\sqrt{\dfrac{9}{169}}\)
\(=\sqrt{\dfrac{3^2}{13^2}}\)
\(=\dfrac{3}{13}\)
f, \(\sqrt{1\dfrac{9}{16}}\)
\(=\sqrt{\dfrac{25}{16}}\)
\(=\sqrt{\dfrac{5^2}{4^2}}\)
\(=\dfrac{5}{4}\)
g, \(\dfrac{\sqrt{2300}}{\sqrt{23}}\)
\(=\dfrac{10\sqrt{23}}{\sqrt{23}}\)
\(=10\)
h, \(\dfrac{\sqrt{12,5}}{\sqrt{0,5}}\)
\(=\dfrac{\dfrac{5\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}}\)
\(=\dfrac{\dfrac{5\sqrt{2}}{2}\cdot2}{\sqrt{2}}\)
\(=\dfrac{5\sqrt{2}}{\sqrt{2}}=5\)
13 tháng 11 2018
1
a,\(\sqrt{\dfrac{36}{121}}=\sqrt{\dfrac{6^2}{11^2}}=\dfrac{6}{11}\)
\(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}=\sqrt{\dfrac{81}{100}}=\sqrt{\dfrac{9^2}{10^2}}=\dfrac{9}{10}\)
29 tháng 6 2023
a: ĐKXĐ: 3x^2+15/-6>=0
=>3x^2+15<=0(vô lý)
b: ĐKXĐ: -81/-x^2-12>=0
=>-x^2-12<0
=>-x^2<12
=>x^2>-12(luôn đúng)
c: ĐKXĐ: 31(x^2+21)/3>=0
=>x^2+21>=0(luôn đúng)
d: ĐKXĐ: -12/x^2+11>=0
=>x^2+11<0(vô lý)
e: ĐKXĐ: 21/-x^2-17>=0
=>-x^2-17>0
=>x^2+17<0(vô lý)
CP
16 tháng 7 2021
a) \(A=\left(1-\dfrac{\sqrt{3}-1}{2}\right):\left(\dfrac{\sqrt{3}-1}{2}+2\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{\sqrt{3}-1}{2}\right):\left(\dfrac{\sqrt{3}-1}{2}+\dfrac{4}{2}\right)\)
\(=\dfrac{2-\left(\sqrt{3}-1\right)}{2}:\dfrac{\left(\sqrt{3}-1\right)+4}{2}\)
\(=\dfrac{3-\sqrt{3}}{2}.\dfrac{2}{\sqrt{3}+3}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(1+\sqrt{3}\right)}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\dfrac{\left(\sqrt{3}-1\right)^2}{2}\)
Vì \(\left\{{}\begin{matrix}\left(\sqrt{3}-1\right)^2>0\\2>0\end{matrix}\right.\) \(\Rightarrow\dfrac{\left(\sqrt{3}-1\right)^2}{2}>0\) hay A>0
=> A có căn bậc 2
Vậy......
b)\(B=\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}\)
\(=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)\left(1+\sqrt{3}\right)}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\sqrt{5}\right):\dfrac{\sqrt{5}+\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\left(\dfrac{\sqrt{2}\left(3-1\right)}{1-3}-\sqrt{5}\right).\dfrac{5-2}{\sqrt{5}+\sqrt{2}}\)
\(=\left(-\sqrt{2}-\sqrt{5}\right).\dfrac{3}{\sqrt{5}+\sqrt{2}}\)
\(=-\left(\sqrt{2}+\sqrt{5}\right).\dfrac{3}{\sqrt{5}+\sqrt{2}}\)
\(=-3\)
Vì -3 < 0 hay B < 0
=> B không có căn bậc 2
Vậy.....
23 tháng 4 2017
Áp dụng quy tắc khai phương một thương, hãy tính :
a) 9169−−−−√ = \(\sqrt{\dfrac{3^2}{13^2}}\) = \(\left|\dfrac{3}{13}\right|\) = \(\dfrac{3}{13}\)
b) 25144−−−−√ = \(\sqrt{\dfrac{5^2}{12^2}}\) = \(\left|\dfrac{5}{12}\right|\) = \(\dfrac{5}{12}\)
c) 1916−−−−√ = \(\sqrt{\dfrac{25}{16}}\) = \(\sqrt{\dfrac{5^2}{4^2}}\) = \(\left|\dfrac{5}{4}\right|\) = \(\dfrac{5}{4}\)
d) 2781−−−−√ = \(\sqrt{\dfrac{169}{81}}\) = \(\sqrt{\dfrac{13^2}{9^2}}\) = \(\left|\dfrac{13}{9}\right|\) = \(\dfrac{13}{9}\)
26 tháng 5 2022
a: \(5\sqrt{2}-8\sqrt{3}+30\sqrt{3}-6\sqrt{3}=5\sqrt{2}+16\sqrt{3}\)
b: \(=14\sqrt{3}-\dfrac{3}{32}\cdot8\sqrt{3}+\dfrac{4}{18}\cdot9\sqrt{3}-\dfrac{1}{10}\cdot10\sqrt{3}\)
\(=14\sqrt{3}-\dfrac{3}{4}\sqrt{3}+2\sqrt{3}-1\sqrt{3}=\dfrac{57}{4}\sqrt{3}\)
c: \(=\dfrac{-1}{2}\cdot6\sqrt{3}+\dfrac{1}{15}\cdot5\sqrt{3}-\dfrac{1}{22}\cdot11\sqrt{3}+2\sqrt{3}\)
\(=-3\sqrt{3}+\dfrac{1}{3}\sqrt{3}-\dfrac{1}{2}\sqrt{3}+2\sqrt{3}=-\dfrac{7}{6}\sqrt{3}\)
d: \(=\dfrac{5}{8}\cdot4\sqrt{3}-\dfrac{1}{33}\cdot11\sqrt{3}+\dfrac{3}{14}\cdot7\sqrt{3}-\dfrac{1}{4}\cdot8\sqrt{3}\)
\(=\dfrac{5}{2}\sqrt{3}-\dfrac{1}{3}\sqrt{3}+\dfrac{3}{2}\sqrt{3}-2\sqrt{3}=\dfrac{5}{3}\sqrt{3}\)
BM
0
HP
17 tháng 12 2020
a, \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)
b, \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3}{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{3}}{\sqrt{3}+\sqrt{2}}\)
\(=\dfrac{\sqrt{6}}{2}+\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=\dfrac{6-\sqrt{6}}{2}\)
c, \(3\sqrt{2}-2\sqrt{3}+2\sqrt{3}+3\sqrt{2}=6\sqrt{2}\)
d, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}+3\right)^2}\)
\(=-\sqrt{6}+3+2\sqrt{6}+3=\sqrt{6}+6\)
e, Ghi đúng đề.
\(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}=\dfrac{a+b-2\sqrt{ab}+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)
27 tháng 7 2021
1) \(\sqrt{1\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{5}{4}\)
2) \(\dfrac{\sqrt{12.5}}{0.5}=\sqrt{\dfrac{12.5}{0.25}}=5\sqrt{2}\)
3) \(\sqrt{\dfrac{25}{64}}=\dfrac{5}{8}\)
4) \(\dfrac{\sqrt{230}}{\sqrt{2.3}}=\sqrt{\dfrac{230}{2.3}}=\sqrt{100}=10\)
5) \(\left(\sqrt{\dfrac{2}{3}}+\sqrt{\dfrac{50}{3}}-\sqrt{24}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{5\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{6\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)
\(=0\cdot\sqrt{6}=0\)
Áp dụng quy tắc chia hai căn bậc hai, hãy tính :
a) 2300−−−−√23−−√ = \(\sqrt{\dfrac{2300}{23}}\) = \(\sqrt{100}\) = 10
b) 12,5−−−−√0,5−−−√ = \(\sqrt{\dfrac{12,5}{0,5}}\) = \(\sqrt{25}\) = 5
c) 192−−−√12−−√ = \(\sqrt{\dfrac{192}{12}}\) = \(\sqrt{16}\) = 4
d) 6–√150−−−√ = \(\sqrt{\dfrac{6}{150}}\) = \(\sqrt{\dfrac{1}{25}}\) = \(\dfrac{1}{5}\)