Áp dung quy tắc khai phương một tích,hãy tính:
a) \(\sqrt{0,09.64}\)
b) \(\sqrt{2^4.\left(-7\right)^2}\)
c) \(\sqrt{12,1.360}\)
d)\(\sqrt{2^2.3^4}\)
K
Khách
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Những câu hỏi liên quan
Q
31 tháng 3 2017
a) ĐS: 2.4.
b) ĐS: 28.
c) HD: Đổi 12,1.360 thành 121.36. ĐS: 66
d) ĐS: 18.
HL
20 tháng 6 2017
a) \(\sqrt{0,09.64}\)
\(=\sqrt{0,09}.\sqrt{64}\)
\(=0,3.8=2,4\)
b) \(\sqrt{2^4.\left(-7\right)^2}\)
\(=\sqrt{2^4}.\sqrt{\left(-7\right)^2}\)
\(=2^2.7=4.7=28\)
c) \(\sqrt{12,1.360}\)
\(=\sqrt{121.36}\)
\(=\sqrt{121}.\sqrt{36}\)
\(=11.6=66\)
d) \(\sqrt{2^2.3^4}\)
\(=\sqrt{2^2}.\sqrt{3^4}\)
\(=2.3^2=2.9=18\)
14 tháng 4 2021
a, \(\sqrt{0.09\cdot64=\sqrt{0.09}\cdot\sqrt{64}=0.3\cdot8=2.4}\)
b, \(\sqrt{2^4\cdot\left(-7\right)^2}=\sqrt{16\cdot49}=\sqrt{16}\cdot\sqrt{49}=4\cdot7=28\)
c, \(\sqrt{121\cdot360}=\sqrt{121\cdot36}=\sqrt{121}\cdot\sqrt{36}=11\cdot6=66\)
d, \(\sqrt{2^2\cdot3^4}=\sqrt{2^2}\cdot\sqrt{3^4}=2\cdot3^2=18\)
13 tháng 5 2021
a)\(\sqrt{0,09}.\sqrt{64}\)=0,3.8=2,4
b)\(\sqrt{2^4}.\sqrt{\left(-7\right)^2}\)=4.7=28
c)\(\sqrt{121.36}\)=\(\sqrt{121}.\sqrt{36}\)=11.6=66
d)\(\sqrt{2^2}.\sqrt{3^4}\)=2.9=18
Y
22 tháng 6 2023
\(i,\sqrt{12,1.360}=\sqrt{12,1}.6\sqrt{10}=6.\sqrt{12,1.10}=6.\sqrt{121}=6.\sqrt{11^2}=6.11=66\)
\(k,\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{\dfrac{64}{25}}=\dfrac{\sqrt{8^2}}{\sqrt{5^2}}=\dfrac{8}{5}\)
\(l,-0,4.\sqrt{\left(-0,4\right)^2}=-0,4.0,4=-0,16\)
\(m,\sqrt{2^4.\left(-7\right)^2}=\sqrt{4^2}.\sqrt{\left(-7\right)^2}=4.7=28\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
i, \(\sqrt{12,1\cdot360}=\sqrt{4356}=\sqrt{66^2}=66\)
k, \(\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4\cdot6,4}=\sqrt{\dfrac{64}{25}}=\sqrt{\dfrac{2^6}{5^2}}=\dfrac{2^3}{5}=\dfrac{8}{5}\)
l, \(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\cdot\left|-0,4\right|=-0,4\cdot0,4=-\dfrac{4}{25}\)
m, \(\sqrt{2^4\cdot\left(-7\right)^2}=2^2\cdot\left|-7\right|=4\cdot7=28\)
AH
Akai Haruma
Giáo viên
26 tháng 6 2021
\(A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}-\sqrt{3+1-2\sqrt{3.1}}\)
\(=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}=|\sqrt{3}+1|-|\sqrt{3}-1|=2\)
\(B=\sqrt{4+5-2\sqrt{4.5}}+\sqrt{4+5+2\sqrt{4.5}}=\sqrt{(\sqrt{4}-\sqrt{5})^2}+\sqrt{(\sqrt{4}+\sqrt{5})^2}\)
\(=|\sqrt{4}-\sqrt{5}|+|\sqrt{4}+\sqrt{5}|=2\sqrt{5}\)
AH
Akai Haruma
Giáo viên
26 tháng 6 2021
\(C\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{7+1+2\sqrt{7.1}}\)
\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}\)
\(=|\sqrt{7}-1|-|\sqrt{7}+1|=-2\Rightarrow C=-\sqrt{2}\)
----------------------------
\(7+4\sqrt{3}=(2+\sqrt{3})^2\Rightarrow 10\sqrt{7+4\sqrt{3}}=10(2+\sqrt{3})\)
\(\Rightarrow \sqrt{48-10\sqrt{7+4\sqrt{3}}}=\sqrt{28-10\sqrt{3}}=\sqrt{(5-\sqrt{3})^2}=5-\sqrt{3}\)
\(\Rightarrow 3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}=3+5(5-\sqrt{3})=28-5\sqrt{3}\)
\(\Rightarrow D=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
18 tháng 9 2021
Bài 1:
a: Ta có: \(\sqrt{3x^2}=\sqrt{12}\)
\(\Leftrightarrow3x^2=12\)
\(\Leftrightarrow x^2=4\)
hay \(x\in\left\{2;-2\right\}\)
b: Ta có: \(\sqrt{\left(x-2\right)^2}=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
15 tháng 7 2019
\(\frac{3\sqrt{128}}{\sqrt{2}}=\frac{\sqrt{9.128}}{\sqrt{2}}=\sqrt{\frac{1152}{2}}=\sqrt{576}=24\)
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−−−−√ = \(\sqrt{\dfrac{169}{81}}\) = \(\sqrt{\dfrac{13^2}{9^2}}\) = \(\left|\dfrac{13}{9}\right|\) = \(\dfrac{13}{9}\)
8 tháng 6 2021
a,\(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5=\left(\sqrt{\dfrac{25}{16}}-\dfrac{3}{4}\right):5=\left(\dfrac{5}{4}-\dfrac{3}{4}\right):5\)
\(=\dfrac{1}{2}:5=\dfrac{1}{10}\)
b,\(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2=\left[\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\right]^2\)
\(=\left[3-4\right]^2=1\)
c,\(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)=11^2-\left(4\sqrt{3}\right)^2\)
\(=121-48=73\)
d,\(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)
\(=2-2\sqrt{2}+1-3+\dfrac{4\sqrt{2}}{5}+\sqrt{\dfrac{36}{25}.2}\)
\(=-2\sqrt{2}+\dfrac{4\sqrt{2}+6\sqrt{2}}{5}\)
\(=-2\sqrt{2}+\dfrac{10\sqrt{2}}{5}=-2\sqrt{2}+2\sqrt{2}=0\)
e,\(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{2021-2\sqrt{2021}.1+1}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{\left(\sqrt{2021}-1\right)^2}\)
\(=\left(1+\sqrt{2021}\right)\left(\sqrt{2021}-1\right)\)
\(=\sqrt{2021}-1+\sqrt{2021^2}-\sqrt{2021}=2020\)
a) \(\sqrt{0,09.64}=\sqrt{\left(0,3\right)^2.8^2}=0,3.8=2,4\)
b) \(\sqrt{2^4.\left(-7\right)^2}=\sqrt{\left(2^2\right)^2.\left(-7\right)^2}=2^2.\left|-7\right|=7.4=28\)
c) \(\sqrt{12,1.360}=\sqrt{12,1.10.36}=\sqrt{121.36}=\sqrt{11^2.6^2}=11.6=66\)
d) \(\sqrt{2^2.3^4}=\sqrt{2^2.\left(3^2\right)^2}=2.3^2=9.2=18\)
a) \(\sqrt{0,09\cdot64}=\sqrt{0,09}\cdot\sqrt{64}=0,3\cdot8=2,4\)
b) \(\sqrt{2^4\cdot\left(-7\right)^2}=\sqrt{2^4}\cdot\sqrt{\left(-7\right)^2}=2^2\cdot7=4\cdot7=28\)
c) \(\sqrt{12,1\cdot360}=\sqrt{12,1\cdot10\cdot36}=\sqrt{121\cdot36}=\sqrt{121}\cdot\sqrt{36}=11\cdot6=66\)
d) \(\sqrt{2^2\cdot3^4}=\sqrt{2^2}\cdot\sqrt{3^4}=2\cdot3^2=2\cdot9=18\)