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a) Ta có: \(\sqrt{0.1}\cdot\sqrt{4000}\)
\(=\sqrt{\frac{1}{10}}\cdot\sqrt{4000}\)
\(=\sqrt{\frac{1}{10}\cdot4000}=\sqrt{400}=20\)
b) Ta có: \(\sqrt{\frac{9}{196}}=\sqrt{\left(\frac{3}{14}\right)^2}\)
\(=\left|\frac{3}{14}\right|\)
\(=\frac{3}{14}\)(Vì \(\frac{3}{14}>0\))
c) Ta có: \(\sqrt{16}\cdot\sqrt{36}-\sqrt{125}:\sqrt{0.01}\)
\(=\sqrt{16\cdot36}-\frac{\sqrt{125}}{\sqrt{\frac{1}{100}}}\)
\(=\sqrt{576}-\sqrt{125:\frac{1}{100}}\)
\(=24-\sqrt{125\cdot100}\)
\(=24-\sqrt{12500}\)
\(=24-50\sqrt{5}\)
d) Ta có: \(\left(\sqrt{112}-\sqrt{63}+\sqrt{7}\right):\sqrt{7}\)
\(=\left(4\sqrt{7}-3\sqrt{3}+\sqrt{7}\right):\sqrt{7}\)
\(=\frac{2\sqrt{7}}{\sqrt{7}}=2\)
e) Ta có: \(\sqrt{2.5}\cdot\sqrt{30}\cdot\sqrt{48}\)
\(=\sqrt{\frac{5}{2}\cdot30\cdot48}=\sqrt{3600}=60\)
a: \(=\sqrt{\dfrac{25}{16}\cdot\dfrac{49}{9}\cdot\dfrac{1}{100}}=\dfrac{5}{4}\cdot\dfrac{7}{3}\cdot\dfrac{1}{10}=\dfrac{35}{120}=\dfrac{7}{24}\)
b: \(=\sqrt{1.44\cdot0.81}=1.2\cdot0.9=1.08\)
c: \(=\sqrt{\dfrac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\dfrac{1}{4}\cdot289}=\dfrac{17}{2}\)
d: \(=\sqrt{\dfrac{\left(149-76\right)\left(149+76\right)}{\left(457-384\right)\left(457+384\right)}}=\sqrt{\dfrac{225}{841}}=\dfrac{15}{29}\)
Cái đầu là tính à?
Ta có: \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}\)
\(=\left(\sqrt{15}\right)^2+2.2\sqrt{3}.\sqrt{15}+\left(2\sqrt{3}\right)^2+12\sqrt{5}\)
\(=15+12\sqrt{5}+12+12\sqrt{5}\)
\(=27+24\sqrt{5}\)
Sau:
Ta thấy: Điều kiện để \(\sqrt{-\left|x+5\right|}\) có nghĩa là \(-\left|x+5\right|\ge0\left(\forall x\right)\)
Mà \(-\left|x+5\right|\le0\left(\forall x\right)\) nên dấu "=" xảy ra khi: \(\left|x+5\right|=0\Rightarrow x=-5\)
Vậy khi x = -5 thì \(\sqrt{-\left|x+5\right|}\) có nghĩa
Làm lại ý 2
\(\sqrt{-\left|x+5\right|}\)có nghĩa
\(\Leftrightarrow-\left|x+5\right|\ge0\)
\(\Leftrightarrow\left|x+5\right|\le0\)
\(\Leftrightarrow x+5\le0\)
\(\Leftrightarrow x\le-5\)
a = 48
b = \(\frac{1}{5}\)
c = 2
d \(\approx0,3904413457\)
a) \(\sqrt{25}+\sqrt{9}-\sqrt{16}\) = \(\sqrt{5^2}+\sqrt{3^2}-\sqrt{4^2}\) = 5 + 3 - 4 = 4
b) \(\sqrt{0,16}+\sqrt{0,01}+\sqrt{0,25}\) = 0,4 + 0,1 + 0,5 = 1
c) \(\left(\sqrt{3^2}\right)-\left(\sqrt{2^2}\right)+\left(\sqrt{5^2}\right)\)
= 3 - 2 + 5 = 6
d) \(\sqrt{4}-\left(-\sqrt{3}\right)^2+\sqrt{49}\) = 2 - 3 + 7 = 6
e) \(\left(2\sqrt{2}\right)^2-\left(3\sqrt{3}\right)^2\)
= \(\left(\sqrt{8}\right)^2-\left(\sqrt{27}\right)^2\) = 8 - 27 = -19
f) \(\left(-2\sqrt{2}\right)^2+\left(3\sqrt{3}\right)^2\) = 8 + 27 = 35
\(\sqrt{0,01}=0,1\)
=0,1