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28 tháng 9 2017
a)\(\left(\sqrt{10}-\sqrt{15}+3\sqrt{3}\right)\sqrt{5}-\sqrt{72}\)
\(=\sqrt{15}-\sqrt{15}+15-6\sqrt{2}\)
\(15-6\sqrt{2}\)
b)\(\dfrac{\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right)}{8\sqrt{10}}\)
\(=\dfrac{\left(15.5\sqrt{2}+5.10\sqrt{2}-3.15\sqrt{2}\right)}{8\sqrt{10}}\)
\(=\dfrac{\left(75\sqrt{2}+50\sqrt{2}-45\sqrt{2}\right)}{8\sqrt{10}}\)
\(=\dfrac{80\sqrt{2}}{8\sqrt{10}}=\dfrac{10\sqrt{2}}{\sqrt{10}}=\sqrt{20}=2\sqrt{5}\)
25 tháng 6 2017
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
= \(2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(2-\sqrt{3}+\sqrt{3}-1\) = \(1\)
b) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
= \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
= \(3-\sqrt{6}+2\sqrt{6}-3\) = \(\sqrt{6}\)
c) \(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)
= \(\dfrac{15\sqrt{200}}{\sqrt{10}}-\dfrac{3\sqrt{450}}{\sqrt{10}}+\dfrac{2\sqrt{50}}{\sqrt{10}}\)
= \(15\sqrt{20}-3\sqrt{45}+2\sqrt{5}\)
= \(30\sqrt{5}-9\sqrt{5}+2\sqrt{5}\) = \(23\sqrt{5}\)
T
22 tháng 9 2019
\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}=\left|2-\sqrt{3}\right|+\sqrt{3-2\sqrt{3}+1}=2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}=2-\sqrt{3}+\left|\sqrt{3}-1\right|=2-\sqrt{3}+\sqrt{3}-1=1\)
13 tháng 7 2018
\(a.\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=3.7-2.\sqrt{7.2.7}+14\sqrt{2}=21\) \(b.\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):10=\left(75\sqrt{2}+50\sqrt{2}-45\sqrt{2}\right).\dfrac{1}{10}=80\sqrt{2}.\dfrac{1}{10}=8\sqrt{2}\) \(c.\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)=\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{2}-3\sqrt{\dfrac{2}{5}}\right)=\left(\sqrt{5}-1\right)\left(2-6\sqrt{\dfrac{1}{5}}\right)\)
HH
25 tháng 6 2018
Giải:
a) \(\left(\sqrt{3}+2\sqrt{5}\right)\sqrt{3}-\sqrt{60}\)
\(=\sqrt{3}.\sqrt{3}+2\sqrt{5}.\sqrt{3}-\sqrt{60}\)
\(=3+2\sqrt{15}-\sqrt{60}\)
\(=3+2\sqrt{15}-2\sqrt{15}\)
\(=3\)
Vậy ...
b) \(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)
\(=\left(15\sqrt{4.50}-3\sqrt{9.50}+2\sqrt{50}\right):\sqrt{10}\)
\(=\left(30\sqrt{50}-9\sqrt{50}+2\sqrt{50}\right):\sqrt{10}\)
\(=23\sqrt{50}:\sqrt{10}\)
\(=\dfrac{23\sqrt{50}}{\sqrt{10}}\)
\(=\dfrac{23\sqrt{5}\sqrt{10}}{\sqrt{10}}\)
\(=23\sqrt{5}\)
Vậy ...
25 tháng 6 2018
\(a\text{) }\left(\sqrt{3}+2\sqrt{5}\right)\sqrt{3}-\sqrt{60}\\ =3+2\sqrt{15}-2\sqrt{15}=3\)
\(b\text{) }\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\\ =15\sqrt{20}-3\sqrt{45}+2\sqrt{5}\\ =30\sqrt{5}-9\sqrt{5}+2\sqrt{5}\\ =\left(30-9+2\right)\sqrt{5}=23\sqrt{5}\)
AH
Akai Haruma
Giáo viên
1 tháng 1 2020
Bài 1:
\((15\sqrt{200}-3\sqrt{450}+2\sqrt{50}):10\)
\(=(15\sqrt{2.10^2}-3\sqrt{2.15^2}+2\sqrt{2.5^2}):10\)
\(=(15.10\sqrt{2}-3.15\sqrt{2}+2.5\sqrt{2}):10\)
\(=115\sqrt{2}:10=\frac{23\sqrt{2}}{2}\)
AH
Akai Haruma
Giáo viên
1 tháng 1 2020
Bài 2:
Ta có:
\(3+\sqrt{20}=3+\sqrt{2^2.5}=3+2\sqrt{5}=(3+\sqrt{5})+\sqrt{5}\)
\(>(3+\sqrt{4})+\sqrt{5}=(3+2)+\sqrt{5}=5+\sqrt{5}\)
\(\Rightarrow \sqrt{3+\sqrt{20}}> \sqrt{5+\sqrt{5}}\)
13 tháng 7 2016
@.@ Trời ơi, nhiều thế ^^
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)=\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-\frac{3\sqrt{2}}{\sqrt{5}}\right)\)
\(=\left(\sqrt{2}.\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}-\frac{3\sqrt{2}}{\sqrt{5}}\right)=2\sqrt{5}-2-6+\frac{6}{\sqrt{5}}=\frac{16\sqrt{5}}{5}-8\)
b) \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}=\frac{75\sqrt{2}+50\sqrt{2}-45\sqrt{2}}{\sqrt{10}}=\frac{80\sqrt{2}}{\sqrt{10}}=\frac{80}{\sqrt{5}}=16\sqrt{5}\)c) \(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}\)
\(=2+\sqrt{2}+2-\sqrt{2}=4\)
d) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)}^2\)
\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
e) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
f)\(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}=1+\sqrt{2}-\sqrt{2}+1=2\)g) \(\sqrt[3]{26+15\sqrt{3}}-\sqrt[3]{26-15\sqrt{3}}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)
\(=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\)
27 tháng 7 2020
a, \(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)-\sqrt{2}\left(\sqrt{3}-1\right)\)
\(=3-1-\sqrt{6}+\sqrt{2}=2+\sqrt{2}-\sqrt{6}\)
b, \(=\sqrt{300.0,04}+2\left|\sqrt{3}-\sqrt{5}\right|\)
\(=2\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)\)
\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)
c, \(=\sqrt{196}-2\sqrt{98}+\sqrt{49}+7\sqrt{8}\)
\(=14-14\sqrt{2}+7+14\sqrt{2}=21\)
d, \(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=15\sqrt{5}+10\sqrt{5}-9\sqrt{5}=16\sqrt{5}\)
27 tháng 7 2020
Bài 1: Rút gọn
a) Ta có: \(\left(\sqrt{3}-\sqrt{2}+1\right)\cdot\left(\sqrt{3}-1\right)\)
\(=\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-1\right)-\sqrt{2}\cdot\left(\sqrt{3}-1\right)\)
\(=3-1-\sqrt{6}+\sqrt{2}\)
\(=2-\sqrt{2}-\sqrt{6}\)
b) Ta có: \(0.2\cdot\sqrt{\left(-10\right)^2\cdot3}+2\cdot\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)
\(=0.2\cdot\sqrt{\left(-10\right)^2}\cdot\sqrt{3}+2\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=0.2\cdot10\cdot\sqrt{3}+2\sqrt{5}-2\sqrt{3}\)
\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}\)
\(=2\sqrt{5}\)
c) Ta có: \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(=\sqrt{196}-2\cdot\sqrt{98}+\sqrt{49}+7\sqrt{8}\)
\(=14-\sqrt{392}+7+\sqrt{392}\)
=21
d) Ta có: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=\sqrt{5}\left(15+5\cdot2-3\cdot3\right)\)
\(=16\sqrt{5}\)
TK:
\(=\left(15\cdot10\sqrt{2}-3\cdot15\sqrt{2}+2\cdot5\sqrt{2}\right):10\\ =\left(150\sqrt{2}-45\sqrt{2}+10\sqrt{2}\right):10\\ =115\sqrt{2}:10=\dfrac{23\sqrt{2}}{2}\)