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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: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=\sqrt{5}\left(15+5\cdot2-3\cdot3\right)\)
\(=16\sqrt{5}\)
T
22 tháng 9 2019
\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}=\left|2-\sqrt{3}\right|+\sqrt{3-2\sqrt{3}+1}=2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}=2-\sqrt{3}+\left|\sqrt{3}-1\right|=2-\sqrt{3}+\sqrt{3}-1=1\)
20 tháng 6 2018
a) \(3\sqrt{8}-4\sqrt{18}+5\sqrt{32}-\sqrt{50}=3\sqrt{4.2}-4\sqrt{9.2}+5\sqrt{16.2}-\sqrt{25.2}=6\sqrt{2}-12\sqrt{2}+20\sqrt{2}-5\sqrt{2}=9\sqrt{2}\)b) \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):10=\left(15\sqrt{50}+5\sqrt{50.4}-3\sqrt{50.9}\right):10=\left(15\sqrt{50}+10\sqrt{50}-9\sqrt{50}\right):10=\dfrac{16\sqrt{50}}{10}=\dfrac{16\sqrt{25.2}}{10}=\dfrac{80\sqrt{2}}{10}=8\sqrt{2}\) c) \(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}=2\sqrt{7.4}+2\sqrt{7.9}-3\sqrt{7.25}+\sqrt{7.16}=4\sqrt{7}+6\sqrt{7}-15\sqrt{7}+4\sqrt{7}=-\sqrt{7}\)
d)
\(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}=14-2.3\sqrt{2.14}+18+6\sqrt{28}=32-6\sqrt{28}+6\sqrt{28}=32\)
27 tháng 10 2019
g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
12 tháng 7 2018
\(1a.2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{548}=2\sqrt{16.5\sqrt{3}}-2\sqrt{\sqrt{75}}-6\sqrt{137}=8\sqrt{\sqrt{75}}-2\sqrt{\sqrt{75}}-6\sqrt{137}=6\sqrt{\sqrt{75}}-6\sqrt{137}\) \(b.\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}=\left(2\sqrt{3}+5\sqrt{3}+3\sqrt{3}\right).\dfrac{1}{\sqrt{15}}=10\sqrt{3}.\dfrac{1}{\sqrt{3}.\sqrt{5}}=2\sqrt{5}\) \(d.\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}=\left(60\sqrt{2}-80\sqrt{2}+105\sqrt{2}\right).\dfrac{1}{\sqrt{10}}=85\sqrt{2}.\dfrac{1}{\sqrt{2}.\sqrt{5}}=17\sqrt{5}\) \(e.\left(\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{\dfrac{9}{7}}\right):\sqrt{7}=\left(\sqrt{\dfrac{1}{7}}-4\sqrt{\dfrac{1}{7}}+3\sqrt{\dfrac{1}{7}}\right).\dfrac{1}{\sqrt{7}}=0\) \(2a.A=\sqrt{3+\sqrt{5+2\sqrt{3}}}.\sqrt{3-\sqrt{5+2\sqrt{3}}}=\sqrt{9-5-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\) \(b.B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}=\sqrt{2}.\sqrt{4-2}=2\)
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) \(\sqrt[3]{26+15\sqrt{3}}-\sqrt[3]{26-15\sqrt{3}}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)
\(=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\)
N
25 tháng 9 2019
\(A=\left(15.\sqrt{200}-4\sqrt{450}+2\sqrt{20}\right):\sqrt{10}-10\sqrt{20}\) \(=\left(15.10\sqrt{2}-4.15\sqrt{2}+2.2\sqrt{5}\right):\sqrt{10}-10.2\sqrt{5}\)
\(=\left(150\sqrt{2}-60\sqrt{2}+4\sqrt{5}\right):\sqrt{10}-20\sqrt{5}\)
\(=\left(90\sqrt{2}+4\sqrt{5}\right):\sqrt{10}-20\sqrt{5}\)
\(=\frac{90\sqrt{2}}{\sqrt{10}}+\frac{4\sqrt{5}}{\sqrt{10}}-20\sqrt{5}\)
\(=18\sqrt{5}+2\sqrt{2}-20\sqrt{5}\)
\(=2\sqrt{2}-2\sqrt{5}\)