\(\sqrt{2}\left(\sqrt{8}-\sqrt{32}+\sqrt{98}\right)\)
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9 tháng 7 2015
\(=\left(4\sqrt{2}+3\sqrt{2}-7\sqrt{2}\right)\left(\sqrt{42}+2\sqrt{5}-\sqrt{32}\right)=0.\left(\right)=0\)
HP
3 tháng 9 2021
b, \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)
\(=\dfrac{2\left(\sqrt{5}-2\right)}{5-4}-\dfrac{2\left(\sqrt{5}+2\right)}{5-4}\)
\(=2\sqrt{5}-4-2\sqrt{5}-4=-8\)
30 tháng 7 2021
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
NT
6 tháng 8 2021
a) \(2\sqrt{98}-3\sqrt{18}+\dfrac{1}{2}\sqrt{32}=14\sqrt{2}-9\sqrt{2}+2\sqrt{2}=7\sqrt{2}\)
b) \(\left(5\sqrt{2}+2\sqrt{5}\right).\sqrt{5}-\sqrt{250}=5\sqrt{10}+10-5\sqrt{10}=10\)
c) \(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}=6-5\sqrt{6}-6=5\sqrt{6}\)
d) \(3\sqrt{48}+2\sqrt{27}-\dfrac{1}{3}\sqrt{243}=12\sqrt{3}+6\sqrt{3}-3\sqrt{3}=15\sqrt{3}\)
e) \(6\sqrt{\dfrac{1}{3}}+\dfrac{9}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}=2\sqrt{3}+3\sqrt{3}=\left(\sqrt{3}+1\right)=4\sqrt{3}-1\)
f) \(4\sqrt{\dfrac{1}{2}}-\dfrac{6}{\sqrt{2}}.\dfrac{2}{\sqrt{2}+1}=2\sqrt{2}-\left(12-6\sqrt{2}\right)=8\sqrt{2}-12\)
AH
Akai Haruma
Giáo viên
20 tháng 7 2020
Lời giải:
a)
$\sqrt{98}-\sqrt{72}+0.5\sqrt{8}=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}$
$=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}$
b)
$\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}$
$=4\sqrt{a}+4\sqrt{10}.\sqrt{a}-9\sqrt{10}.\sqrt{a}$
$=(4+4\sqrt{10}-9\sqrt{10})\sqrt{a}=(4-5\sqrt{10}).\sqrt{a}$
c)
$(2\sqrt{3}+\sqrt{5})\sqrt{3}-\sqrt{60}=2.3+\sqrt{15}-2\sqrt{15}$
$=6-\sqrt{15}$
d)
$(\sqrt{99}-\sqrt{18}-\sqrt{11})\sqrt{11}+3\sqrt{32}$
$=\sqrt{99}.\sqrt{11}-\sqrt{18}.\sqrt{11}-11+3\sqrt{32}$
$=\sqrt{9}.\sqrt{11}.\sqrt{11}-3\sqrt{2}.\sqrt{11}-11+12\sqrt{2}$
$=3.11+\sqrt{2}(12-3\sqrt{11})-11$
$=22+\sqrt{2}(12-3\sqrt{11})$
21 tháng 6 2019
\(A=10-\left(\sqrt{32}-\sqrt{8}-\sqrt{27}\right)\left(\sqrt{8}-\sqrt{32}-\sqrt{27}\right)\)
\(A=10-\left[-\sqrt{27}+\left(\sqrt{32}-\sqrt{8}\right)\right]\left[-\sqrt{27}-\left(\sqrt{32}-\sqrt{8}\right)\right]\)
\(A=10-\left[\left(-\sqrt{27}\right)^2-\left(\sqrt{32}-\sqrt{8}\right)^2\right]\)
\(A=10-\left(27-8\right)\)
\(A=-9\)
-18
\(\sqrt{2}\left(\sqrt{8}-\sqrt{32}+\sqrt{98}\right)\)
= \(\sqrt{2}\left(\sqrt{4.2}-\sqrt{16.2}+\sqrt{49.2}\right)\)
= \(\sqrt{2}\left(2\sqrt{2}-4\sqrt{2}+7\sqrt{2}\right)\)
= \(\sqrt{2}\left(5\sqrt{2}\right)\)
= \(5.4=20\)