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\(=\left(4\sqrt{2}+3\sqrt{2}-7\sqrt{2}\right)\left(\sqrt{42}+2\sqrt{5}-\sqrt{32}\right)=0.\left(\right)=0\)
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)
$(\sqrt{99}-\sqrt{18}-\sqrt{11})\sqrt{11}+3\sqrt{32}$
$=\sqrt{99}.\sqrt{11}-\sqrt{18}.\sqrt{11}-11+3\sqrt{32}$
$=\sqrt{9}.\sqrt{11}.\sqrt{11}-3\sqrt{2}.\sqrt{11}-11+12\sqrt{2}$
$=3.11+\sqrt{2}(12-3\sqrt{11})-11$
$=22+\sqrt{2}(12-3\sqrt{11})$
a/ \(\left(3-a\right)^2-\sqrt{\frac{180a^2}{5}}=a^2-6a+9-6\left|a\right|\)
Nếu \(a\ge0\) thì \(a^2-6a+9-6\left|a\right|=a^2-12a+9\)
Nếu \(a< 0\) thì \(a^2-6a+9-6\left|a\right|=a^2+9\)
b/ \(\sqrt{150}-3\sqrt{98}+2\sqrt{8}+3\sqrt{32}-5\sqrt{18}\)
\(=5\sqrt{6}-21\sqrt{2}+4\sqrt{2}+12\sqrt{2}-15\sqrt{2}\)
\(5\sqrt{6}-20\sqrt{2}=5\sqrt{2}\left(\sqrt{3}-4\right)\)
c/ Bạn viết lại đề nhé :)
\(C=\sqrt{\sqrt{\left(5+2\right)^2}}-\sqrt{5}\\ =\sqrt{5+2}-\sqrt{5}=\sqrt{7}-\sqrt{5}\)
\(D=\sqrt{8}+\sqrt{18}-\sqrt{32}\\ =\sqrt{2}\left(\sqrt{4}+\sqrt{9}-\sqrt{16}\right)\\ =\sqrt{2}\left(2+3-4\right)\\ =\sqrt{2}\)
\(E=\sqrt{9-4\sqrt{5}}-\sqrt{5}\\ =\sqrt{4+5-4\sqrt{5}}-\sqrt{5}\\ =\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\\ =\sqrt{5}-2-\sqrt{5}\\ =-2\)
\(F=2\sqrt{8}-\sqrt{50}+\sqrt{\left(\sqrt{2}+1\right)^2}\\ =2\sqrt{8}-\sqrt{50}+\sqrt{2}+1\\ =\sqrt{2}\left(2\sqrt{4}-\sqrt{25}+1\right)+1\\ =\sqrt{2}\left(4-5+1\right)+1=1\)
-18
\(\sqrt{2}\left(\sqrt{8}-\sqrt{32}+\sqrt{98}\right)\)
= \(\sqrt{2}\left(\sqrt{4.2}-\sqrt{16.2}+\sqrt{49.2}\right)\)
= \(\sqrt{2}\left(2\sqrt{2}-4\sqrt{2}+7\sqrt{2}\right)\)
= \(\sqrt{2}\left(5\sqrt{2}\right)\)
= \(5.4=20\)