\(\left(\sqrt{1\frac{9}{16}}-\sqrt{\frac{9}{16}}\right)\)
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\(b,\left(\sqrt{1\frac{9}{16}-\sqrt{\frac{9}{16}}}\right):5\)
\(=\left(\sqrt{\frac{25}{16}-\frac{3}{4}}\right):5\)
\(=\sqrt{\frac{13}{16}}:5\)
\(=\frac{\sqrt{13}}{4}:5\)
\(=\frac{\sqrt{13}}{20}\)
\(\sqrt{\frac{25}{4}}+\left(\sqrt{\frac{1}{2}}\right)^2:\left(\frac{-\sqrt{9}}{4}\right).\sqrt{\frac{16}{81}}-4^2-\left(-2\right)^3\)
\(=\frac{5}{2}+\frac{1}{2}:\frac{-3}{4}.\frac{4}{9}-16+8\)
\(=\frac{5}{2}-\frac{8}{27}-8\)
\(=\frac{-313}{54}\)
\(0,5\sqrt{100}-\sqrt{\frac{4}{25}}=0,5.10-\frac{\sqrt{4}}{\sqrt{25}}=5-\frac{2}{5}=\frac{23}{5}=\frac{138}{30}\)
\(\left(\sqrt{1\frac{1}{9}-\sqrt{\frac{9}{16}}}\right):5=\left(\sqrt{\frac{10}{9}-\frac{3}{4}}\right):5=\sqrt{\frac{13}{36}}:5=\frac{\sqrt{13}}{6}:5=\frac{\sqrt{13}}{30}\)
Vì 13 < 138 nên \(\sqrt{13}< 138\Rightarrow\frac{\sqrt{13}}{30}< \frac{138}{30}\)
Vậy \(0,5\sqrt{100}-\sqrt{\frac{4}{25}}>\left(\sqrt{1\frac{1}{9}-\sqrt{\frac{9}{16}}}\right):5\).
1. a) 3+2=5
b) 0,5-0,1=0,4
c) 4/5-1/9=31/45
d) 2-0,6=1,4
2. a) 8-4+3=7
b) 11+5-3=13
c) 3/2-4/6-7-37/6
d) 4+5-6=3
Bài 1:
a) Ta có: \(\frac{12\sqrt{50}-8\sqrt{200}+7\sqrt{450}}{\sqrt{10}}\)
\(=\frac{12\cdot\sqrt{5}\cdot\sqrt{10}-8\cdot\sqrt{20}\cdot\sqrt{10}+7\cdot\sqrt{45}\cdot\sqrt{10}}{\sqrt{10}}\)
\(=\frac{\sqrt{10}\left(12\sqrt{5}-8\sqrt{20}+7\sqrt{45}\right)}{\sqrt{10}}\)
\(=12\sqrt{5}-8\sqrt{20}+7\sqrt{45}\)
\(=\sqrt{5}\left(12-16+21\right)\)
\(=17\sqrt{5}\)
b) Ta có: \(\frac{\frac{\sqrt{1}}{7}-\sqrt{\frac{16}{7}}+\sqrt{\frac{9}{7}}}{\sqrt{7}}\)
\(=\left(\frac{1}{\sqrt{7}}-\frac{4}{\sqrt{7}}+\frac{3}{\sqrt{7}}\right)\cdot\frac{1}{\sqrt{7}}\)
\(=0\cdot\frac{1}{\sqrt{7}}=0\)
\(\left(\sqrt{1\frac{9}{16}}-\sqrt{\frac{9}{16}}\right)\)
\(=\left(\sqrt{\frac{25}{16}}-\sqrt{\frac{9}{16}}\right)\)
\(=\left(\sqrt{\left(\frac{5}{4}\right)^2}-\sqrt{\left(\frac{3}{4}\right)^2}\right)\)
\(=\left(\frac{5}{4}-\frac{3}{4}\right)=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}\)
...Vậy ...................
\(\sqrt{1\frac{9}{16}}-\sqrt{\frac{9}{16}}=\sqrt{\frac{25}{16}}-\sqrt{\frac{9}{16}}\)
\(=\sqrt{\left(\frac{5}{4}\right)^2}-\sqrt{\left(\frac{3}{4}\right)^2}=\frac{5}{4}-\frac{3}{4}=\frac{2}{4}=\frac{1}{2}\)
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