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\(a,\sqrt{81}=9\)
\(b.\sqrt{8100}=90\)
\(c,\sqrt{64}=8\)
\(d,\sqrt{25}=5\)
\(e,\sqrt{0,64}=0,8\)
\(f,\sqrt{10000}=100\)
\(g,\sqrt{0,01}=0,1\)
\(h,\sqrt{\frac{49}{100}}=\frac{7}{10}\)
\(i,\sqrt{\frac{0,09}{121}}=\frac{0,3}{11}\)
\(j,\sqrt{\frac{4}{25}}=\frac{2}{5}\)
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a) 9
b) 90
c) 8
d) 5
e) 0,8
f) 100
g) 0,1
h) \(\frac{7}{10}\)
i) \(\frac{0,3}{11}\)
j) 0,4.
\(\sqrt{81}=9\)
\(\sqrt{0,64}=0,8\)
\(\sqrt{\frac{49}{100}}=\frac{7}{10}\)
\(\sqrt{8100}=90\)
\(\sqrt{100=}10\)
\(\sqrt{0,01}=0,1\)
\(\sqrt{\frac{4}{25}}=\frac{2}{5}\)
\(\sqrt{\frac{0,09}{121}}=\frac{0,3}{11}\)
\(\sqrt{81}=9\);\(\sqrt{0,64}=0,8\);\(\sqrt{\frac{49}{100}}=\frac{7}{10}\);\(\sqrt{8100}=90\); \(\sqrt{100}=10\); \(\sqrt{0,01}=0,1\); \(\sqrt{\frac{4}{25}}=\frac{2}{5}\); \(\sqrt{\frac{0,09}{121}}=\frac{0,3}{11}=\frac{3}{110}\)
=1-2+3-4+5-6+...+19-20
=-1-1-1-1-1-1-...-1
20 so-1
=-1.20=-20
Câu a)
\(A=\sqrt{20+1}+\sqrt{40+2}+\sqrt{60+3}\)
\(=\sqrt{1\left(20+1\right)}+\sqrt{2\left(20+1\right)}+\sqrt{3\left(20+1\right)}\)
\(=\sqrt{20+1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
\(B=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{20}+\sqrt{40}+\sqrt{60}\)
\(=1\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{1}\cdot\sqrt{20}+\sqrt{2}\cdot\sqrt{20}+\sqrt{3}\cdot\sqrt{20}\right)\)
\(=\sqrt{1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\sqrt{20}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
\(=\left(\sqrt{20}+\sqrt{1}\right)\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
Ta thấy: \(\hept{\begin{cases}\left(\sqrt{20+1}\right)^2=20+1\\\left(\sqrt{20}+\sqrt{1}\right)^2=20+1+2\sqrt{20}\end{cases}}\)
\(\Rightarrow\left(\sqrt{20+1}\right)^2< \left(\sqrt{20}+\sqrt{1}\right)^2\Rightarrow\sqrt{20+1}< \sqrt{20}+\sqrt{1}\)
Vậy A < B.
a, Ta có: \(\sqrt{36}=6\)
Vì \(36>35\Rightarrow\sqrt{36}>\sqrt{35}\) hay \(6>\sqrt{35}\)
\(\sqrt{8100=90}\) \(\sqrt{3136=56}\)
\(\sqrt{3364=58}\) \(\sqrt{722500=850}\)
\(\sqrt{900=30}\)
\(\sqrt{3969=63}\)