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a) \(\sqrt{125}+\sqrt{\left(-14\right)^2}-\sqrt{225}=5\sqrt{5}+14-15=-1+5\sqrt{5}\)
b) \(\sqrt{\frac{9}{49}}.\sqrt{\left(\frac{-1}{3}\right)^2}+\sqrt{\frac{4}{9}}=\frac{3}{7}.\frac{1}{3}+\frac{2}{3}=\frac{17}{21}\)
Bài 1:
a) Ta có: \(\left(0.125\right)\cdot\left(-3\cdot7\right)\cdot\left(-2\right)^3\)
\(=\frac{1}{8}\cdot\left(-21\right)\cdot\left(-8\right)\)
\(=\frac{1}{8}\cdot168\)
\(=21\)
b) Ta có: \(\sqrt{36}\cdot\sqrt{\frac{25}{16}}+\frac{1}{4}\)
\(=\sqrt{36\cdot\frac{25}{16}}+\frac{1}{4}\)
\(=\sqrt{\frac{225}{4}}+\frac{1}{4}\)
\(=\frac{15}{2}+\frac{1}{4}\)
\(=\frac{31}{4}\)
c) Ta có: \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)
\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)
\(=\frac{2}{5}-\frac{7}{5}=-1\)
d) Ta có: \(0,1\cdot\sqrt{225}\cdot\sqrt{\frac{1}{4}}\)
\(=0,1\cdot15\cdot\frac{1}{2}=\frac{3}{4}\)
cảm ơn bạn nhé 
\(A=\frac{1}{\sqrt{2.1}\left(\sqrt{2}+\sqrt{1}\right)}+\frac{1}{\sqrt{2.3}\left(\sqrt{3}+\sqrt{2}\right)}+\frac{1}{\sqrt{3.4}\left(\sqrt{4}+\sqrt{3}\right)}+...+\frac{1}{\sqrt{999.1000}\left(\sqrt{1000}+\sqrt{999}\right)}\)
\(A=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2.1}\left(2-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2.3}\left(3-2\right)}+\frac{\sqrt{4}-\sqrt{3}}{\sqrt{3.4}\left(4-3\right)}+...+\frac{\sqrt{1000}-\sqrt{999}}{\sqrt{999.1000}\left(1000-999\right)}\)
\(A=\frac{\sqrt{2}}{\sqrt{2.1}}-\frac{\sqrt{1}}{\sqrt{2.1}}+\frac{\sqrt{3}}{\sqrt{2.3}}-\frac{\sqrt{2}}{\sqrt{2.3}}+\frac{\sqrt{4}}{\sqrt{3.4}}-\frac{\sqrt{3}}{\sqrt{3.4}}+...+\frac{\sqrt{1000}}{\sqrt{999.1000}}-\frac{\sqrt{999}}{\sqrt{1000.999}}\)
\(A=\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{999}}-\frac{1}{\sqrt{1000}}\)
\(A=\frac{1}{1}-\frac{1}{\sqrt{1000}}=\frac{\sqrt{1000}-1}{\sqrt{1000}}=\frac{10\sqrt{10}-1}{10\sqrt{10}}\)
Sorry mới lớp 6 chưa học
thông cảm
no chửi
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vào bài toán ta được
\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{225\sqrt{224}+224\sqrt{225}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{224}}-\frac{1}{\sqrt{225}}\)
\(=1-\frac{1}{\sqrt{225}}=1-\frac{1}{15}=\frac{14}{15}\)