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a) \(\sqrt{\frac{25}{81}\cdot\frac{16}{49}\cdot\frac{169}{9}}\\ =\sqrt{\left(\frac{5}{9}\right)^2\cdot\left(\frac{4}{7}\right)^2\cdot\left(\frac{13}{3}\right)^2}\\ =\sqrt{\left(\frac{5}{9}\cdot\frac{4}{7}\cdot\frac{13}{3}\right)^2}\\ =\frac{5}{9}\cdot\frac{4}{7}\cdot\frac{13}{3}\\ =\frac{260}{189}\)
b) \(\sqrt{3\frac{1}{6}\cdot2\frac{14}{25}\cdot2\frac{34}{81}}\\ =\sqrt{\frac{19}{6}\cdot\frac{64}{25}\cdot\frac{196}{81}}\\ =\sqrt{\frac{19}{6}\cdot\left(\frac{8}{5}\right)^2\cdot\left(\frac{14}{9}\right)^2}\\ =\sqrt{\frac{19}{6}\cdot\left(\frac{8}{5}\cdot\frac{14}{9}\right)^2}\\ =\sqrt{\frac{19}{6}\cdot\frac{112}{45}}\\ =\sqrt{\frac{1064}{135}}\)
Bổ sung câu b :
\(\sqrt{3\frac{1}{16}.2\frac{14}{25}.2\frac{34}{81}}=\sqrt{\frac{49}{16}.\frac{64}{25}.\frac{196}{81}}=\sqrt{\frac{49}{16}}.\sqrt{\frac{64}{25}}.\sqrt{\frac{196}{81}}=\frac{7}{4}.\frac{8}{5}.\frac{14}{9}=\frac{196}{45}\)
Ta có:
\(\frac{n\left(n+2\right)}{\left(n+1\right)^2}=1-\frac{1}{\left(n+1\right)^2}>1-\frac{1}{n\left(n+2\right)}=1+\frac{1}{2}.\left(\frac{1}{n+2}-\frac{1}{n}\right)\)
Thế vô bài toán ta được
\(B=\frac{2.4}{3^2}+\frac{4.6}{5^2}+...+\frac{200.202}{201^2}\)
\(>1+1+...+1+\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{2}+\frac{1}{6}-\frac{1}{4}+...+\frac{1}{202}-\frac{1}{200}\right)\)
\(=100+\frac{1}{2}.\left(\frac{1}{202}-\frac{1}{2}\right)=\frac{10075}{101}>99,75\)
Ta có đánh giá sau:\(\frac{n\left(n+2\right)}{\left(n+1\right)^2}=1-\frac{1}{\left(n+1\right)^2}\)
\(>1-\frac{1}{x\left(x+2\right)}=1-\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)\)
Suy ra \(B=\frac{2\cdot4}{3^2}+\frac{4\cdot6}{5^2}+\frac{6\cdot8}{7^2}+...+\frac{200\cdot202}{201^2}\)
\(>1-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)+1-\frac{1}{2}\left(\frac{1}{4}-\frac{1}{6}\right)+...+1-\frac{1}{2}\left(\frac{1}{200}-\frac{1}{202}\right)\)
\(=100-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{200}-\frac{1}{202}\right)\)
\(=100-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{202}\right)\)\(=100-\frac{1}{2}\cdot\frac{50}{101}\)
\(>100-\frac{1}{2}\cdot\frac{50}{100}=100-0,25=99,75\)
Tức là \(B>99,75\)