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Bài 1:
ta có: \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
\(B=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{100^2}\)
\(B=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)
\(\Rightarrow B< \frac{1}{4}\)
Bài 2:
ta có: \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Học tốt nhé bn !!
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Ta có:
B=\(\frac{4^2-2^2}{2^2\times4^2}+\frac{6^2-4^2}{4^2\times6^2}+...+\frac{98^2-96^2}{96^2\times98^2}+\frac{100^2-98^2}{98^2\times100^2}\)
=\(\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
= \(\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)
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Quy đồng 2 phân số về dạng mẫu số chung . Ta có :
\(\frac{-64}{85}=\frac{-64.81}{85.81}=\frac{-5184}{6885}\)
\(\frac{-73}{81}=\frac{-73.85}{81.85}=\frac{-6205}{6885}\)
Ta thấy : (-5184) > (-6205) \(\left(\frac{-5184}{6885}>\frac{-6205}{6885}\right)\)
Vậy \(\frac{-64}{85}>\frac{-73}{81}\)
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1. Ta có : Dùng phân số \(\frac{84}{71}\)làm phân số trung gian ta được : \(\frac{84}{81}>\frac{73}{71}\)
2. Ta có : \(\left[\frac{3}{8}\right]^5=\left[\frac{3}{2^3}\right]^5=\frac{243}{2^{15}};\left[\frac{5}{243}\right]^3=\left[\frac{5}{3^5}\right]^3=\frac{125}{3^{15}}\)
Chọn phân số \(\frac{243}{3^{15}}\)làm phân số trung gian để so sánh ta được : \(\frac{243}{2^{15}}>\frac{125}{3^{15}}\)
Vậy : \(\left[\frac{3}{8}\right]^5>\left[\frac{5}{243}\right]^3\)
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Bài 1:
Ta có:
\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)
\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)
Lại có:
\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)
\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)
Bài 2:
Ta có:
\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)
\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\Rightarrow A>B\)
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a) Ta có:
\(\frac{3}{14}=\frac{15}{70}\)
\(\frac{5}{16}=\frac{15}{48}\)
Vì \(48< 70\)
\(\Rightarrow\frac{15}{48}>\frac{15}{70}\)
\(\Rightarrow\frac{5}{16}>\frac{3}{14}\)
b) Vì \(11< 12\)
\(\Rightarrow\frac{7}{11}>\frac{7}{12}\)
\(\frac{388}{6045}\)>\(\frac{16}{81}\)
bạn có thể giải thích vì sao được không