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\(\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3.3}< \frac{1}{2.3}\)
......
\(\frac{1}{100.100}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{100.100}< \frac{1}{1.2}+..+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2.2}+..+\frac{1}{100.100}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2.2}+..+\frac{1}{100.100}< 1-\frac{1}{100}< 1\).Suy ra điều phải chứng minh. câu b tương tự. bấm đúng cho mình nha
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Vì \(\frac{1}{33}>\frac{1}{34}>\frac{1}{35}>\frac{1}{36}\)
\(\Rightarrow M>\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\)\(\)
\(\Rightarrow M>\frac{4}{36}=\frac{1}{9}\)
Mà \(\frac{1}{9}>\frac{1}{10}\)
\(\Rightarrow\)\(M>\frac{1}{9}>\frac{1}{10}\)
Vậy : M > N
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Ta có\(A=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}\right)\)\(>1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+4\times\frac{1}{8}+4\times\frac{1}{12}+4\times\frac{1}{16}\)
\(=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(=1+2\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
\(>1+2\times\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\right)=1+2=3=B\)
\(\Rightarrow A>B\)
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a)\(3\frac{4}{10}>2\frac{9}{10}\)
b)\(3\frac{4}{10}< 3\frac{9}{10}\)
c)\(5\frac{1}{10}>2\frac{9}{10}\)
d)\(3\frac{4}{10}=3\frac{2}{5}\)
Hok tốt!~
Toán 6 nha mn
Toán lớp 6 nha mn