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1.\(\frac{456}{461}va\frac{123}{128}\)
Ta có: \(\frac{456}{461}+\frac{5}{461}=1\)
\(\frac{123}{128}+\frac{5}{128}=1\)
vì \(\frac{5}{461}< \frac{5}{128}\)nên \(\frac{456}{461}>\frac{123}{128}\)
2.\(\frac{53}{57}va\frac{531}{571}\)
Vì \(\frac{53}{57}< 1\)
\(\Rightarrow\frac{53}{57}=\frac{530}{570}< \frac{530+1}{570+1}=\frac{531}{571}\)
\(\Rightarrow\frac{53}{57}< \frac{531}{571}\)
Tương tự như câu này mình làm giúp ĐỖ KHÁNH LINH rồi. Cậu nhớ tìm hiểu rồi làm nhé! Chỉ cần vận dụng vào đó thôi.
Ta có: \(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}+\frac{1}{90}\)
\(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(B=1-\frac{1}{10}\)
\(B=\frac{10}{10}-\frac{1}{10}\)
\(B=\frac{9}{10}\)
Vậy: \(B=\frac{9}{10}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(B=1-\frac{1}{10}\)
\(B=\frac{9}{10}\)
Vì \(\frac{9}{10}< 1\)nên B < 1
Vậy B < 1
2A=\(\frac{2}{1\cdot2\cdot3}\)+\(\frac{2}{2\cdot3\cdot4}\)+\(\frac{2}{3\cdot4\cdot5}\)+...+\(\frac{2}{2014\cdot2015\cdot2016}\)
2A=\(\frac{1}{1\cdot2}\)-\(\frac{1}{2\cdot3}\)+\(\frac{1}{2\cdot3}\)-\(\frac{1}{3\cdot4}\)+\(\frac{1}{3\cdot4}\)-\(\frac{1}{4\cdot5}\)+...+\(\frac{1}{2014\cdot2015}\)-\(\frac{1}{2015\cdot2016}\)
2A=\(\frac{1}{2}\)-\(\frac{1}{2015\cdot2016}\)
A=(\(\frac{1}{2}\)-\(\frac{1}{2015\cdot2016}\)):2
A=\(\frac{1}{2}\):2-\(\frac{1}{2015\cdot2016}\):2
A=\(\frac{1}{4}\)-\(\frac{1}{2015\cdot2016\cdot2}\)<\(\frac{1}{4}\)
Vậy A<\(\frac{1}{4}\)
Bài 1 :
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}< 1\left(1\right)\)
\(B=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)\)\(>\frac{1}{10}+\frac{1}{100}.90=1\left(2\right)\)
Từ (1) và ( 2) ta có \(A< 1\) \(B>1\)NÊN \(A< B\)
Bài 2:
\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(=\frac{\left(a+b+c\right)-\left(b+c\right)}{b+c}+\)\(\frac{\left(a+b+c\right)-\left(c+a\right)}{c+a}\)\(+\frac{\left(a+b+c\right)-\left(a+b\right)}{a+b}\)
\(=\frac{7-\left(b+c\right)}{b+c}+\frac{7-\left(c+a\right)}{c+a}+\frac{7-\left(a+b\right)}{a+b}\)
\(=7.\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)
\(=7.\frac{7}{10}-3\)\(=\frac{49}{10}-3=\frac{19}{10}\)
\(S=\frac{19}{10}>\frac{19}{11}=1\frac{8}{11}\)
Chúc bạn học tốt ( -_- )
Bài 1:
ta có: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}< 1\)
\(\Rightarrow A< 1\)(1)
ta có: \(\frac{1}{11}>\frac{1}{100};\frac{1}{12}>\frac{1}{100};...;\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\) ( có 90 số 1/100)
\(=\frac{90}{100}=\frac{9}{10}\)
\(\Rightarrow B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{10}+\frac{9}{10}=1\)
\(\Rightarrow B>1\)(2)
Từ (1);(2) => A<B
\(A=\frac{1\cdot1}{1\cdot2}\cdot\frac{2\cdot2}{2\cdot3}\cdot\frac{3\cdot3}{3\cdot4}\cdot\frac{4\cdot4}{4\cdot5}=\frac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot3\cdot4}\cdot\frac{1\cdot2\cdot3\cdot4}{2\cdot3\cdot4\cdot5}=\frac{1}{5}\)
xin lỗi em mới học có lớp 5 thôi
chị kb với em nha!
1/1-1/2=1/2
1/1.2=1/2
=>1/1.2=1/1-1/2