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Khách
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TP
3
14 tháng 2 2022
\(\dfrac{1}{2}=\dfrac{6}{12};\dfrac{2}{3}=\dfrac{8}{12};\dfrac{7}{12}=\dfrac{7}{12}\)
14 tháng 2 2022
Chắc là quy đồng nhỉ :)
\(\dfrac{1\times6}{2\times6}=\dfrac{6}{12};\dfrac{2\times4}{3\times4}=\dfrac{8}{12};\dfrac{7}{12}\)
TT
4
K
26 tháng 2 2022
a) \(\dfrac{5}{4}+\dfrac{3}{5}=\dfrac{25}{20}+\dfrac{12}{20}=\dfrac{37}{20}\)
b) \(\dfrac{7}{8}+\dfrac{9}{24}=\dfrac{21}{24}+\dfrac{9}{24}=\dfrac{30}{24}=\dfrac{15}{12}\)
c) \(\dfrac{2}{36}+\dfrac{1}{6}+\dfrac{5}{12}=\dfrac{2}{36}+\dfrac{6}{36}+\dfrac{15}{36}=\dfrac{23}{36}\)
TV
1
6 tháng 11 2021
\(a,=7xy\left(x^2-2xy+y^2\right)=7xy\left(x-y\right)^2\\ b,=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\\ c,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
1 tháng 2
c: Đặt \(A=\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\)
Ta có: \(\frac{1}{51}>\frac{1}{75};\frac{1}{52}>\frac{1}{75};..;\frac{1}{75}=\frac{1}{75}\)
Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+\cdots+\frac{1}{75}=\frac{25}{75}=\frac13\) (2)
Ta có: \(\frac{1}{76}>\frac{1}{100};\frac{1}{77}>\frac{1}{100};\ldots;\frac{1}{100}=\frac{1}{100}\)
Do đó: \(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\cdots+\frac{1}{100}=\frac{25}{100}=\frac14\) (1)
Từ (1),(2) suy ra \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}+\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}>\frac13+\frac14\)
=>\(A>\frac{7}{12}\)
=>\(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}>\frac{7}{12}\)
=>\(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}>\frac{7}{12}+\frac{1}{50}>\frac{7}{12}\) (ĐPCM)
a: \(\frac{1}{26}>\frac{1}{50};\frac{1}{27}>\frac{1}{50};\ldots;\frac{1}{50}=\frac{1}{50}\)
Do đó: \(\frac{1}{26}+\frac{1}{27}+\cdots+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{25}{50}=\frac12\) (1)
Ta có: \(\frac{1}{26}<\frac{1}{25};\frac{1}{27}<\frac{1}{25};\ldots;\frac{1}{50}<\frac{1}{25}\)
Do đó: \(\frac{1}{26}+\frac{1}{27}+\cdots+\frac{1}{50}<\frac{1}{25}+\frac{1}{25}+\cdots+\frac{1}{25}=\frac{25}{25}=1\) (2)
Từ (1),(2) suy ra \(\frac12<\frac{1}{26}+\frac{1}{27}+\cdots+\frac{1}{50}<1\) (ĐPCM)
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