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Câu 2:
230=23.10=(23)10
320=32.10=(32)10
Vì 23 = 8 ; 32 = 9 => 23 < 32 =>(23)10 < (32)10 hay 230 < 320
Vậy...
Đặt \(A=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
\(A>\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}=\frac{3}{14}.5=\frac{15}{14}>1\)
\(A< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{3}{10}.5=\frac{15}{10}=\frac{3}{2}< 2\)
Vậy \(1< A< 2\)
- Ta có:\(\frac{3}{10}>\frac{3}{15};\frac{3}{11}>\frac{3}{15};\frac{3}{12}>\frac{3}{15};\frac{3}{13}>\frac{3}{15};\frac{3}{14}>\frac{3}{15}\)
=>\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}\)
mà \(\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{15}{15}=1\)
=>\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>1\)(1)
- Ta có:\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}\)
mà \(\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{15}{10}< \frac{20}{10}=2\)
=>\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< 2\)(2)
Từ (1) và (2) => \(1< \frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< 2\)
Cách 1:
\(\dfrac{3}{4}=\dfrac{9}{12}\)
\(\dfrac{4}{3}=\dfrac{16}{12}\)
Do đó \(\dfrac{3}{4}< \dfrac{4}{3}\)
Cách 2:
\(\dfrac{3}{4}< 1\)
\(1< \dfrac{4}{3}\)
Do đó \(\dfrac{3}{4}< \dfrac{4}{3}\)
\(-------\)
Cách 1:
\(\dfrac{11}{8}=\dfrac{55}{40}\)
\(\dfrac{7}{10}=\dfrac{28}{40}\)
Do đó \(\dfrac{11}{8}>\dfrac{7}{10}\)
Cách 2:
\(\dfrac{11}{8}>1\)
\(1>\dfrac{7}{10}\)
Do đó \(\dfrac{11}{8}>\dfrac{7}{10}\)
\(\frac{1}{3}< \frac{3}{4}\)
\(\frac{9}{10}< \frac{10}{11}\)
\(\frac{1}{3}< \frac{3}{4}\)
\(\frac{9}{10}< \frac{10}{11}\)
nha