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a. 6/7 = 6x9/7x9=54/63
7/9=7x7/9x7=49/63
do 54/73>49/63 => 6/7>7/9
b.8/11=8x9/11x9=72/99
15/9=15x11/9x11=165/99
do 72/99<165/99=>8/11<15/9
`a,`
`5/6=1-1/6`
`7/8=1-1/8`
Mà `1/6>1/8 -> 5/6<7/8`
`b,`
`9/5=(9 \times 2)/(5 \times 2)=18/10`
`3/2=(3 \times 5)/(2 \times 5)=15/10`
`18/10 > 15/10 -> 9/5 > 3/2`
`c,`
`2017/2018 = 1-1/2018`
`2019/2020=1-1/2020`
`1/2018 > 1/2020 -> 2017/2018 < 2019/2020`
`d,`
`2018/2017 = 1+1/2017`
`2020/2019 = 1+1/2019`
`1/2017 > 1/2019 -> 2018/2017>2020/2019`
a) \(\dfrac{12}{14}=\dfrac{1200}{1400}=\dfrac{1400-200}{1400}=1-\dfrac{200}{1400}\)
\(\dfrac{1212}{1414}=\dfrac{1414-200}{1414}=1-\dfrac{200}{1414}\)
vì \(\dfrac{200}{1414}< \dfrac{200}{1400}\)
Nên \(1-\dfrac{200}{1400}< 1-\dfrac{200}{1414}\)
Vậy \(\dfrac{12}{14}< \dfrac{1212}{1414}\)
Các bài sau tương tự
\(\frac{4}{5};\frac{5}{6};\frac{7}{8};\frac{6}{5}\)
Đúng 100%
Đúng 100%
Đúng 100%
`3/7-2/5`
`=1/35>0`
`=>3/7>2/5`
`b,9>8`
`=>1/9<1/8`
`=>5/9<5/8`
`d,8/7>1`
`7/8<1`
`=>8/7>7/8`
\(\frac{75}{100}=\frac{3}{4}=\frac{6}{8}< \frac{7}{8}\)
\(\frac{7}{8}< 1< \frac{3}{2}\)
\(\frac{60}{108}=\frac{5}{9}=\frac{15}{27}>\frac{15}{37}\)
\(\frac{15}{37}=\frac{30}{74}< \frac{31}{74}< \frac{31}{54}\)
\(\frac{0}{16}=\frac{0}{21}\)
Xét \(1-\frac{1965}{1967}=\frac{2}{1967}>\frac{2}{1975}=1-\frac{1973}{1975}\Rightarrow\frac{1965}{1967}< \frac{1973}{1975}\)
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 \(\dfrac{11}{8}>\dfrac{7}{10}\)
