\(\dfrac{37}{-49}\) so sánh với\(\dfrac{-12}{-35}\) bằng cách so sánh với số trung gian
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a: \(\dfrac{12}{49}< \dfrac{13}{49}< \dfrac{13}{47}\)
b: \(\dfrac{12}{47}>\dfrac{19}{47}>\dfrac{19}{77}\)
`a)75<77`
`=>1/75>1/77`
`=>37/75>37/77>35/77`
`b)7779>7569`
`=>1/7779<1/7569`
`=>3734/7569>3734/7779>2099/7779`
\(a,\dfrac{11}{49}< \dfrac{11}{46};\dfrac{11}{46}< \dfrac{13}{46}\\ Nên:\dfrac{11}{49}< \dfrac{13}{46}\\ b,\dfrac{62}{85}< \dfrac{62}{80};\dfrac{62}{80}< \dfrac{73}{80}\\ Nên:\dfrac{62}{85}< \dfrac{73}{80}\\ c,\dfrac{n}{n+3}< \dfrac{n}{n+2};\dfrac{n}{n+2}< \dfrac{n+1}{n+2}\\ Nên:\dfrac{n}{n+3}< \dfrac{n+1}{n+2}\)
c) \(\dfrac{27}{26}\)và\(\dfrac{38}{37}\)
Ta có: \(\dfrac{27}{26}=1+\dfrac{1}{26}\); \(\dfrac{38}{37}=1+\dfrac{1}{37}\)
Vì \(\dfrac{1}{26}>\dfrac{1}{37}\) nên \(\dfrac{27}{26}>\dfrac{38}{37}\)
c: \(\dfrac{27}{26}-1=\dfrac{1}{26}\)
\(\dfrac{38}{37}-1=\dfrac{1}{37}\)
mà \(\dfrac{1}{26}>\dfrac{1}{37}\)
nên \(\dfrac{27}{26}>\dfrac{38}{37}\)
\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)
\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)
\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)
Áp dụng t/c dtsbn:
\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)
Ta có: 37/-49 < 0; -12/-35 > 0
Suy ra 37/-49 < -12/-35
Ta có:
\(\dfrac{37}{-49}< 0;\dfrac{-12}{-35}=\dfrac{12}{35}>0\)
\(\Rightarrow\dfrac{37}{-49}< \dfrac{-12}{-35}\)
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