So sánh hai câu này cho mik với 27/31 và 2727/3131 , 11/31 và 111/311 chi tiết nhé để mik làm bài
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`#3107.101107`
`a)`
Ta có:
\(\dfrac{2727}{3131}=\dfrac{2727\div27}{3131\div31}=\dfrac{27}{31}\)
Vì \(\dfrac{27}{31}=\dfrac{27}{31}\)
\(\Rightarrow\dfrac{27}{31}=\dfrac{2727}{3131}\)
`b)`
Ta có:
\(\dfrac{11}{31}=1-\dfrac{20}{31}=1-\dfrac{200}{310}\)
\(\dfrac{111}{311}=1-\dfrac{200}{311}\)
Vì \(\dfrac{200}{310}>\dfrac{200}{311}\)
\(\Rightarrow1-\dfrac{200}{310}< 1-\dfrac{200}{311}\)
\(\Rightarrow\dfrac{11}{31}< \dfrac{111}{311}.\)
\(a,\dfrac{2727}{3131}=\dfrac{2727:101}{3131:101}=\dfrac{27}{31}\\ Vậy:\dfrac{27}{31}=\dfrac{2727}{3131}\)
\(\dfrac{2727}{3131}=\dfrac{27.101}{31.101}=\dfrac{27}{31}\)
⇒\(\dfrac{27}{31}=\dfrac{2727}{3131}\)
\(\dfrac{27}{31}=\dfrac{27\times101}{31\times101}=\dfrac{2727}{3131}\)
\(\dfrac{11}{31}\) và \(\dfrac{111}{311}\)
\(\dfrac{11}{31}\) = \(\dfrac{11\times10}{31\times10}\) = \(\dfrac{110}{310}\) = 1 - \(\dfrac{200}{310}\)
\(\dfrac{111}{311}\) = 1 - \(\dfrac{200}{311}\)
Vì \(\dfrac{200}{310}\) > \(\dfrac{200}{311}\)
Nên \(\dfrac{11}{31}\) < \(\dfrac{111}{311}\)
\(\dfrac{11}{31}=\dfrac{11x311}{31x311}=\dfrac{3421}{31x311}\)
\(\dfrac{111}{311}=\dfrac{111x31}{31x311}=\dfrac{3441}{31x311}\)
mà \(\dfrac{3421}{31x311}< \dfrac{3441}{31x311}\)
\(\Rightarrow\dfrac{11}{31}< \dfrac{111}{311}\)
a)1714>1614=256>3211=222>3111
b)102330<102430=2300<2305=3261<3361
c)8217>8117=368>363=2721>2621
\(A=\frac{19^{30}+5}{19^{31}+5}=>19A=\frac{19^{31}+95}{19^{31}+5}=1+\frac{90}{19^{31}+5}\left(1\right)\)
\(B=\frac{19^{31}+5}{19^{32}+5}=>19B=\frac{19^{32}+95}{19^{32}+5}=1+\frac{90}{19^{32}+5}\left(2\right)\)
từ (1) and (2)
=>19A>19B
=>A>B
Ta có:
19A=19^31+95/19^31+5
19A= (19^31+5)+90/19^31+5
19A=1+90/19^31+5
19B=19^32+95/19^32+5
19B=(19^32+5)+90/19^32+5
19B=1+90/19^32+5
Vì: 90/19^31+5>90/19^31+5 nên 19A>19B hay A>B
27/31<2727/3131
11/31<111/311