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\(\dfrac{4}{17}=\dfrac{16}{68}\\ Vì:\dfrac{16}{68}< \dfrac{16}{63}\Rightarrow\dfrac{4}{17}< \dfrac{16}{63}\\ ---\\ \dfrac{1007}{1009}=1-\dfrac{2}{1009};\dfrac{1005}{1007}=1-\dfrac{2}{1007}\\ Vì:\dfrac{2}{1009}< \dfrac{2}{1007}\Rightarrow1-\dfrac{2}{1009}>1-\dfrac{2}{1007}\\ \Rightarrow\dfrac{1007}{1009}>\dfrac{1005}{1007}\)
a: 4/17=16/68
16/68<16/63
=>4/17<16/63
b: 19/53<20/53
20/53<20/50(Vì 53>50)
=>19/53<20/50=2/5
mà 2/5=30/75<30/73
nên 19/53<30/73
c: 1007/1009=1-2/1009
1005/1007=1-2/1007
1009>1007
=>2/1009<2/1007
=>-2/1009>-2/1007
=>1007/1009>1005/1007
a ) Ta có :
\(1-\frac{41}{91}=\frac{50}{91}\) \(=\frac{500}{910}\) ; \(1-\frac{411}{911}=\frac{500}{911}\)
Vì \(\frac{500}{910}>\frac{500}{911}\)nên \(\frac{41}{91}< \frac{411}{911}\)
b ) Ta có :
\(1-\frac{113}{115}=\frac{2}{115}\) ; \(1-\frac{93}{95}=\frac{2}{95}\)
Vì \(\frac{2}{115}< \frac{2}{95}\)nên \(\frac{113}{115}>\frac{93}{95}\).
c ) Quy đồng TS ta có :
\(\frac{13}{53}=\frac{143}{583}\) ; \(\frac{11}{30}=\frac{143}{390}\)
Vì \(\frac{143}{583}< \frac{143}{390}\)nên \(\frac{13}{53}< \frac{11}{30}\).
1) (-25).8.(-4).(-125)
= [(-25).4].[(-125).8]
=(-100).(-1000)
=100000
2)30.(5-10)
=30.5-30.10
=30.5-300
=150-300
=(-150)
3) 43.(53-81)+53.(81-43)
=43.53-43.81+53.81-53.43
=(43.53-53.43)+(43.81-53.81)
=0+81.(43-53)
=(-810)
4) Mk chưa biết làm nha
1) (-25).8.(-4).(-125)
= [ (-25) . (-4) ] . [ 8 . (-125) ]
= 100 . (-1000)
= -100 000
2) 30.(-5-10)
= 30 . (-15)
= -450
3) 43.(53-81)+53.(81-43)
= 43 . 53 - 43 . 81 + 53 . 81 - 53.43
= -43 . 81 + 53 . 81
= 81 . ( -43 + 53)
= 81 . 10
= 810
4) (-6)².(-17)+23.(-36)-72.30
= 36 . (-17) + 23 . (-36) - 72. 30
= (-36) . 17 + 23 . (-36) - 72 . 30
= (-36) . ( 17 + 23 ) - 72 . 30
= (-36) . 40 - 72.30
= (-1440 )- 2160
= -3600
# HOK TỐT #
\(\dfrac{4}{7}v\text{à }\dfrac{16}{63}\\ \dfrac{4}{7}=\dfrac{4\cdot9}{7\cdot9}=\dfrac{36}{63}\\ \dfrac{36}{63}>\dfrac{16}{63}\\ \Rightarrow\dfrac{4}{7}>\dfrac{16}{36}\)
\(\dfrac{4}{17}\) và \(\dfrac{16}{63}\)
\(\dfrac{4}{63}>\dfrac{16}{63}\)
\(=>\dfrac{4}{17}>\dfrac{16}{63}\)
\(\dfrac{5}{29}\) và \(\dfrac{7}{33}\)
\(\dfrac{5}{33}< \dfrac{7}{33}\)
\(=>\dfrac{5}{29}< \dfrac{7}{33}\)
\(\dfrac{44}{57}\) và \(\dfrac{89}{99}\)
\(\dfrac{44}{99}< \dfrac{89}{99}\)
\(=>\dfrac{44}{57}< \dfrac{89}{99}\)
\(\dfrac{19}{53}\) và \(\dfrac{30}{73}\)
\(\dfrac{19}{73}>\dfrac{30}{73}\)
\(=>\dfrac{19}{53}>\dfrac{30}{73}\)
\(A=\frac{53^{30}}{53^{30}-6}=\frac{53^{30}-6+6}{53^{30}-6}=1+\frac{6}{53^{30}-6}\)
\(B=\frac{53^{30}-4}{53^{30}-2}=\frac{53^{30}-2+6}{53^{30}-2}=1+\frac{6}{53^{30}-2}\)
Vì 5330-6<5330-2 nên \(\frac{6}{53^{30}-6}>\frac{6}{53^{30}-2}\)
=> A>B
Ta có:
A=\(\frac{53^{30}-6+6}{53^{30}-6}\)=\(1-\frac{6}{5^{30}-6}\)
B=\(\frac{53^{30}-2+6}{53^{30}-2}\)=\(1-\frac{6}{5^{30}-2}\)
Vì 5330-6<5330-2 nên \(\frac{6}{53^{30}-6}\)>\(\frac{6}{53^{30}-2}\)
=> \(1-\frac{6}{53^{30}-6}\)\(< 1-\frac{6}{5^{30}-2}\)
Vậy A<B