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\(\frac{3737}{4747}\)\(=\frac{37}{47}\)
\(\frac{3737:11=37}{4747:11=47}\)
Hok tốt
ó: 1- 13/27= 14/27
1- 27/41= 14/41
So sánh: 14/27>14/41(cóa chung tử số) => 13/27<27/41
(A<B <=> 1-A>1-B)
\(\frac{3737}{5353}=\frac{37}{53}\)
\(\frac{373737}{535353}=\frac{37}{53}\)
Vì \(\frac{37}{53}=\frac{37}{53}\)nên \(\frac{3737}{5353}=\frac{373737}{535353}\)
Ta có:
\(\frac{3737}{5353}=\frac{37x101}{53x101}=\frac{37}{53}\)
\(\frac{373737}{535353}=\frac{37x10101}{53x10101}=\frac{37}{53}\)
Vì \(\frac{37}{53}=\frac{37}{53}\)
=> \(\frac{3737}{5353}=\frac{373737}{535353}\)
ỦNG HỘ NHA
Ta có: \(\frac{27}{26}\)= \(\frac{27\times37}{26\times37}\)= \(\frac{999}{962}\).
\(\frac{38}{37}\)= \(\frac{38\times26}{37\times26}\)= \(\frac{988}{962}\).
Mà \(\frac{999}{962}\)> \(\frac{988}{962}\).
=> \(\frac{27}{26}\)> \(\frac{38}{37}\).
a) Ta có: \(\dfrac{15}{7}>1\) (tử lớn hơn mẫu)
\(\dfrac{9}{14}< 1\) (tử nhỏ hơn mẫu)
Vậy: \(\dfrac{15}{7}>\dfrac{9}{14}\)
b) Ta có:
\(\dfrac{899}{900}=1-\dfrac{1}{900}\)
\(\dfrac{1235}{1236}=1-\dfrac{1}{1236}\)
Mà: \(\dfrac{1}{900}>\dfrac{1}{1236}\)
Vậy: \(\dfrac{1235}{1236}>\dfrac{899}{900}\)
c) Ta có:
\(\dfrac{77}{75}=1+\dfrac{2}{75}\)
\(\dfrac{37}{35}=1+\dfrac{2}{35}\)
Mà: \(\dfrac{2}{75}< \dfrac{2}{35}\)
Vậy: \(\dfrac{37}{35}>\dfrac{77}{75}\)
\(\left\{{}\begin{matrix}\dfrac{15}{7}=\dfrac{30}{14}\\\dfrac{9}{14}< \dfrac{30}{14}\end{matrix}\right.\Rightarrow\dfrac{15}{7}>\dfrac{9}{14}\)
\(\left\{{}\begin{matrix}\dfrac{899}{900}=\dfrac{899.1236}{900.1236}=\dfrac{\text{1111164}}{900.1236}\\\dfrac{1235}{1236}=\dfrac{1235.900}{900.1236}=\dfrac{\text{1111500}}{900.1236}>\dfrac{\text{1111164}}{900.1236}\end{matrix}\right.\Rightarrow\dfrac{1235}{1236}>\dfrac{899}{900}\)
\(\left\{{}\begin{matrix}\dfrac{77}{75}=\dfrac{539}{525}\\\dfrac{37}{35}=\dfrac{555}{525}>\dfrac{539}{525}\end{matrix}\right.\Rightarrow\dfrac{77}{73}< \dfrac{37}{35}\)
12/18<13/17
vì 12<13 nên 12/18<1/17
13/41<25/77
vì 13<25 nên 13/41<25/77