75+12
36+21
94+6
36+24
36+64
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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: \(\dfrac{37}{35}>\dfrac{77}{75}\)
\(\left\{{}\begin{matrix}\dfrac{15}{7}=\dfrac{30}{14}\\\dfrac{9}{14}< \dfrac{30}{14}\end{matrix}\right.\Rightarrow\dfrac{15}{7}>\dfrac{9}{14}\)
\(\left\{{}\begin{matrix}\dfrac{899}{900}=\dfrac{899.1236}{900.1236}=\dfrac{\text{1111164}}{900.1236}\\\dfrac{1235}{1236}=\dfrac{1235.900}{900.1236}=\dfrac{\text{1111500}}{900.1236}>\dfrac{\text{1111164}}{900.1236}\end{matrix}\right.\Rightarrow\dfrac{1235}{1236}>\dfrac{899}{900}\)
\(\left\{{}\begin{matrix}\dfrac{77}{75}=\dfrac{539}{525}\\\dfrac{37}{35}=\dfrac{555}{525}>\dfrac{539}{525}\end{matrix}\right.\Rightarrow\dfrac{77}{73}< \dfrac{37}{35}\)
Dễ
S=(-2194)x(21952195+10001)+(2194+1)x21942194
S=(-2194x21942194)+(-2194x10001)+(21942194x1)+(21942194x2194)
S=(-2194x21942194+2194x21942194)+(10001x(-2194)+1x21942194)
S= 0 + 10001x(-2194)+1x21942194
S= 10001x(-2194)+21942194
S= 10001x(-2194)+10001x2194
S=10001x(-2194x2194)
S=10001x0
S=0
S=2194*21942194+2195*21952195=2194*2194*10001+2195*2195*10001=(21942+21952)*10001
S=-2194.2195.10001+2195.2014.10001
S=2195.10001(-2194+2194)
S=2195.10001.0
S=0
bằng
87
57
100
60
100 nha
ht
75+12=87
36+21=57
94+6=100
36+24=60
36+64=100
chúc bạn học tốt