So sánh 9^15 và 35^15
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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: \(\dfrac{37}{35}>\dfrac{77}{75}\)
\(\left\{{}\begin{matrix}\dfrac{15}{7}=\dfrac{30}{14}\\\dfrac{9}{14}< \dfrac{30}{14}\end{matrix}\right.\Rightarrow\dfrac{15}{7}>\dfrac{9}{14}\)
\(\left\{{}\begin{matrix}\dfrac{899}{900}=\dfrac{899.1236}{900.1236}=\dfrac{\text{1111164}}{900.1236}\\\dfrac{1235}{1236}=\dfrac{1235.900}{900.1236}=\dfrac{\text{1111500}}{900.1236}>\dfrac{\text{1111164}}{900.1236}\end{matrix}\right.\Rightarrow\dfrac{1235}{1236}>\dfrac{899}{900}\)
\(\left\{{}\begin{matrix}\dfrac{77}{75}=\dfrac{539}{525}\\\dfrac{37}{35}=\dfrac{555}{525}>\dfrac{539}{525}\end{matrix}\right.\Rightarrow\dfrac{77}{73}< \dfrac{37}{35}\)
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a: 6^9=(6^3)^3=216^3>15^3
B: 6^36=(6^2)^18=36^18>35^18
c: 7^18=(7^2)^9=49^9>30^9
d: 3^500=243^100
7^300=343^100
=>3^500<7^300
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a) 35 : 5 và (35 x 4) : (5 x 4)
35:5= 7
(35 x 4) : (5 x 4) = 140 : 20 =7
=> 35 : 5 = (35 x 4) : (5 x 4)
b) 105 : 15 và (105 : 5) : (15 : 5)
105:15 =7
(105 : 5) : (15 : 5)= 7 : 3= 2 (dư 1)
=>105 : 15 > (105 : 5) : (15 : 5)
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a) − 1 5 + 4 − 5 = − 1 < 1 nên − 1 5 + 4 − 5 < 1
b) 2 3 + − 1 5 = 7 15 ; 3 5 = 9 15 mà 7 15 < 9 15 nên 3 5 > 2 3 + − 1 5
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21/35=3/5; 12/20=3/5
=>21/35=12/20
24/35<1<21/7
35/49=5/7; 40/64=5/8
=>35/49>40/64
2424/4848=1/2
9/11=1-2/11
13/15=1-2/15
mà 2/11>2/15
nên 9/11<13/15
19/15>11/15
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9 ^ 15 và 35 ^ 15
Vì 9 < 35
=> 9 ^ 15 < 35 ^ 15
Vậy 9 ^ 15 < 35 ^ 15
de thay 9<15 suy ra 9^15<35^15