[Đề đại số lớp 7] Lũy thừa

Y

yukiko1012

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Q

quynhphamdq

a. Ta có :
[TEX]A =\frac{3}{8^3}+\frac{7}{8^4}=\frac{3.8}{8^4}+\frac{7}{8^4}=\frac{31}{8^4}[/TEX]
Và [TEX]B=\frac{7}{8^3}+\frac{3}{8^4}=\frac{7.8}{8^4}+\frac{3}{8^4}=\frac{59}{8^4}[/TEX]
Vì [TEX]\frac{31}{8^4} < \frac{59}{8^4}[/TEX]
[TEX]\Rightarrow A< B [/TEX]
b.Tương tự.
 
I

iceghost

a)$A = \dfrac3{8^3} + \dfrac7{8^4} \\
= \dfrac{24}{8^4} + \dfrac7{8^4} \\
= \dfrac{24+7}{8^4} = \dfrac{31}{8^4} \\
~\\
B = \dfrac7{8^3} + \dfrac3{8^4} \\
= \dfrac{56}{8^4} + \dfrac3{8^4} \\
= \dfrac{56+3}{8^4} = \dfrac{59}{8^4}$

$\implies$ $A < B ( \dfrac{31}{8^4} < \dfrac{59}{8^4} $
 
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