122 + 123 + 124 + 125
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A = \(\dfrac{3^{123}+1}{3^{125}+1}\) Vì 3123 + 1 < 2125 + 1 Nên A = \(\dfrac{3^{123}+1}{3^{125}+1}\)< \(\dfrac{3^{123}+1+2}{3^{125}+1+2}\)
A < \(\dfrac{3^{123}+3}{3^{125}+3}\) = \(\dfrac{3.\left(3^{122}+1\right)}{3.\left(3^{124}+1\right)}\) = \(\dfrac{3^{122}+1}{3^{124}+1}\) = B
Vậy A < B
(1/2+2/3+3/4+4/5+...+122/123+123/124).(125-5.25)
=(1/2+2/3+3/4+4/5+...+122/123+123/124).(125-125)
=(1/2+2/3+3/4+4/5+...+122/123+123/124).0=0
\(A=\dfrac{3^{123}+1}{3^{125}+1}\Leftrightarrow3^2A=\dfrac{3^{125}+9}{3^{125}+1}\)
\(9A=\dfrac{3^{125}+1}{3^{125}+1}+\dfrac{8}{3^{125}+1}=1+\dfrac{8}{3^{125}+1}\)
\(B=\dfrac{3^{122}+1}{3^{124}+1}\Leftrightarrow3^2B=\dfrac{3^{124}+9}{3^{124}+1}\)
\(9B=\dfrac{3^{124}+1+8}{3^{124}+1}+\dfrac{3^{124}+1}{3^{124}+1}+\dfrac{8}{3^{124}+1}=1+\dfrac{8}{3^{124}+1}\)
\(9A< 9B\Leftrightarrow A< B\)
Phep so sanh nao dung?
A. 123/124 > 124/125
B. 123/124=124/125
C. 123/124<124/125
D. 123/124 = 124-1/125
122+123+124+125
=245+124+125
=369+125
=494
122 + 123 + 124 + 125
= 245 + 124 + 125
= 369 + 125
= 494