toan nang cao

S

shinxun

A = $\frac{9999999 + 1}{9999999 - 1}$ = $\frac{9999999 - 1 + 2}{9999999 - 1}$ = 1 + $\frac{2}{9999999-1}$ = 1 + $\frac{2}{9999998}$

B = $\frac{9999999 - 3}{9999999 - 1}$ = $\frac{9999999 - 1 - 2}{9999999 - 1}$ = 1 - $\frac{2}{9999999-1}$ = 1 - $\frac{2}{9999998}$

\Rightarrow A > B
 
S

su10112000a

cách khác:
ta có:
9999999+1>9999999-3
nên $\frac{9999999+1}{9999999-1}$>$\frac{9999999-3}{9999999-1}$ (vì 9999999-1>0)
\RightarrowA>B
 
Top Bottom