cau 1 A=100^100+1 ,B=100^101+1 ———————- —————- 100^99+1 100^100+1

Question

cau 1 A=100^100+1 ,B=100^101+1 ———————- —————- 100^99+1 100^100+1

Question

cau 1
A=100^100+1 ,B=100^101+1
———————- —————-
100^99+1 100^100+1

in progress 0

Môn Toán Cherry 2 years 2021-04-22T22:20:05+00:00 2021-04-22T22:20:05+00:00 2 Answers 24 views 0

Answers ( )

phucdien · Answer 1 · 2021-04-22T22:21:48+00:00

Đáp án:

– `A <B`

Giải thích các bước giải:

$\text{B = $\dfrac{100^{101} + 1}{100^{100}}$ + 1 > 1}$

`=>` $\text{B = $\dfrac{100^{101} + 1 + 99}{100^{100} + 1 + 99}$}$

$\text{= $\dfrac{100^{101} + 100}{100^{100} + 100}$}$

$\text{= $\dfrac{(100^{100} + 1) . 100}{(100^{99} + 1) .100}$}$

$\text{= $\dfrac{100^{100} + 1}{100^{99} + 1}$ = A}$

Vậy `A < B`

dangkhoa · Answer 2 · 2021-04-22T22:21:50+00:00

Đáp án:

. Có: $\frac{100^{101} + 1}{100^{100} + 1} > 1 \Rightarrow \frac{100^{101} + 1}{100^{100} + 1} > \frac{100^{101} + (1 + 99)}{100^{100} + (1 + 99)}$

$\Rightarrow B > \frac{100^{101} + 100}{100^{100} + 100} \Rightarrow B > \frac{100 (100^{100} + 1)}{100 (100^{99} + 1)} \Rightarrow B > \frac{100^{100} + 1}{100^{99} + 1} = A \Leftrightarrow A < B$

Vậy A < B

b. Có: $\frac{13^{16} + 1}{13^{17} + 1} < 0 \Rightarrow \frac{13^{16} + 1}{13^{17} + 1} < \frac{13^{16} + (1 + 12)}{13^{17} + (1 + 12)}$

$\Rightarrow B < \frac{13^{16} + 13}{13^{17} + 13} \Rightarrow B < \frac{13 (13^{15} + 1)}{13 (13^{16} + 1)} \Rightarrow B < \frac{13^{15} + 1}{13^{16} + 1} = A \Leftrightarrow A > B$

Vậy A > B

Giải thích các bước giải: