Đáp án:
Giải thích các bước giải:
Bài 9:
a. $\frac{1}{2.3}$ + $\frac{1}{3.4}$ + ... + $\frac{1}{49.50}$
= $\frac{1}{2}$ - $\frac{1}{3}$ + $\frac{1}{3}$ - $\frac{1}{4}$ + ... + $\frac{1}{49}$ - $\frac{1}{50}$
= $\frac{1}{2}$ - $\frac{1}{50}$
= $\frac{25}{50}$ - $\frac{1}{50}$
= $\frac{24}{50}$
= $\frac{12}{25}$
Bài 10:
$\frac{3^2}{20.23}$ + $\frac{3^2}{23.26}$ + ... + $\frac{3^2}{77.80}$
= 3.$\frac{3}{20.23}$ + 3.$\frac{3}{23.26}$ + ... + 3.$\frac{3}{77.80}$
= 3.($\frac{3}{20.23}$ + $\frac{3}{23.26}$ + ... + $\frac{3}{77.80}$)
= 3.($\frac{1}{20}$ - $\frac{1}{23}$ + $\frac{1}{23}$ - $\frac{1}{26}$ + ... + $\frac{1}{77}$ - $\frac{1}{80}$)
= 3.($\frac{1}{20}$ - $\frac{1}{80}$)
= 3.($\frac{4}{80}$ - $\frac{1}{80}$)
= 3. $\frac{3}{80}$
= $\frac{9}{80}$
mà $\frac{9}{80}$ < $\frac{80}{80}$ = 1
⇒ $\frac{9}{80}$ < 1
