Question

[Trần Lê Nhật Minh]

(\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+...+\(\dfrac{1}{8.9.10}\)).x=\(\dfrac{23}{45}\)

Answer 1

[Akai Haruma]

Lời giải:

Gọi tổng trong ngoặc là $A$
$2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+....+\frac{10-8}{8.9.10}$

$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}$

$=\frac{1}{1.2}-\frac{1}{9.10}=\frac{1}{2}-\frac{1}{90}=\frac{22}{45}$

Vậy $\frac{22}{45}x=\frac{23}{45}$

$\Rightarrow x=\frac{23}{45}: \frac{22}{45}=\frac{23}{22}$