Lời giải:

$\frac{1}{4}< \frac{1}{1.2}$

$\frac{1}{9}< \frac{1}{2.3}$

$\frac{1}{16}< \frac{1}{3.4}$

....

$\frac{1}{2500}< \frac{1}{49.50}$

Cộng theo vế:

$A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1$

Ta có đpcm.

=(1-1/3)(1-1/4)(1-1/5)*...*(1-1/50)(1+1/3)(1+1/4)*...*(1+1/50)

=2/3*3/4*...*49/50*4/3*5/4*...*51/50

=2/50*51/3=17*1/25=17/25

\(\left(1-\dfrac{1}{9}\right)\cdot\left(1-\dfrac{1}{16}\right)\cdot\left(1-\dfrac{1}{25}\right)\cdot...\cdot\left(1-\dfrac{1}{2500}\right)\)

\(=\left(\dfrac{9}{9}-\dfrac{1}{9}\right)\cdot\left(\dfrac{16}{16}-\dfrac{1}{16}\right)\cdot...\cdot\left(\dfrac{2500}{2500}-\dfrac{1}{2500}\right)\)

\(=\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{2499}{2500}\)

\(=\dfrac{8\cdot15\cdot24\cdot...\cdot2499}{9\cdot16\cdot25\cdot...\cdot2500}\)

\(=\dfrac{\left(2\cdot4\right)\cdot\left(3\cdot5\right)\cdot\left(4\cdot6\right)\cdot....\cdot\left(49\cdot51\right)}{\left(3\cdot3\right)\cdot\left(4\cdot4\right)\cdot\left(5\cdot5\right)\cdot...\cdot\left(50\cdot50\right)}\)

\(=\dfrac{\left(2\cdot3\cdot4\cdot5\cdot...\cdot49\right)\left(4\cdot5\cdot6\cdot...\cdot51\right)}{\left(2\cdot3\cdot4\cdot...\cdot50\right)\left(2\cdot3\cdot4\cdot...\cdot50\right)}\)

\(=\dfrac{1\cdot51}{50\cdot2}\)

\(=\dfrac{51}{100}\)

Đề bài yêu cầu làm j z bn

quy luật +3+5+7+9+11+........

bai toan nay kho

bài toán này khó duyệt đi