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\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right)\)
\(-A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{121}\right)\)
\(-A=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{120}{121}\)
\(-A=\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot10\cdot12}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot11\cdot11}\)
\(-A=\frac{\left(1\cdot2\cdot3\cdot...\cdot10\right)\left(3\cdot4\cdot5\cdot...\cdot12\right)}{\left(2\cdot3\cdot4\cdot...\cdot11\right)\left(2\cdot3\cdot4\cdot...\cdot11\right)}\)
\(-A=\frac{1\cdot12}{11\cdot2}=\frac{6}{11}\)
\(A=-\frac{6}{11}\)
\(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{37}-\frac{1}{38}\)
\(B=1-\frac{1}{38}=\frac{37}{38}\)
\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+..+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}=\frac{1}{1}-\frac{1}{2}-\frac{1}{38}+\frac{1}{39}=\frac{370}{741}\)
A = 1 - 1/2 + 1/3 - 1/4 + ..... + 1/19 - 1/20
= 1 + 1/2 + 1/3 + 1/4 +............+ 1/19 + 1/20 - 2(1/2 + 1/4 + 1/6 +...........+ 1/20)
= 1 + 1/2 + 1/3 + 1/4 +............+ 1/19 + 1/20 - ( 1 + 1/2 + 1/3 + .............. + 1/10)
= 1/11 + 1/12 + ...... + 1/20
=> A / B = 1
bài này mình đã học rồi nhưng nó nâng cao , bạn nhé !
Chúc bạn học tốt ! 
\(A< \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{49.50.51}.\)
\(2A< \frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{49.50.51}\)
\(2A< \frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{51-49}{49.50.51}\)
\(2A< \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{49.50}-\frac{1}{50.51}\)
\(2A< \frac{1}{2}-\frac{1}{50.51}< \frac{1}{2}\Rightarrow A< \frac{1}{4}< \frac{1}{2}\)