Đáp án: $A=\dfrac6{11}$
Giải thích các bước giải:
Ta có:
$A=(\dfrac14-1)(\dfrac19-1)(\dfrac1{16}-1)...(\dfrac1{100}-1)(\dfrac{1}{121}-1)$
$\to A=(\dfrac1{2^2}-1)(\dfrac1{3^2}-1)(\dfrac1{4^2}-1)...(\dfrac1{10^2}-1)(\dfrac{1}{11^2}-1)$
$\to A=(1-\dfrac{1}{2^2})(1-\dfrac{1}{3^2})(1-\dfrac{1}{4^2})....(1-\dfrac{1}{10^2})(1-\dfrac{1}{11^2})$
$\to A=\dfrac{2^2-1}{2^2}.\dfrac{3^2-1}{3^2}.\dfrac{4^2-1}{4^2}....\dfrac{10^2-1}{10^2}.\dfrac{11^2-1}{11^2}$
$\to A=\dfrac{(2-1)(2+1)}{2^2}.\dfrac{(3-1)(3+1)}{3^2}.\dfrac{(4-1)(4+1)}{4^2}....\dfrac{(10-1)(10+1)}{10^2}.\dfrac{(11-1)(11+1)}{11^2}$
$\to A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}....\dfrac{9.11}{10^2}.\dfrac{10.12}{11^2}$
$\to A=\dfrac{1.2.3...9.10}{2.3.4...11}.\dfrac{3.4.5...11.12}{2.3.4....11}$
$\to A=\dfrac{1}{11}.\dfrac{12}{2}$
$\to A=\dfrac{6}{11}$
