Toán tính A=(2^4+4).(6^4+4).(10^4+4)…(82^4+4)/(4^4+4).(8^4+4).(12^4+4)…(84^4+4) 02/12/2021 By Arianna tính A=(2^4+4).(6^4+4).(10^4+4)…(82^4+4)/(4^4+4).(8^4+4).(12^4+4)…(84^4+4)
Đáp án: $A=\dfrac{1^2+1}{85^2+1}$ Giải thích các bước giải: Ta có: $n^4+4=(n^4+4n^2+4)-4n^2=(n^2+2)^2-(2n)^2=(n^2+2-2n)(n^2+2+2n)=((n-1)^2+1)((n+1)^2+1)$ $\to A=\dfrac{((2-1)^2+1)((2+1)^2+1)\cdot ((6-1)^2+1)((6+1)^2+1)\cdot ((10-1)^2+1)((10+1)^2+1)….\cdot ((82-1)^2+1)((82+1)^2+1)}{((4-1)^2+1)((4+1)^2+1)\cdot ((8-1)^2+1)((8+1)^2+1)\cdot ((12-1)^2+1)((12+1)^2+1)….\cdot ((84-1)^2+1)((84+1)^2+1)}$ $\to A=\dfrac{(1^2+1)(3^2+1)\cdot (5^2+1)(7^2+1)\cdot (9^2+1)(11^2+1)….\cdot (81^2+1)(83^2+1)}{(3^2+1)(5^2+1)\cdot (7^2+1)(9^2+1)\cdot (11^2+1)(13^2+1)….\cdot (83^2+1)(85^2+1)}$ $\to A=\dfrac{1^2+1}{85^2+1}$ Trả lời
Đáp án: $A=\dfrac{1^2+1}{85^2+1}$
Giải thích các bước giải:
Ta có:
$n^4+4=(n^4+4n^2+4)-4n^2=(n^2+2)^2-(2n)^2=(n^2+2-2n)(n^2+2+2n)=((n-1)^2+1)((n+1)^2+1)$
$\to A=\dfrac{((2-1)^2+1)((2+1)^2+1)\cdot ((6-1)^2+1)((6+1)^2+1)\cdot ((10-1)^2+1)((10+1)^2+1)….\cdot ((82-1)^2+1)((82+1)^2+1)}{((4-1)^2+1)((4+1)^2+1)\cdot ((8-1)^2+1)((8+1)^2+1)\cdot ((12-1)^2+1)((12+1)^2+1)….\cdot ((84-1)^2+1)((84+1)^2+1)}$
$\to A=\dfrac{(1^2+1)(3^2+1)\cdot (5^2+1)(7^2+1)\cdot (9^2+1)(11^2+1)….\cdot (81^2+1)(83^2+1)}{(3^2+1)(5^2+1)\cdot (7^2+1)(9^2+1)\cdot (11^2+1)(13^2+1)….\cdot (83^2+1)(85^2+1)}$
$\to A=\dfrac{1^2+1}{85^2+1}$