Question

[Thương Thương]

1) Tính:

a) \(\left(\dfrac{1}{9}-1\right).\left(\dfrac{1}{10}-1\right)....\left(\dfrac{1}{2004}-1\right).\left(\dfrac{1}{2005}-1\right)\)

b) \(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+3}}}\)

2) Cho A = \(x.\left(x-\dfrac{4}{9}\right)\). Tìm x, để:

a) A = 0; b) A > 0; c) A < 0

3) Cho \(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=...=\dfrac{a_{n-1}}{a_n}=\dfrac{a_n}{a_1}\)

\(a_1+a_2+...+a_n\ne0;a_1=-\sqrt{15}\)

Tính \(a_2;a_3;...;a_n\).

4) Tìm một số có 3 chữ số biết số đó chia hết cho 18 và các số của nó tỉ lệ với 1; 2; 3

Answer 1

[Mỹ Duyên]

Bài 1:

a) \(\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{10}-1\right)......\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-1\right)\)

= \(\dfrac{-8}{9}.\dfrac{-9}{10}.......\dfrac{-2003}{2004}.\dfrac{-2004}{2005}\) = \(\dfrac{-8}{2005}\)

b) \(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+3}}}\) = \(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{1}}}\)

= \(-2+\dfrac{1}{-2+\dfrac{1}{-1}}\) = \(-2+\dfrac{1}{-3}\) = \(\dfrac{-7}{3}\)

Answer 2

[Trần Quốc Lộc]

\(\text{Câu 1 : }\) Tính

\(\text{a) }\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{10}-1\right)...\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-1\right)\\ =\left(1-\dfrac{9}{9}\right)\left(\dfrac{1}{10}-\dfrac{10}{10}\right)...\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-\dfrac{2005}{2005}\right)\\ =\dfrac{-8}{9}\cdot\dfrac{-9}{10}\cdot...\cdot\dfrac{-2003}{2004}\cdot\dfrac{-2004}{2005}\\ =\dfrac{\left(-8\right)\cdot\left(-9\right)\cdot..\cdot\left(-2003\right)\cdot\left(-2004\right)}{9\cdot10\cdot...\cdot2004\cdot2005}\\ =-\dfrac{8\cdot9\cdot...\cdot2003\cdot2004}{9\cdot10\cdot...\cdot2004\cdot2005}\\ =-\dfrac{8}{2005}\)

\(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+3}}}\\ =-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{1}}}\\ =-2+\dfrac{1}{-2+\dfrac{1}{-1}}\\ =-2+\dfrac{1}{-3}\\ =-2+\dfrac{-1}{3}=-\dfrac{7}{3}\)