Question

[Hòa An Nguyễn]

Cho 9 số \(A_1\) ; \(A_2\) ; ... ; \(A_9\) thỏa mãn \(A_1\) + \(A_2\) + ... + \(A_9\) = 90

và \(\dfrac{A_1-1}{9}\) = \(\dfrac{A_2-2}{8}\) = \(\dfrac{A_7-3}{7}\) = ... = \(\dfrac{A_8-8}{2}\) = \(\dfrac{A_9-9}{1}\)

Tính \(A_1\) ; \(A_2\) ; ... ; \(A_9\)

Giúp mình với mọi người !!! T_T

Answer 1

[Ái Nữ]

\(\dfrac{A1-1}{9}=\dfrac{A2-A}{8}=\dfrac{A3-3}{7}=....=\dfrac{A8-8}{2}=\dfrac{A9-9}{1}\)(1)

Từ (1) => \(\dfrac{A1-1}{9}=\dfrac{A2-A}{8}=\dfrac{A3-3}{7}=....=\dfrac{A8-8}{2}=\dfrac{A9-9}{1}\)

=\(\dfrac{A1-1+A2-2+A3-3+...+A9-9}{9+8+7+.......+1}\)

=\(\dfrac{\left(A1+A2+A3+...+A9\right)-\left(1+2+3+....+9\right)}{9+8+7+..+1}\)

=\(\dfrac{90-45}{45}=\dfrac{45}{45}=1\)

=>\(\dfrac{A1-1}{9}=1\Rightarrow A1=10\)

TƯƠNG TỰ TÍNH : A2=A3=......=A9=10

vẬY A1=A2=A3=.....=A9=10