Question

Tìm \(a_1,a_2,a_3,...,a_9\)  biết \(\frac{a_1-1}{9}=\frac{a_2-2}{8}=\frac{a_3-3}{7}=...=\frac{a_9-9}{1}\) và \(a_1+a_2+a_3+...+a_9=90\)

Answer 1

\(\frac{a_1-1}{9}=\frac{a_2-2}{8}=...=\frac{a_9-9}{1}=\frac{a_1-1+a_2-2+...+a_9-9}{9+8+...+1}\)

\(=\frac{\left(a_1+a_2+...+a_9\right)-\left(1+2+...+9\right)}{45}\)

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

=> a₁ - 1 = 9 => a₁ = 10

     a₂ - 2 = 8 => a₂ = 10

.........................

     a₉ - 9 = 1 => a₉ = 10

KL: a₁ = a₂ =.......= a₉ = 10

Answer 2

a_{1 = 2}

a2 = 3

a₃ = 4

a₄ = 5

a₅ = 6

a₆ = 7

a₇ = 8

a₈ = 9

a₉ = 10