Question

Tìm \(x_1;x_2;x_3....;x_9,\)biết rằng:

\(\dfrac{x_1-1}{9}=\dfrac{x_2-2}{8}=\dfrac{x_3-3}{7}=.....=\dfrac{x_9-9}{1}\) và \(x_1+x_2+x_3+....+x_9\)= 90

Answer 1

Theo bài ra ta có : \(\dfrac{x1-1}{9}=\dfrac{x2-2}{8}=\dfrac{x3-3}{7}=......=\dfrac{x9-9}{1}\)

= \(\dfrac{\left(x1-1\right)+\left(x2-2\right)+\left(x3-3\right)+....+\left(x9-9\right)}{9+8+7+....+2+1}\)

=\(\dfrac{\left(x1+x2+x3+....+x9\right)-\left(1+2+3+...+9\right)}{9+8+7+...+1}\)

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

=> \(x1=9.1+1=10\)

\(x2=8.1+2=10\)

\(x3=7.1+3=10\)

\(x4=6.1+4=10\)

\(x5=5.1+5=10\)

\(x6=4.1+6=10\)

\(x7=3.1+7=10\)

\(x8=2.1+8=10\)

\(x9=1.1+9=10\)

Vậy \(x1,x2,x3,x4,x5,...,x9\) tất cả đều bằng 10