Question

Tìm các số a,b,c,...,i biết  \(\frac{a-1}{9}=\frac{b-2}{8}=\frac{c-3}{7}=....=\frac{i-9}{1}\)và a + b + c + ... + i = 90

Answer 1

Áp dụng t/c của dãy tỉ số bằng nhau ta có:

\(\frac{a-1}{9}=\frac{b-2}{8}=\frac{c-3}{7}=....=\frac{i-9}{1}=\frac{\left(a-1\right)+\left(b-2\right)+\left(c-3\right)+...+\left(i-9\right)}{9+8+7+...+1}=\frac{\left(a+b+c+..+i\right)-\left(1+2+3+...+9\right)}{1+2+3+...+9}\)

=> \(\frac{a-1}{9}=\frac{b-2}{8}=\frac{c-3}{7}=....=\frac{i-9}{1}=\frac{90-45}{45}=1\)

=> a - 1 = 9 ; b - 2 = 8; c - 3 = 7; d- 4 = 6; e - 5 = 5; f - 6 = 4; ...; i - 9 = 1

=> a = 10; b = 10; c = 10= d = ..= i 

Answer 2

\(\frac{a-1}{9}=\frac{b-2}{8}=\frac{c-3}{7}=...=\frac{i-9}{1}=\frac{\left(a-1\right)+\left(b-2\right)+\left(c-3\right)+...+\left(i-9\right)}{9+8+7+...+1}=\frac{\left(a+b+c+...+i\right)-\left(1+2+3+...+9\right)}{9+8+7+...+1}\)\(=\frac{90-\frac{9.10}{2}}{\frac{9.10}{2}}=\frac{90-45}{45}=\frac{45}{45}=1\)

=> a = 9 + 1 = 10

     b = 8 + 2 = 10

     c = 7 + 3 = 10

     ....

     i = 1 + 9  = 10

Vậy a = b = c = ... = i = 10