Question

tìm tỉ số \(\frac{a+b}{b+c}\)biết rằng \(\frac{b}{a}\)=2 và \(\frac{c}{b}\)=3

Answer 1

Với \(b+c;a;c\ne0\)

=> Khi \(\frac{b}{a}=2\Rightarrow b=2a;\)

Khi\(\frac{c}{b}=3\Rightarrow c=3b\)

Khi đó \(\frac{a+b}{b+c}=\frac{a+2a}{b+3b}=\frac{3a}{4b}=\frac{3a}{4.2a}=\frac{3a}{8a}=\frac{3}{8}\)

Vậy khi \(\frac{b}{a}=2;\frac{c}{b}=3\)thì \(\frac{a+b}{b+c}=\frac{3}{8}\)

Answer 2

\(\frac{a+b}{b+c}=\frac a b +\frac b c =\frac 1 2 + \frac 1 3 = \frac 5 6\)

Answer 3

thank you