Question

Biết \(\dfrac{a}{a'}=\dfrac{b}{b'}=\dfrac{c}{c'}=4\)

tính \(\dfrac{a-3b+2c}{a'-3b'-+2c'}\)

Answer 1

Từ giả thiết \(\Rightarrow a=4a';b=4b';c=4c'\)

Nên \(\dfrac{a+b+c}{a'+b'+c'}=\dfrac{4\left(a'+b'+c'\right)}{a'+b'+c'}=4\)

\(\dfrac{a-3b+2c}{a'-3b'+2c'}=\dfrac{4\left(a'-3b'+2c'\right)}{a'-3b'+2c'}=4\)

Answer 2

@Phạm Ngân Hà mk ko biet cach nay co dung ko ban xem giup mk nhe :v

\(\dfrac{b}{b'}=\dfrac{3b}{3b'};\dfrac{c}{c'}=\dfrac{2c}{2c'}\)

de bai: \(\dfrac{a}{a'}=\dfrac{b}{b'}=\dfrac{c}{c'}\Leftrightarrow\dfrac{a}{a'}=\dfrac{3b}{3b'}=\dfrac{2c}{2c'}=\dfrac{a-3b+2c}{a'-3b'+2c'}=4\)(TCDTSBN)