Question

Tìm a, b, c biết

a) a:b:c=2:4:5 và 2a-b+c=7

b) \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\) và a²-b²+2c²=108

Answer 1

a)a:b:c=2:4:5 =>\(\dfrac{a}{2}=\dfrac{b}{4}=\dfrac{c}{5}\Rightarrow\dfrac{2a}{4}=\dfrac{b}{4}=\dfrac{c}{5}=\dfrac{2a-b+c}{4-4+5}=\dfrac{7}{5}\)

=>a=\(2\cdot\dfrac{7}{5}=\dfrac{14}{5}\)

\(b=4\cdot\dfrac{7}{5}=\dfrac{28}{5}\)

\(c=5\cdot\dfrac{7}{5}=7\)

Vậy...

Answer 2

b)\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)

\(\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)

=>a²=16 b²=36 c²=64

=>a=4 b=6 c=8 hoặc a=-4 b=-6 c=-8

Answer 3

b) Ta có : \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\) =>\(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\) => \(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{2c^2}{32}\)

Áp dụng...........

\(\dfrac{a^2-b^2+c^2}{4-9+32}\)=\(\dfrac{108}{27}\)=4

Khi đó

a/2=4=>a=8

b/3=4=>b=12

c/4=4=>c=16