Question

Tinh gia tri bieu thuc   N=\(\frac{3a^2+6b^2-5c^2}{2a^2-4b^2+3c^2}\)biet 6a=4b=3c                                       

Answer 1

6a = 4b  = 3c => \(\frac{6a}{12}=\frac{4b}{12}=\frac{3c}{12}\Rightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=t\)

=> a = 2t ; b = 3t ; c = 4t thay vào N ta có \(\frac{3.\left(2t\right)^2+6.\left(3t\right)^2-5\left(4t\right)^2}{2.\left(2t\right)^2-4.\left(3t\right)^2+3\left(4t\right)^2}=\frac{3.4.t^2+6.9.t^2-5.16.t^2}{2.4.t^2-4.9.t^2+3.16.t^2}=\frac{-14t^2}{20t^2}=-\frac{7}{10}\)