Question

Tính giá trị của biểu thức N=\(\dfrac{3a^2+6b^2-5c^2}{2a^2-4b^2+3c^2}\) biết 6a=4b=3c

Answer 1

Ta có:

6a = 4b = 3c

=> \(\dfrac{6a}{12}=\dfrac{4b}{12}=\dfrac{3c}{12}\)

=> \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)

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

Đặt \(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)= k

=>\(\left\{{}\begin{matrix}a^2=4k\\b^2=9k\\c^2=16k\end{matrix}\right.\)

Thay \(\left\{{}\begin{matrix}a^2=4k\\b^2=9k\\c^2=16k\end{matrix}\right.\)vào biểu thức N ta được:

N = \(\dfrac{3a^2+6b^2-5c^2}{2a^2-4b^2+3c^2}\)

N = \(\dfrac{3.4k+6.9k-5.16k}{2.4k-4.9k+3.16k}\)

N = \(\dfrac{12k+54k-80k}{8k-36k+48k}\)

N = \(\dfrac{-14k}{20k}\)

N = \(\dfrac{-7}{10}\)