Giải:
Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\)
\(\Rightarrow a=2k,b=3k,c=4k\)
Ta có: \(\frac{a^2+b^2+2c^2}{a^2-4b^2+c^2}\)
\(=\frac{\left(2k\right)^2+\left(3k\right)^2+2\left(4k\right)^2}{\left(2k\right)^2-4\left(3k\right)^2+\left(4k\right)^2}\)
\(=\frac{2^2.k^2+3^2.k^2+2.4^2.k^2}{2^2.k^2-4.3^2.k^2+4^2.k^2}\)
\(=\frac{4.k^2+9.k^2+32.k^2}{4.k^2-36.k^2+16.k^2}\)
\(=\frac{k^2.\left(4+9+32\right)}{k^2.\left(4-36+16\right)}\)
\(=\frac{45}{-16}\)
\(A=\frac{a^2+b^2+2c^2}{a^2-4b^2+c^2}\)
Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\Rightarrow a=2k;b=3k;c=4k\)
Suy ra \(A=\frac{\left(2k\right)^2+\left(3k\right)^2+2\left(4k\right)^2}{\left(2k\right)^2-4\left(3k\right)^2+\left(4k\right)^2}=\frac{4k^2+9k^2+2\cdot16k^2}{4k^2-4\cdot9k^2+16k^2}\)
\(=\frac{k^2\left(4+9+32\right)}{k^2\left(4-36+16\right)}=\frac{45}{-16}=-\frac{45}{16}\)
