Question

cho pt \(x^2-4\sqrt{3}x+8=0\) có 2 nghiệm x1,x2. ko giải pt, tính Q = \(\frac{6x_1^2+10x_1x_2+6x_2^2}{5x_1x_2^3+5x_1^3x_2}\)

Answer 1

\(\left\{{}\begin{matrix}x_1+x_2=4\sqrt{3}\\x_1x_2=-8\end{matrix}\right.\)

\(Q=\frac{6\left(x_1+x_2\right)^2-2x_1x_2}{5x_1x_2\left(x_1^2+x_2^2\right)}=\frac{6\left(x_1+x_2\right)^2-2x_1x_2}{5x_1x_2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}\)

\(Q=\frac{6.\left(4\sqrt{3}\right)^2-2.\left(-8\right)}{5.\left(-8\right).\left[\left(4\sqrt{3}\right)^2-2.\left(-8\right)\right]}\)

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