Question

Cho x-2y=6. Tính

\(P=\dfrac{x-y}{x+6}\left(x\ne y;x\ne-6\right)\)

\(Q=\dfrac{2x+6}{3x-y}+\dfrac{2y-6}{4y-x}\left(3x\ne2y;4y\ne x\right)\)

Answer 1

x-2y=6

nên x=2y+6

\(P=\dfrac{2y+6-y}{2y+6+6}=\dfrac{y+6}{2y+12}=\dfrac{1}{2}\)

\(Q=\dfrac{2x+6}{3x-2y}+\dfrac{2y-6}{4y-x}\)

\(=\dfrac{2\left(2y+6\right)}{3\left(2y+6\right)-2y}+\dfrac{2y-6}{4y-\left(2y+6\right)}\)

\(=\dfrac{4y+12}{6y+18-2y}+\dfrac{2y-6}{4y-2y-6}\)

\(=\dfrac{4y+12}{4y+18}+\dfrac{2y-6}{2y-6}\)

\(=\dfrac{8y^2-24y+24y-48+8y^2+36y-24y-108}{\left(4y+18\right)\left(2y-6\right)}\)

\(=\dfrac{16y^2+12y-156}{4\left(2y+9\right)\cdot\left(y-3\right)}\)

\(=\dfrac{4y^2+3y-34}{\left(2y+9\right)\left(y-3\right)}\)