Tính giá trị biểu thức?

C

chonhoi110

Ta có: $\dfrac{9}{2}y^2+ \dfrac{1}{2}x^2-3xy-18y+6x+18=0$

$\leftrightarrow 9y^2+x^2-6xy-36y+12x+36=0$

$\leftrightarrow (3y-x-6)^2=0$

$\leftrightarrow 3y-x-6=0$

$\rightarrow x=3y-6$

Ta có: $Q=\dfrac{2x}{y-2}+\dfrac{4x-6y}{x-6}$

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

$=\dfrac{6(y-2)}{y-2}+\dfrac{6y-24}{3y-12}$

$=6+2=8$
 
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