Помогите найти знач выраж. 2x^2 - 4xy^2 + 3xy- 6y^3 если x=1//4 (одна четвёртая) y=1//6
2x²-4xy²+3xy-6y³ x=1/4 y=1/6
2x²-4xy²+3xy-6y³=(2x²+3xy)-(4xy²+6y³)=x(2x+3y)-2y²(2x+3y)=(x-2y²)*(2x+3y)
(x-2y²)*(2x+3y)=(1/4-2*1/36)*(2*1/4+3*1/6)=(1/4-1/18)*(1/2+1/2)=1/4-1/18=
9/36-2/36=7/36
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2x²-4xy²+3xy-6y³ x=1/4 y=1/6
2x²-4xy²+3xy-6y³=(2x²+3xy)-(4xy²+6y³)=x(2x+3y)-2y²(2x+3y)=(x-2y²)*(2x+3y)
(x-2y²)*(2x+3y)=(1/4-2*1/36)*(2*1/4+3*1/6)=(1/4-1/18)*(1/2+1/2)=1/4-1/18=
9/36-2/36=7/36