當(dāng)x=2時(shí),多項(xiàng)式ax的三次方+bx+1的值為-17,那么當(dāng)x=-1時(shí),多項(xiàng)式12ax=3bx的三次方-5的值為多少拜托各位
當(dāng)x=2時(shí),多項(xiàng)式ax的三次方+bx+1的值為-17,那么當(dāng)x=-1時(shí),多項(xiàng)式12ax=3bx的三次方-5的值為多少拜托各位
數(shù)學(xué)人氣:419 ℃時(shí)間:2020-10-01 14:01:44
優(yōu)質(zhì)解答
∵當(dāng)x=2時(shí),多項(xiàng)式ax^3+bx+1的值為-17,∴8a+2b+1=-17 ∴4a+b=-9 ∴當(dāng)x=-1時(shí),12ax+3bx^3-5 =-12a-3b-5 =-3×(4a+b)-5 =-3×(-9)-5 =27-5 =22
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