二十二代數(shù)式(第3課時(shí))1.在式子5x2-x,x2y,x2,a+b中,單項(xiàng)式有 A.1個(gè) B.2個(gè) C.3個(gè) D.4個(gè)2.單項(xiàng)式-3πa2bA.-35和7 B.-3C.-3π5和6 D.-3π3.(2024·東莞期中)多項(xiàng)式a4-2ab+b的次數(shù)和二次項(xiàng)系數(shù)分別為 (D)A.2,2 B.2,-2 C.4,2 D.4,-24.(2024·重慶期中)下列說法正確的是 (B)A.-3π2a3的次數(shù)是5B.-5π7x2的系數(shù)是-C.52a2-3aD.-3x3y-2x2y+x3的次數(shù)是35.已知(m-1)a|m+1|b3是關(guān)于a,b的五次單項(xiàng)式,則m=-3.

6.若多項(xiàng)式4x2y|m|-(m-1)y2+1是關(guān)于x,y的三次三項(xiàng)式,則常數(shù)m=-1.

7.(2024·衡陽期中)若單項(xiàng)式12xym-1與單項(xiàng)式-5x2y2的次數(shù)相同,則m=48.多項(xiàng)式3a2-kab+b2-1中不含ab項(xiàng),則k=0.

9.(2024·沈陽期中)觀察下列各式:-ab,-a2,2a,a+b,a2+a回答下列問題:(1)單項(xiàng)式分別為:-ab,-a2(2)多項(xiàng)式分別為:a+b,a2+a-1;

(3)整式有4個(gè);

(4)-ab的系數(shù)為-1;

(5)次數(shù)最高的多項(xiàng)式為a2+a-1.

10.已知多項(xiàng)式2x3y-4y3+5x2-1,按要求解答下列問題:(1)指出該多項(xiàng)式的項(xiàng);(2)該多項(xiàng)式的次數(shù)是________,三次項(xiàng)的系數(shù)是________;

(3)若|x+1|+|y-2|=0,求該多項(xiàng)式的值.【解析】(1)多項(xiàng)式2x3y-4y3+5x2-1各項(xiàng)依次為:2x3y,-4y3,5x2,-1;(2)該多項(xiàng)式的次數(shù)是4,三次項(xiàng)的系數(shù)是-4;答案:4-4(3)由題意可知,x+1=0,y-2=0,解得x=-1,y=2,把x=-1,y=2代入多項(xiàng)式,得2x3y-4y3+5x2-1=2×(-1)3×2-4×23+5×(-1)2-1=2×(-1)×2-4×8+5×1-1=-4-32+5-1=-32.11.(創(chuàng)新挑戰(zhàn)題·模型觀念、抽象能力、運(yùn)算能力)歷史上的數(shù)學(xué)巨人歐拉最先把關(guān)于x的多項(xiàng)式用記號(hào)f(x)的形式來表示,把x等于某數(shù)a時(shí)的多項(xiàng)式的值用f(a)來表示,例如x=-1時(shí),多項(xiàng)式f(x)=x2+3x-5的值記為f(-1),則f(-1)=-7.已知f(x)=ax5+bx3+3x+c,且f(0)=-1.(1)c=________.

(2)若f(1)=2,求a+b的值.(3)若f(2)=9,求f(-2)的值.【解析】(1)因?yàn)閒(x)=ax5+bx3+3x+c,且f(0)=-1,所以c=-1.答案:-1(2)因?yàn)閒(1)=2,c=-1,所以a+b+3-1=2,所以a+b=0.(3)因?yàn)閒(2)=9,c