化簡與求值:
(1)3a2-2a-4a2-7a;
(2)2(7m2n-5mn2)-(4m2n-5mn2);
(3)先化簡再求值:(2x2y-4xy2)-(-3xy2+x2y),其中x=-1,y=2;
(4)已知一個多項(xiàng)式與x2-2x+1的和是3x-2,求這個多項(xiàng)式.
解:
(1)原式=-a2-9a;
(2)原式=14m2n-10mn2-4m2n+5mn2=10m2n-5mn2;
(3)原式=2x2y-4xy2+3xy2-x2y=x2y-xy2,
當(dāng)x=-1,y=2時(shí),原式=2+4=6;
(4)由題意得:這個多項(xiàng)式表示為3x-2-(x2-2x+1)=3x-2-x2+2x-1=-x2+5x-3.
分析:(1)直接合并同類項(xiàng)計(jì)算;
(2)先去括號,再合并同類項(xiàng)計(jì)算;
(3)先去括號,合并同類項(xiàng),再代入求值;
(4)由減法是加法的逆運(yùn)算知,這個多項(xiàng)式表示為3x-2-(x2-2x+1),然后去括號,合并同類項(xiàng)即可.
點(diǎn)評:此題考查了整式的加減運(yùn)算.注意去括號法則:++得+,--得+,-+得-,+-得-;合并同類項(xiàng)法則:把同類項(xiàng)的系數(shù)相加減,字母和字母指數(shù)的部分不變.