计算:(1)(3x2+4-5x3)-(x3-3+3x2);(2)(3x2-xy-2y2)-2(x2+xy-2y2).
题型:解答题难度:一般来源:不详
计算: (1)(3x2+4-5x3)-(x3-3+3x2); (2)(3x2-xy-2y2)-2(x2+xy-2y2). |
答案
(1)原式=3x2+4-5x3-x3+3+-3x2(去括号) =(-5x3-x3)+(3x2-3x2)+(4+3)(合并同类项) =-6x3+7;
(2) 原式=3x2-xy-2y2-2x2-2xy+4y2(去括号) =(3x2-2y2)+(-xy-2xy)+(-2y2+4y2)(合并同类项) =x2-3xy+2y2 故答案为:(1)-6x3+7;(2)x2-3xy+2y2. |
举一反三
先化简,再求值: (1)(4a2-3a)-(2a2+a-1)+(2-a2+4a),其中a=2. (2)已知A=3a2-5ab-b2,B=-2a2+3ab+b2,其中a=-1,b=,求-3(A-B)+2(B+3A)的值. |
先化简,再求值:-(3a2-4ab)+[a2-2(a+2ab)],其中a=-2,b=. |
求值:5a2-[a2-(2a-5a2)-2(a2-3a)],其中a=-2. |
(1)若|a-1|+(b-2)2=0,A=3a2-6ab+b2,B=-a2-5,求A-B的值. (2)试说明:无论x,y取何值时,代数式. (x3+3x2y-5xy2+6y3)+(y3+2xy2+x2y-2x3)-(4x2y-x3-3xy2+7y3)的值是常数. |
最新试题
热门考点