解答题目.因式分解(要有步骤)1。2ax+3bx+4ay+6by?
解答题目.因式分解(要有步骤) 1。2ax+3bx+4ay+6by 2.(x-2y)²+4xy-8y+4 3。已知实数a.b满足(a+b)²=1,(a-b)²=25,求证a²+b²+ab的值。 3。用公式计算 998*1002 2008-4016-3
1. 2ax+3bx+4ay+6by =(2ax+4ay)+(3bx+6by) =2a(x+2y)+3b(x+2y) =(x+2y)(2a+3b) 2. (x-2y)²+4xy-8y+4 =(x²-4xy+4y²)+4xy-8y+4 =x²+4y²-8y+4 =x²+4(y-1)².......在实数范围内无法分解,是不是题目抄错了? 3.已知实数a.b满足(a+b)²=1,求证a²+b²+ab的值。 (a+b)²=a²+2ab+b²=1..........① (a-b)²=a²-2ab+b²=25.........② (①×3+②)/ 4 得 [(3a²+6ab+3b²)+(a²-2ab+b²)]/4 =(4a²+4b²+4ab)/4 =a²+b²+ab =(1×3+25)/4 =7 3. 998*1002 =(1000-2)×(1000+2) =1000²-2² =00-4 =999996 2008-4016-3 =2008-2×2008-3 =(1-2)×2008-3 =-2008-3 =-2011