(x+2)(x+3)(x+4)(x+5)+k
=(x+2)(x+5)*(x+3)(x+4)+k
=(x²+7x+10)(x²+7x+12)+k
=(x²+7x)²+22(x²+7x)+120+k
=(x²+7x+11)²+k-1
k=1
完全平方公式求值若(x+2)(x+3)(x+4)(x+5)+k是完全平方公式 则k的值为
(x+2)(x+3)(x+4)(x+5)+k
=(x+2)(x+5)*(x+3)(x+4)+k
=(x²+7x+10)(x²+7x+12)+k
=(x²+7x)²+22(x²+7x)+120+k
=(x²+7x+11)²+k-1
k=1