比较下列代数式的大小(1)(a+3)(a-5)与(a+2)(a-4)(2) (2x-3)(x+1)与(x+2)(x-3)
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比较下列代数式的大小(1)(a+3)(a-5)与(a+2)(a-4)(2) (2x-3)(x+1)与(x+2)(x-3)
比较下列代数式的大小
(1)(a+3)(a-5)与(a+2)(a-4)
(2) (2x-3)(x+1)与(x+2)(x-3)
比较下列代数式的大小(1)(a+3)(a-5)与(a+2)(a-4)(2) (2x-3)(x+1)与(x+2)(x-3)
采用作差的方式:
1、[(a+3)(a-5)]-[(a+2)(a-4)]
=[a²-2a-15]-[a²-2a-8]
=-7(x+2)(x-3)
1.小于
2.大于
(1)(a+3)(a-5)
=a²-2a-15
(a+2)(a-4)
=a²-2a-8
(a²-2a-15)-(²-2a-8)=﹣7<0所以(a+3)(a-5)<(a+2)(a-4)
(2)(2x-3)(x+1)
=2x²+2x-3x-3
=2x²-x-3
(x+2)(x-3...
全部展开
(1)(a+3)(a-5)
=a²-2a-15
(a+2)(a-4)
=a²-2a-8
(a²-2a-15)-(²-2a-8)=﹣7<0所以(a+3)(a-5)<(a+2)(a-4)
(2)(2x-3)(x+1)
=2x²+2x-3x-3
=2x²-x-3
(x+2)(x-3)
=x²-x-6
(2x²-x-3)-(x²-x-6)=x²+3
因为x²≥0,所以x²+3 >0
所以(2x-3)(x+1)>(x+2)(x-3)
收起
先将代数式化简:
(a+3)(a-5)=a^2-2*a-15 (a+2)(a-4)=a^2-2*a-8
显然前者小
(2x-3)(x+1)=2*x^2-x-3 (x+2)(x-3)=x^2-x-6
这个难说,你试一试在坐标图中将两个曲线画出来比比就知道了。
大的在上面,小的在下面。
不明白,欢迎追问!
(1)展开求解即可
(2)做差比较
(1)(a+3)(a-5)与(a+2)(a-4)
(a+3)(a-5)=a^2+2a-5a-15=a^2-3a-15
(a+2)(a-4)=a^2+2a-4a-8=a^2-2a-8
(a+3)(a-5)<(a+2)(a-4)
(2) (2x-3)(x+1)与(x+2)(x-3)
(2x-3)(x+1)=2x^2-3x+2x-3=2x^2-x-3
(x+2)(x-3)=x^2+2x-3x-6=x^2-x-6
(2x-3)(x+1)>(x+2)(x-3)
(1)(a+3)(a-5)-(a+2)(a-4)
=a^2-2a-15-a^2+2a+8
=-7 <0
所以(a+3)(a-5)<(a+2)(a-4)
(2) (2x-3)(x+1)-(x+2)(x-3)
=2x^2-x-3-x^2+x+6
=x^2+3>0
所以 (2x-3)(x+1)>(x+2)(x-3)