Question

高一简单数学题

比较下列各题中两式值的大小 过程写下哦..
1.x(x-y)与y(x-y) (x不等于Y)
2.(3a+1)(a+1)与2(a+1)平方-3
3.(t+1)(t-5)=(t-2)平方.

Answer 1

1. x(x-y) - y(x-y) = (x-y)(x-y) = (x-y)^2 > 0 (因x不等于y)
x(x-y) 〉 y(x-y)

2. (3a+1)(a+1) -[2(a+1)^2 -3]
= 3a^2 + 4a + 1 -(2a^2 + 4a +2 -3)
= a^2 + 2 > 0
(3a+1)(a+1) > 2(a+1)^2 -3

3. (t+1)(t-5) - (t-2)^2
= t^2 -4t -5 -(t^2 -4t +4)
= -9 < 0
(t+1)(t-5) < (t-2)^2

Answer 2

1、x(x-y)-y(x-y)
=x^2-xy-xy-y^2
=(x-y)^2>0，所以x(x-y)>y(x-y)
2、(3a+1)(a+1)-[2(a+1)^2-3]
=3a^2+4a+1-2a^2-4a+1
=a^2+2>0，所以(3a+1)(a+1)>2(a+1)^2-3
3、(t+1)(t-5)-(t-2)^2
=t^2-4t-5-t^2+4t-4
=-9<0，所以(t+1)(t-5)<(t-2)平方

Answer 3

1、x(x-y)-y(x-y) =x^2+2xy+y^2=(x+y)^2>=0;因为x≠y;所以（x+y)^2>0,即x(x-y)>y(x-y)
23a+1)(a+1) -[2(a+1)^2 -3]= 3a^2 + 4a + 1 -(2a^2 + 4a +2 -3)= a^2 + 2 > 0
所以([3a+1)(a+1)]>=[2(a+1)平方-3]
3、(t+1)(t-5)-(t-2)平方=-9<0所以(t+1)(t-5)=(t-2)平方.
比较大小，可以作差，作商