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Mathematics problems in eighth grade
1。 (a+b+c)^2-(a-b-c)^2

=[(a+b+c)+(a-b-c)][(a+b+c)-(a-b-c)]

=2a(2b+2c)

=4a(b+c)

=4ab+4ac

2.x^5-(x^3)(y^2)

=x^3(x^2-y^2)

=x^3(x+y)(x-y)

3.(x^2)(x-y)-(y^2)(x-y)

=(x-y)(x^2-y^2)

=(x-y)(x+y)(x-y)

=(x-y)^2(x+y)

Because x is not equal to y.

So (x-y) 2 > 0.

Because both x and y are positive numbers, x+y >; 0

So (x-y) 2 (x+y) > 0.

(x^2)(x-y)-(y^2)(x-y)>; 0

(x^2)(x-y)>; (y^2)(x-y)

4。 (2x+3y)(2x-3y)= 3 1 = 1 * 3 1

Because x and y are positive integers.

So 2x+3y > 2x-3y, 2x+3y > 0.

So 2x+3y=3 1.

2x-3y= 1

Plus 4x=32

x=8

y=(3 1-2x)/3=5