题目

1:
(x+1)(x+3)(x+5)(x+7)+15
2:
x^2-2xy+y^2-13x+13y+30
3:
先把4(x+2z)(x-2z)+12xy+9y^2分解因式,在求值,其中x＝1/2,y＝2/3,z＝3/4.
4:
已知a,b,c为三角形ABC三边,且a,b,c满足(a+b+c)^2=3(a^2+b^2+c^2)．问三角形ABC是哪种三角形?

提问时间：2020-10-19

答案

1.
(x+1)(x+3)(x+5)(x+7)+15
=[(x+1)(x+7)][(x+3)(x+5)]+15
=(x^2+8x+7)(x^2+8x+15)+15
=(x^2+8x+7)[(x^2+8x+7)+8]+15
=(x^2+8x+7)^2+8(x^2+8x+7)+15
=[(x^2+8x+7)+3][(x^2+8x+7)+5]
=(x^2+8x+10)(x^2+8x+12)
=(x^2+8x+10)(x+2)(x+6)
2.
x^2-2xy+y^2-13x+13y+30
=(x-y)^2-13(x-y)+30
=(x-y-3)(x-y-10)
3:
先把4(x+2z)(x-2z)+12xy+9y^2分解因式,在求值,其中x＝1/2,y＝2/3,z＝3/4.
原式=4(x^2-4z^2)+12xy+9y^2=(4x^2+12xy+9y^2)-16z^2
=(2x+3y)^2-16z^2
=(2x+3y+4z)(2x+3y-4z)
=(1+2+3)(1+2-3)
=0
4:
已知a,b,c为三角形ABC三边,且a,b,c满足(a+b+c)^2=3(a^2+b^2+c^2)．问三角形ABC是哪种三角形?
(A+B+C)^2=3(A^2+B^2+C^2)
A^2+B^2+C^2+2AB+2AC+2BC=3A^2+3B^2+3C^2
2AB+2AC+2BC=2A^2+2B^2+2C^2
A^2-2AB+B^2+A^2-2AC+C^2+B^2-2BC+C^2=0
(A-B)^2+(A-C)^2+(B-C)^2=0
A-B=0,A-C=0,B-C=0
A=B=C
等边三角形