1:(x+1)(x+3)(x+5)(x+7)+152:x^2-2xy+y^2-13x+13y+303:先把4(x+2z)

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/04 23:58:48

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?

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

1:(x+1)(x+3)(x+5)(x+7)+152:x^2-2xy+y^2-13x+13y+303:先把4(x+2z) 1、(-7x^y)(2x^y-3xy^3+xy) 2、((x-y)^6)/((y-x)^3)/(x-y) {5x-3y+z=2{5x+2y-4z=3{-5x+y-z=2 {x-y-z=-1{3x+5y+7z{4x-y+2x=- 2X+Y+Z=10,X+2Y+Z=-6,X+Y+2Z=8 5X+4Y-3Y=13,2X-3Y+7Z=19,3X-Y=2X 化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(xy-2z)(y+z-2x)+(x- 解下列方程组（1）3x-y+z=3 2x+y-3z=11 x+y+z=12（2）5x-4y+4z=13 2x+7y-3z 计算(x^8-x^7y+x^5y^3-x^4y^4+x^3y^5-x^y^7+y^8)(x^2+xy+y^2) x^8-x^7y+x^6y^2-x^5y^3+x^4y^4-x^3y^5+x^2y^-xy^7+y^8 通分:(1)x/(x^2+2xy+y^2)与y/(x+y) (2)(x+4)/(x^2+4x)与(x-5)/x(x-5) 1、x(x-y)(x+y)-x(x+y)^2 1.解下列方程（1）（x+y=-2（x+z=13（y+z=1（2）（x-y-z=-1（3x+5y+7z=11（4x-y+ 三元二次方程组(XY+X)/(X+Y+1)=2(XZ+2X)/(X+Z+2)=3(Y+1)(Z+2)/(Z+Y+3)=4