几个关于因式分解的数学问题
1.因式分解(1)a²+4a+4-c²(2)X^4+4(3)3y²-4y+1(4)4xy+1-4x²-y²2.用配方法因...
1.因式分解
(1)a²+4a+4-c²
(2)X^4 +4
(3)3y²-4y+1
(4)4xy+1-4x²-y²
2.用配方法因式分解
(1)x²-6x-27
(2)a²-3a-28
3.已知x²+y²-4x+6y+13=0
求3x-2y=?
4.解不等式
(3x+4)(3x-4)<9(x-2)(x+3)
一定要有过程哦 展开
(1)a²+4a+4-c²
(2)X^4 +4
(3)3y²-4y+1
(4)4xy+1-4x²-y²
2.用配方法因式分解
(1)x²-6x-27
(2)a²-3a-28
3.已知x²+y²-4x+6y+13=0
求3x-2y=?
4.解不等式
(3x+4)(3x-4)<9(x-2)(x+3)
一定要有过程哦 展开
展开全部
1.因式分解
(1)a²+4a+4-c²
=(a+2)²-c²
=(a+2+c)(a+2-c)
(2)X^4 +4
额 应该是 x^4-4 吧
x^4-4
=(x²+2)(x²-2)
=(x²+2)(x+√2)(x-√2)
(3)3y²-4y+1
=(y-1)(3y-1)
(4)4xy+1-4x²-y²
=-(4x²-4xy+y²-1)
=-[(2x-y)²-1]
=-(2x-y+1)(2x-y-1)
2.用配方法因式分解
(1)x²-6x-27
= x²-6x+9-36
=(x-3)²-36
=(x-3+6)(x-3-6)
=(x+3)(x-9)
(2)a²-3a-28
=a²-3a+9/4-121/4
=(a-3/2)²-121/4
=(a-3/2+11/2)(a-3/2-11/2)
=(a+4)(a-7)
3.已知x²+y²-4x+6y+13=0
求3x-2y=?
解:x²+y²-4x+6y+13=0
则 x²-4x+4+y²+6y+9=0
则 (x-2)²+(y+3)²=0
两个非负数的和为0,则这两个数都为0
即 x-2=0 y+3=0
即 x=2 y=-3
所以 3x-2y=2×3-2×(-3)=6+6=12
4.解不等式
(3x+4)(3x-4)<9(x-2)(x+3)
解:原不等式变形得
9x²-16<9(x²+x-6)
9x²-16<9x²+9x-54
则 9x>38
x>38/9
(1)a²+4a+4-c²
=(a+2)²-c²
=(a+2+c)(a+2-c)
(2)X^4 +4
额 应该是 x^4-4 吧
x^4-4
=(x²+2)(x²-2)
=(x²+2)(x+√2)(x-√2)
(3)3y²-4y+1
=(y-1)(3y-1)
(4)4xy+1-4x²-y²
=-(4x²-4xy+y²-1)
=-[(2x-y)²-1]
=-(2x-y+1)(2x-y-1)
2.用配方法因式分解
(1)x²-6x-27
= x²-6x+9-36
=(x-3)²-36
=(x-3+6)(x-3-6)
=(x+3)(x-9)
(2)a²-3a-28
=a²-3a+9/4-121/4
=(a-3/2)²-121/4
=(a-3/2+11/2)(a-3/2-11/2)
=(a+4)(a-7)
3.已知x²+y²-4x+6y+13=0
求3x-2y=?
解:x²+y²-4x+6y+13=0
则 x²-4x+4+y²+6y+9=0
则 (x-2)²+(y+3)²=0
两个非负数的和为0,则这两个数都为0
即 x-2=0 y+3=0
即 x=2 y=-3
所以 3x-2y=2×3-2×(-3)=6+6=12
4.解不等式
(3x+4)(3x-4)<9(x-2)(x+3)
解:原不等式变形得
9x²-16<9(x²+x-6)
9x²-16<9x²+9x-54
则 9x>38
x>38/9
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
解:1:a²+4a+4-c²
=(a+2)²-c²
=(a+2-c)(a+2+c)
a^4+4
=a^4+4a²+4-4a²
=(a²+2)²-(2a)²
=(a²-2a+2)(a²+2a+2)
3y²-4y+1
=(3y-1)(y-1)
4xy+1-4x²-y²
=1-(4x²-4xy+y²)
=1-(2x-y)²
=(1-2x+y)(1+2x-y)
2: x²-6x-27
=x²-6x+9-36
=(x-3)²-6²
=(x-9)(x+3)
a²-3a-28
=a²-3a+(3/2)²-121/4
=(a-3/2)²-(11/2)²
=(a-7)(a+4)
3:x²+y²-4x+6y+13=0
(x-2)² +(y+3)²=0
∵(x-2)²≥0, (y+3)²≥0
∴ x-2=0, x=2
y+3=0, y=-3
3x-2y=3×2-2×(-3)=12
4:(3x+4)(3x-4)<9(x-2)(x+3)
9x²<9x²+9x-54
9x>54
x>6
=(a+2)²-c²
=(a+2-c)(a+2+c)
a^4+4
=a^4+4a²+4-4a²
=(a²+2)²-(2a)²
=(a²-2a+2)(a²+2a+2)
3y²-4y+1
=(3y-1)(y-1)
4xy+1-4x²-y²
=1-(4x²-4xy+y²)
=1-(2x-y)²
=(1-2x+y)(1+2x-y)
2: x²-6x-27
=x²-6x+9-36
=(x-3)²-6²
=(x-9)(x+3)
a²-3a-28
=a²-3a+(3/2)²-121/4
=(a-3/2)²-(11/2)²
=(a-7)(a+4)
3:x²+y²-4x+6y+13=0
(x-2)² +(y+3)²=0
∵(x-2)²≥0, (y+3)²≥0
∴ x-2=0, x=2
y+3=0, y=-3
3x-2y=3×2-2×(-3)=12
4:(3x+4)(3x-4)<9(x-2)(x+3)
9x²<9x²+9x-54
9x>54
x>6
本回答被提问者采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
1.(1)=<(a+2)^2-c^2>
=(a+2-c)(a+2+c)
3……x^2+y^2-4x+6y+13=0
x^2-4x+4+y^2+6y+9=0
(x-2)^2+(y+3)^3=0
x=2
y=-3
=(a+2-c)(a+2+c)
3……x^2+y^2-4x+6y+13=0
x^2-4x+4+y^2+6y+9=0
(x-2)^2+(y+3)^3=0
x=2
y=-3
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询