求m的值(1)(x+4)(x+9)=x^2+mx+36(2)(x-2)(x-18)=x^2+mx+36(3)(x+3)(x+p)=x^2+mx+36(4)(x-6)(x-p)=x^2mx+36(5)(x+p)(x+q)=x^2+mx+36,p,q为正整数(x^2表示x的平方)惨了惨了...后面的x^2全部改为m^2(1)(x+4)(x+9)=m^2+mx+36(2)(x-2)(x-18)=m^2+mx+3
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![求m的值(1)(x+4)(x+9)=x^2+mx+36(2)(x-2)(x-18)=x^2+mx+36(3)(x+3)(x+p)=x^2+mx+36(4)(x-6)(x-p)=x^2mx+36(5)(x+p)(x+q)=x^2+mx+36,p,q为正整数(x^2表示x的平方)惨了惨了...后面的x^2全部改为m^2(1)(x+4)(x+9)=m^2+mx+36(2)(x-2)(x-18)=m^2+mx+3](/uploads/image/z/4099912-16-2.jpg?t=%E6%B1%82m%E7%9A%84%E5%80%BC%281%29%28x%2B4%29%28x%2B9%29%3Dx%5E2%2Bmx%2B36%282%29%28x-2%29%28x-18%29%3Dx%5E2%2Bmx%2B36%283%29%28x%2B3%29%28x%2Bp%29%3Dx%5E2%2Bmx%2B36%284%29%28x-6%29%28x-p%29%3Dx%5E2mx%2B36%285%29%28x%2Bp%29%28x%2Bq%29%3Dx%5E2%2Bmx%2B36%2Cp%2Cq%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%28x%5E2%E8%A1%A8%E7%A4%BAx%E7%9A%84%E5%B9%B3%E6%96%B9%29%E6%83%A8%E4%BA%86%E6%83%A8%E4%BA%86...%E5%90%8E%E9%9D%A2%E7%9A%84x%5E2%E5%85%A8%E9%83%A8%E6%94%B9%E4%B8%BAm%5E2%281%29%28x%2B4%29%28x%2B9%29%3Dm%5E2%2Bmx%2B36%282%29%28x-2%29%28x-18%29%3Dm%5E2%2Bmx%2B3)
求m的值(1)(x+4)(x+9)=x^2+mx+36(2)(x-2)(x-18)=x^2+mx+36(3)(x+3)(x+p)=x^2+mx+36(4)(x-6)(x-p)=x^2mx+36(5)(x+p)(x+q)=x^2+mx+36,p,q为正整数(x^2表示x的平方)惨了惨了...后面的x^2全部改为m^2(1)(x+4)(x+9)=m^2+mx+36(2)(x-2)(x-18)=m^2+mx+3
求m的值
(1)(x+4)(x+9)=x^2+mx+36
(2)(x-2)(x-18)=x^2+mx+36
(3)(x+3)(x+p)=x^2+mx+36
(4)(x-6)(x-p)=x^2mx+36
(5)(x+p)(x+q)=x^2+mx+36,p,q为正整数
(x^2表示x的平方)
惨了惨了...后面的x^2全部改为m^2
(1)(x+4)(x+9)=m^2+mx+36
(2)(x-2)(x-18)=m^2+mx+36
(3)(x+3)(x+p)=m^2+mx+36
(4)(x-6)(x-p)=m^2mx+36
(5)(x+p)(x+q)=m^2+mx+36,p,q为正整数
不好意思...白费大家的心思了 因此再+50分..对不起
..最后的..不然我就关闭问题了
算你们狠..我关闭问题..
求m的值(1)(x+4)(x+9)=x^2+mx+36(2)(x-2)(x-18)=x^2+mx+36(3)(x+3)(x+p)=x^2+mx+36(4)(x-6)(x-p)=x^2mx+36(5)(x+p)(x+q)=x^2+mx+36,p,q为正整数(x^2表示x的平方)惨了惨了...后面的x^2全部改为m^2(1)(x+4)(x+9)=m^2+mx+36(2)(x-2)(x-18)=m^2+mx+3
(1)(x+4)(x+9)=x^2+mx+36
x^2+13x+36=x^2+mx+36
m=13
(2)(x-2)(x-18)=x^2+mx+36
x^2-20x+36=x^2+mx+36
m=-20
(3)(x+3)(x+p)=x^2+mx+36
x^2+(3+p)x+3p=x^2+mx+36
3+p=m
3p=36
解得:
p=12
m=15
(4)(x-6)(x-p)=x^2mx+36
x^2-(6+p)x+6p=x^2+mx+36
6+p=-m
6p=36
解得:
p=6
m=-12
(5)(x+p)(x+q)=x^2+mx+36,p,q为正整数
x^2+(p+q)x+pq=x^2+mx+36
qp=36
p+q=m
p,q为正整数,
所以
1)p,q分别为1,36,m=37
2)p,q分别为2,18,m=20
3)p,q分别为3,12,m=15
4)p,q分别为4,9,m=13
5)p,q分别为6,6,m=12
(1)m=13
(2)m=-20
(3)m=15
(4)m=-12
(5)m=37,20,13,12 ,15
(x+a)(x+b)=x^2+(a+b)x+ab
1、m=4+9=13
2、m=-2-18=-20
3、m=3+p
4、m=-6-p
5、m=p+q
(1)(x+4)(x+9)=x^2+mx+36
(x+4)(x+9)
=x^2+13x+36
m=13
(2)(x-2)(x-18)=x^2+mx+36
(x-2)(x-18)
=x^2-20x+36
m=-20
(3)(x+3)(x+p)=x^2+mx+36
(x+3)(x+p)
=x^2+(p+3)x+3p
全部展开
(1)(x+4)(x+9)=x^2+mx+36
(x+4)(x+9)
=x^2+13x+36
m=13
(2)(x-2)(x-18)=x^2+mx+36
(x-2)(x-18)
=x^2-20x+36
m=-20
(3)(x+3)(x+p)=x^2+mx+36
(x+3)(x+p)
=x^2+(p+3)x+3p
3p=36
p=12
m=p+3=12+3=15
(4)(x-6)(x-p)=x^2mx+36
(x-6)(x-p)
=x^2-(p+6)x+6p
6p=36
p=6
m=-(p+6)=0
(5)(x+p)(x+q)=x^2+mx+36,p,q为正整数
(x+p)(x+q)
=x^2+(p+q)x+pq
pq=36
pq是正整数,
所以p=1,2,3,4,6,9,12,18,36
对应q=36,18,12,9,6,3,2,1
则m=37,20,15,13,12
收起
(1)(x+4)(x+9)=(x+4)x+(x+4) 9=x^2+4x+9x+36=x^2+13x+36所以m=13
(2)(x-2)(x-18)=(x-2)x-(x-2)18=x^2-2x-18x+36=x^2-20x+36所以m=-20
(3)(x+3)(x+p)=(x+3)x+(x+3)p=x^2+3x+px+3p
3p=36所以p=12,3+p=15所以m=15
全部展开
(1)(x+4)(x+9)=(x+4)x+(x+4) 9=x^2+4x+9x+36=x^2+13x+36所以m=13
(2)(x-2)(x-18)=(x-2)x-(x-2)18=x^2-2x-18x+36=x^2-20x+36所以m=-20
(3)(x+3)(x+p)=(x+3)x+(x+3)p=x^2+3x+px+3p
3p=36所以p=12,3+p=15所以m=15
(4)(x-6)(x-p)=x^2-(6+p)x+6p 因为6p=36.
所以p=6; -(6+p)=m, m=-6-p=-12
(5)(x+p)(x+q)=x^2+(p+q)x+pq
因为pq=36,pq为正整数
p=1, q=36时m=37;
p=2, q=18时m=20
p=3, q=12时,m=15
p=4, q=9时, m=13
p=6, q=6时,m=12
收起
1.(x+4)(x+9)=x^2+13x+39 m=13
2.(x-2)(x-18)=x^2-20x+36 m=-20
3.(x+3)(x+p)=x^2+(3+p)x+3p 所以3p=36, p=12; (3+p)=15, m=15
4.(x-6)(x-p)=x^2-(6+p)x+6p 所以6p=36.
p=6; (6+p)=-m, m=-6-p=-12
5...
全部展开
1.(x+4)(x+9)=x^2+13x+39 m=13
2.(x-2)(x-18)=x^2-20x+36 m=-20
3.(x+3)(x+p)=x^2+(3+p)x+3p 所以3p=36, p=12; (3+p)=15, m=15
4.(x-6)(x-p)=x^2-(6+p)x+6p 所以6p=36.
p=6; (6+p)=-m, m=-6-p=-12
5(x+p)(x+q)=x^2+(p+q)x+pq
所以pq=36,又因为pq为正整数
所以:1. p=1, q=36; 2. p=2, q=18; 3. p=3, q=12, 4. p=4, q=9, 5.p=6, q=6
你可以根据不同情况算出不同m。
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1 m=9+4=13
2 m=-2-18=-20
3 3*p=36 p=12 m=3+12=15
4 6p=36 p=6 m=-6-6=-12
5 pq=36
p=1 q=36 m=37
p=2 q=18 m=20
p=3 q=12 m=15
p=4 q=9 m=13
p=6 q=6 m=12
(1)x^2+13x+36=x^2+mx+36
13x=mx
X=13
(2)x^2-20x+36=x^2+mx+36
-20X=mx
x=-20
(3)x^2+(P+3)x+3p=x^2+mx+36
(P+3)x+3p=mx+36
所...
全部展开
(1)x^2+13x+36=x^2+mx+36
13x=mx
X=13
(2)x^2-20x+36=x^2+mx+36
-20X=mx
x=-20
(3)x^2+(P+3)x+3p=x^2+mx+36
(P+3)x+3p=mx+36
所以3p=36
p=12
所以(12+3)x=mx
m=15
(4) x^2-(p+6)x+6p=x^2+mx+36
-(p+6)x+6p=mx+36
所以6P=36
P=6
所以-(p+6)x=mx
m=-12
(5)x^2+(p+q)x+pq=x^2+mx+36
(p+q)x+pq=mx+36
所以pq=36
p=36/q
所以(p+q)x=mx
m=36/q+q=36+q^2
第5提不确定,其他都确定
收起
(1)(x+4)(x+9)=x^2+4x+9x+36=x^2+13x+36∴m=13
(2)(x-2)(x-18)=x^2-2x-18x+36=x^2-20x+36∴m=-20
(3)(x+3)(x+p)=x^2+3x+px+36∵3p=36∴p=12
(4)(x-6)(x-p)=x^2-6x-px+36∵6p=36∴p=6
(5)(x+p)(x+q)=x^2+px...
全部展开
(1)(x+4)(x+9)=x^2+4x+9x+36=x^2+13x+36∴m=13
(2)(x-2)(x-18)=x^2-2x-18x+36=x^2-20x+36∴m=-20
(3)(x+3)(x+p)=x^2+3x+px+36∵3p=36∴p=12
(4)(x-6)(x-p)=x^2-6x-px+36∵6p=36∴p=6
(5)(x+p)(x+q)=x^2+px+qx+36,p,q为正整数,pq=
36∴当p得1、2、3、4、6、8、9、12、18、36时
q可得36、18、12、9、8、6、4、3、2、1。
收起
1)(x+4)(x+9)=x^2+mx+36
m=4+9=13
(2)(x-2)(x-18)=x^2+mx+36
m=(-2)+(-18)=-20
3)(x+3)(x+p)=x^2+mx+36
3p=36,p=12,m=3+p=3+12=15
(4)(x-6)(x-p)=x^2+mx+36
(-6)(-p)=36,p=6,
m=(-6...
全部展开
1)(x+4)(x+9)=x^2+mx+36
m=4+9=13
(2)(x-2)(x-18)=x^2+mx+36
m=(-2)+(-18)=-20
3)(x+3)(x+p)=x^2+mx+36
3p=36,p=12,m=3+p=3+12=15
(4)(x-6)(x-p)=x^2+mx+36
(-6)(-p)=36,p=6,
m=(-6)+(-p)=-12
(5)(x+p)(x+q)=x^2+mx+36,p,q为正整数
因为pq=36=1x36=2x18=3x12=4x9=6x6
所以,m=p+q=1+36=37,或m=2+18=20,
或m=3+12=15,或m=4+9=13,或m=6+6=12
以上解题过程均根据韦达定理,即根与系数的关系
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