因式分解难度中等的题目~ 求100道因式分解计算题的题目+答案、要中等难度
\u56e0\u5f0f\u5206\u89e3\u96be\u5ea6\u4e2d\u7b49\u7684\u4e00\u4e9b\u9898(1)\u539f\u5f0f=(abc+ab)+(ac+a)+(bc+b)+(c+1)
=ab(c+1)+a(c+1)+b(c+1)+(c+1)
=(c+1)(ab+a+b+1)=(a+1)(b+1)(c+1)
(2)\u539f\u5f0f=x^3-2x^2+4x^2-8x+3x-6
=(x^3-2x^2)+(4x^2-8x)+(3x-6)
=x^2(x-2)+4x(x-2)+3(x-2)
=(x-2)(x^2+4x+3)=(x-2)(x+1)(x+3)
(3)\u539f\u5f0f=x^3+y^3+x^2y+y^2x+xy+xy
=....\u786e\u5b9a\u9898\u76ee\u6ca1\u6284\u9519\u5417\uff1f\u5e94\u8be5\u540e\u9762\u8fd8\u6709\u9879\u6f0f\u6284\u4e86\u3002\u3002\u5982\u679c\u6ca1\u9519\uff0c\u8fd9\u6837\u5c31\u662f\u6700\u7b80\u5f0f\u4e86
(4)\u539f\u5f0f=x^4-x^3+x^3-x^2+x^2-x-3x+6
=(x^4-x^3)+(x^3-x^2)+(x^2-x)-(3x-6)=(x-1)(x^3+x^2+x-3)
=(x-1)[(x^3-1)+(x^2-1)+(x-1)]
=(x-1)^2(x^2+2x+3)
(5)\u539f\u5f0f=m^4+64n^4+16m^2n^2-16m^2n^2
=(m^2+8n^2)^2-(4mn)^2
=(m^2+8n^2+4mn)(m^2+8n^2-4mn)
\u5206\u89e3\u56e0\u5f0f\u7684\u601d\u8def\u4e00\u822c\u90fd\u662f\u80fd\u914d\u65b9\u7684\u5c31\u914d\u65b9\uff0c\u63d0\u53d6\u516c\u56e0\u5f0f\uff01
\u4f60\u8fd8\u4e0d\u5982\u53bb\u4e70\u672c\u4e66\u5462\uff0c\u73b0\u5728\u53c2\u8003\u8d44\u6599\u5f88\u8be6\u7ec6 = = \uff0c100\u9053\u9898\u6253\u5b8c\u624b\u90fd\u6b8b\u4e86
1.f(x)=x^3-3x^2-13x+15f(x)=x^3-x^2-2x^2+2x-15x+15
f(x)=(x-1)x^2-2x(x-1)-15(x-1)
f(x)=(x^2-2x-15)(x-1)
f(x)=(x-5)(x+3)(x-1)
2.f(x)=3x^5-3x^4-13x^3-11x^2-10x-6
f(x)=3x^5+3x^4-6x^4-6x^3-7x^3-7x^2-4x^2-4x-6x-6
f(x)=(3x^4-6x^3-7x^2-4x-6)(x+1)
f(x)=(3x^4+3x^3-9x^3-9x^2+2x^2+2x-6x-6)(x+1)
f(x)=(x+1)^2(3x^3-9x^2+2x-6)
f(x)=(x+1)^2(x-3)(3x^2+2)
3.f(x,y,z)=x^2(y-z)+y^2(z-x)+z^2(x-y)
f(x,y,z)=x^2y-x^2z+y^2z-y^2x+z^2(x-y)
f(x,y,z)=xy(x-y)-z(x^2-y^2)+z^2(x-y)
f(x,y,z)=(x-y)(xy+z^2-xz-yz)
f(x,y,z)=(x-y)((xy-xz)-(yz-z^2)
f(x,y,z)=(x-y(y-z)(x-z)
4.f(a,b,c)=(a+b+c)(ab+bc+ca)-abc
=2abc+a2b+a2c+b2a+b2c+c2a+c2b
=(2abc+a2c+b2c)+(a2b+b2a)+c2(a+b)
=c(a+b)^2+ab(a+b)+c^2(a+b)
=(a+b)(ac+bc+ab+c^2)
=(a+b)(a+c)(b+c)
5.问：当k为何值时，关于x,y的多项式kx^2-2xy-3y^2+3x-5y+2可以分解为两个一次因式的乘积？
设这两个一次因式的乘积为
(ax+by+c)(a'x+b'y+c')=kx^2-2xy+3y^2+3x-5y+2
当x=0时,原式=(by+c)(b'y+c')=-3y^2-5y+2=-(3y-1)(y+2)
(ax-3y+1)(a'x+y+2)
=a'ax^2+(-3a'+a)xy-3y^2+(2a+a')x-5y+2=kx^2-2xy-3y^2+3x-5y+2
-3a'+a=-2,
2a'+a=3,a'=1,a=1,
k=aa'=1
所以k=1
6.证明：多项式x^2-xy+y^2+x+y不能分解为两个一次因式的乘积
如果能分解为两个一次因式乘积,设两个因式为(x+my+a)和(x+ny+b), a,b,m,n是实数
(x+my+a)(x+ny+b)
=x^2+mny^2+(m+n)xy+a(x+ny)+b(x+my)+ab
由此得出以下等式关系
m+n=-1
mn=1
a+b=1
an+bm=1
ab=0
a和b有一个为0,另一个为1
由an+bm=1可知n,m中有一个是1
根据mn=1可知,m,n均为1
则可推出m+n=2,与m+n=-1矛盾
所以不能分解为两个一次因式的乘积
1.x^3-3x^2-13x+15
=x^3-3x^2+2x-15x+15
=x(x^2-3x+2)-15(x-1)
=x(x-2)(x-1)-15(x-1)
=(x-1)[x(x-2)-15]
=(x-1)[x^2-2x-15]
=(x-1)(x-5)(x+3)
2.我这不会做。希望对您有帮助，谢谢！祝您成功！请采纳我吧，谢谢！
3.x^2(y-z)+y^2(z-x)+z^2(x-y)
=x^2y-x^2z+y^2z-y^2x+z^2x-z^2y
=y(x^2-z^2)-xz(x-z)-y^2(x-z)
=y(x+z)(x-z)-xz(x-z)-y^2(x-z)
=(x-z)(xy+yz-xz-y^2)
=(x-z)[x(y-z)-y(y-z)]
=(x-z)(y-z)(x-y)
4.(a+b+c)(ab+bc+ca)-abc
=a^2b+2abc+ca^2+ab^2+b^2c+bc^2+c^2a
=(a^2b+ab^2)+(bc^2+ac^2)+(2cab+ca^2+cb^2)
=ab(a+b)+c^2(a+b)+c(a+b)^2
=(a+b)(ab+c^2+ac+cb)
=(a+b)[(ab+bc)+(c^2+ac)]
=(a+b)(a+c)(b+c)
5.(ax+by+c)(dx+ey+f)=kx^2-2xy+3y^2+3x-5y+2
设x=0,(by+c)(ey+f)=3y^2-5y+2=(3y-2)(y-1)
(ax+3y-2)(dx+y-1)=kx^2-2xy+3y^2+3x-5y+2
3d+a=-2,
-2d-a=3,d=-1,a=1,
k=ad=-1
6.反证法：
因为x^2和y^2前面的系数都是1，
所以设 x^2-xy+y^2+x+y = （x+y+a)(x+y+b)
（x+y+a)(x+y+b)=x^2+xy+bx+xy+y^2+by+ax+ay+ab
=x^2+2xy+y^2+(a+b)x+(a+b)y+ab
用系数待定法,两式相减：-xy+x+y=2xy+(a+b)x+(a+b)y+ab
有ab=0,a+b=1,a和b有解，但是xy的系数-1显然不等于2
所以（x+y+a)(x+y+b)不成立，即如题
1.x^3-3x^2-13x+15
=x^3-3x^2+2x-15x+15
=x(x^2-3x+2)-15(x-1)
=x(x-2)(x-1)-15(x-1)
=(x-1)[x(x-2)-15]
=(x-1)[x^2-2x-15]
=(x-1)(x-5)(x+3)
2.f(x)=3x^5-3x^4-13x^3-11x^2-10x-6
f(x)=3x^5+3x^4-6x^4-6x^3-7x^3-7x^2-4x^2-4x-6x-6
f(x)=(3x^4-6x^3-7x^2-4x-6)(x+1)
f(x)=(3x^4+3x^3-9x^3-9x^2+2x^2+2x-6x-6)(x+1)
f(x)=(x+1)^2(3x^3-9x^2+2x-6)
f(x)=(x+1)^2(x-3)(3x^2+2)
3.f(x,y,z)=x^2(y-z)+y^2(z-x)+z^2(x-y)
f(x,y,z)=x^2y-x^2z+y^2z-y^2x+z^2(x-y)
f(x,y,z)=xy(x-y)-z(x^2-y^2)+z^2(x-y)
f(x,y,z)=(x-y)(xy+z^2-xz-yz)
f(x,y,z)=(x-y)((xy-xz)-(yz-z^2)
f(x,y,z)=(x-y(y-z)(x-z)
4.(a+b+c)(ab+bc+ca)-abc
=a^2b+2abc+ca^2+ab^2+b^2c+bc^2+c^2a
=(a^2b+ab^2)+(bc^2+ac^2)+(2cab+ca^2+cb^2)
=ab(a+b)+c^2(a+b)+c(a+b)^2
=(a+b)(ab+c^2+ac+cb)
=(a+b)[(ab+bc)+(c^2+ac)]
=(a+b)(a+c)(b+c)
5.(ax+by+c)(dx+ey+f)=kx^2-2xy+3y^2+3x-5y+2
设x=0,(by+c)(ey+f)=3y^2-5y+2=(3y-2)(y-1)
(ax+3y-2)(dx+y-1)=kx^2-2xy+3y^2+3x-5y+2
3d+a=-2,
-2d-a=3,d=-1,a=1,
k=ad=-1
6.反证法：
因为x^2和y^2前面的系数都是1，
所以设 x^2-xy+y^2+x+y = （x+y+a)(x+y+b)
（x+y+a)(x+y+b)=x^2+xy+bx+xy+y^2+by+ax+ay+ab
=x^2+2xy+y^2+(a+b)x+(a+b)y+ab
用系数待定法,两式相减：-xy+x+y=2xy+(a+b)x+(a+b)y+ab
有ab=0,a+b=1,a和b有解，但是xy的系数-1显然不等于2
所以（x+y+a)(x+y+b)不成立，即如题
1.(x-1)(x+3)(x-5)
2.(x+1)^2(x-3)(3x^2+2) (学过复数的话再分解一步)
3.-(x-y)(y-z)(z-x)
4.(a+b)(b+c)(c+a)
5.k=1 (提示：先把y和常数项摆平)
6.没学过复数的话说二次项Δ<0即可。学过复数的话先分解二次项，然后两个常数项要满足三个方程，容易说明无解。
我发好了，lz撒花吧
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