因式分解(1)3m2-10mn-
1)3m^2-10mn-12n^2
=3(m^2-10mn/3)-12n^2
=3[m^2-2m(5n/3)+25m^2/9-25n^2/9]-12n^2
=3(m-5n/3)^2-61n^2/3
=3[m-(5-√61)n/3]*[m-(5+√60)n/3]
2)m^2+(m+1)^2+(m^2+m)^2
=m^2+(m^2+2m+1)+(m^2+m)^2
=1+2(m^2+m)+(m^2+m)^2
=(m^2+m+1)^2
3)设x^2-mx-4=(x+1)(x+n)=x^2+(n+1)x+n
比较两边的同类项的系数得到方程组
-m=n+1,n=-4
--->m=3,n=-4
所以另一...全部
1)3m^2-10mn-12n^2
=3(m^2-10mn/3)-12n^2
=3[m^2-2m(5n/3)+25m^2/9-25n^2/9]-12n^2
=3(m-5n/3)^2-61n^2/3
=3[m-(5-√61)n/3]*[m-(5+√60)n/3]
2)m^2+(m+1)^2+(m^2+m)^2
=m^2+(m^2+2m+1)+(m^2+m)^2
=1+2(m^2+m)+(m^2+m)^2
=(m^2+m+1)^2
3)设x^2-mx-4=(x+1)(x+n)=x^2+(n+1)x+n
比较两边的同类项的系数得到方程组
-m=n+1,n=-4
--->m=3,n=-4
所以另一个多项式是x-4。
4)|P+2|+q^2-8q+16=0--->||p+2|+(q-4)^2=0
因为|P+2|>=0,(q-4)^2>=0所以|p+2|+(q-4)^2>=0，现在“=”成立所以|p+2|=0并且(q-4)^2=0
所以p=-2，q=4
因此x^2+y^2-pxy-q=(x^2+2xy+y^2)-4=(x+y)^2-2^2=(x+y-2)(x+y_2)。
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