X2-(m2+n2)x=mn(m2-n2)

熱心網友

解:由x^2-(m^2+n^2)x=mn(m^2-n^2)得:x^2-(m^2+n^2)x-mn(m+n)(m-n)=0∴[x-m(m+n)][x+n(m-n)]=0∴原方程的兩根為:x1=m(m+n),x2=n(n-m).

熱心網友

X2-(m2+n2)x=mn(m2-n2)X2-(m2+n2)x-mn(m2-n2)=0mn(m2-n2)=mn*(m+n)*(m-n)(x-m*(m+n))*(x+n*(m-n))=0x1=m(m+n)x2=-n(m-n)