2x^2+3x-5 . phân tích thành đa thức

Giải:

Từ \(a+b+c+d=0\Leftrightarrow a+c=-\left(b+d\right)\)

\(\Leftrightarrow\left(a+c\right)^3=-\left(b+d\right)^3\)

\(\Leftrightarrow a^3+c^3+3ac\left(a+c\right)=-\left[b^3+d^3+3bd\left(b+d\right)\right]\)

\(\Leftrightarrow VT=a^3+b^3+c^3+d^3=-3bd\left(b+d\right)-3ac\left(a+c\right)\)

\(=-3bd\left(b+d\right)+3ac\left(b+d\right)=3\left(ac-bd\right)\left(b+d\right)=VP\) (Đpcm)

help me please

ab=cd

a=b=c=d=0

P khong xac dinh

de thieu

Ta có \(a+b+c+d=0\Leftrightarrow a+c=-\left(b+d\right)\Leftrightarrow\left(a+c\right)^3=\left[-\left(b+d\right)\right]^3\Leftrightarrow a^3+3a^2c+3ac^2+c^3=-b^3-3b^2d-3bd^2-d^3\Leftrightarrow a^3+b^3+c^3+d^3=-3a^2c-3ac^2-3b^2d-3bd^2\Leftrightarrow a^3+b^3+c^3+d^3=-3ac\left(a+c\right)-3bd\left(b+d\right)\Leftrightarrow a^3+b^3+c^3+d^3=3ac\left(b+d\right)-3bd\left(b+d\right)\Leftrightarrow a^3+b^3+c^3+d^3=3\left(b+d\right)\left(ac-bd\right)\)Vậy \(a+b+c+d=0\) thì \(a^3+b^3+c^3+d^3=3\left(b+d\right)\left(ac-bd\right)\)

P=\(\frac{\left(a+c\right)\left(a+d\right)\left(b+c\right)\left(b+d\right)}{\left(a+b+c+d\right)^2}\)=\(\frac{\left(a^2+ad+ac+cd\right)\left(b^2+bd+bc+cd\right)}{\left(a+b+c+d\right)^2}\)

=\(\frac{\left(a^2+ac+ad+ab\right)\left(b^2+bc+bd+ab\right)}{\left(a+b+c+d\right)^2}\) (do ab=cd)

=\(\frac{a\left(a+b+c+d\right)b\left(a+b+c+d\right)}{\left(a+b+c+d\right)^2}\)

=\(\frac{ab\left(a+b+c+d\right)^2}{\left(a+b+c+d\right)^2}\)=ab