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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)
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
\(a+b+c+d=0\Rightarrow a+b=-\left(c+d\right)\)
\(\Rightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Rightarrow\left(a+b\right)^3+\left(c+d\right)^3=0\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)+c^3+d^3+3cd\left(c+d\right)=0\)
\(\Rightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
\(\Rightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\) (do \(a+b=-\left(c+d\right)\)
\(\Rightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\)
a ) Ta có : \(a+b+c=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+ac+bc\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+ac+bc\right)^2\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4\left(a^2b^2+b^2c^2+c^2a^2+2ab^2c+2a^2bc+2c^2ab\right)\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)+8abc\left(a+b+c\right)\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+a^2c^2\right)+8abc.0\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
Lại có : \(\dfrac{\left(a^2+b^2+c^2\right)^2}{2}=\dfrac{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)}{2}\)
\(=\dfrac{a^4+b^4+c^4+a^4+b^4+c^4}{2}=\dfrac{2\left(a^4+b^4+c^4\right)}{2}\)
\(=a^4+b^4+c^4\left(đpcm\right)\)
b ) \(a+b+c+d=0\)
\(\Leftrightarrow a+b=-\left(c+d\right)\)
\(\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow\left(a+b\right)^3+\left(c+d\right)^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+d^3+3a^2b+3b^2a+3c^2d+3d^2c=0\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3a^2b-3b^2a-3c^2d-3d^2c\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(-a^2b-b^2a-c^2d-d^2c\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left[-ab\left(a+b\right)-cd\left(c+d\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left[ab\left(c+d\right)-cd\left(c+d\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\left(đpcm\right)\)
\(A=\dfrac{-\left(ac+bc+ad+bd\right)-\left(cd-ca-bd+ba\right)}{\left(ab+bc+cd+ad\right)\cdot abcd}\)
\(=\dfrac{-ac-bc-ad-bd-cd+ca+bd-ba}{\left(ab+bc+cd+ad\right)\cdot abcd}\)
\(=\dfrac{-bc-ad-cd-ba}{\left(ab+bc+cd+ad\right)\cdot abcd}=-\dfrac{1}{abcd}\)