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Theo đề, a+b+c+d=0
\(\Rightarrow a+b=-\left(c+d\right)\)
Ta có: \(VT=\left(a+b\right)\left(a^2-ab+b^2\right)+\left(c+d\right)\left(c^2-cd+d^2\right)\)
\(\Leftrightarrow VT=\left(c+d)\left(c^2-cd+d^2-a^2+ab-b^2\right)\right)\)
Để có ĐPCM ta xét hiệu: \(c^2-cd+d^2-a^2+ab-b^2-3\left(ab+cd\right)=c^2-4cd+d^2-a^2-2ab-b^2=c^2-4cd+d^2-\left(a+b\right)^2=c^2-4cd+d^2-\left(c+d\right)^2=-6cd\)
S nó ko = 0 ta:::xem lại đề..Hay mk lm sai j đó
\(a.a^3+b^3+c^3=3abc\)
⇔ \(a^3+b^3+c^3-3abc=0\)
⇔ \(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
⇔ \(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
⇔\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
⇔ \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Với : a + b + c = 0 thì dễ thấy đẳng thức trên đúng .
Từ đó suy ra : đpcm .
\(b.a+b+c+d=0\)
⇔ \(a+b=-\left(c+d\right)\)
⇔ \(\left(a+b\right)^3=-\left(c+d\right)^3\)
⇔ \(a^3+b^3+3a^2b+3ab^2=-\left(c^3+3c^2d+3cd^2+d^3\right)\)
⇔ \(a^3+b^3+c^3+d^3=-3c^2d-3cd^2-3a^2b-3ab^2\)
⇔ \(a^3+b^3+c^3+d^3=-3cd\left(c+d\right)-3ab\left(a+b\right)\)
⇔ \(a^3+b^3+c^3+d^3=-3cd\left(c+d\right)+3ab\left(c+d\right)\)
⇔ \(a^3+b^3+c^3+d^3=3\left(c+d\right)\left(ab-cd\right)\) ( đpcm)
Ta có: a+b+c+d=0
⇔\(a+d=-\left(b+c\right)\)
\(\Leftrightarrow\left(a+d\right)^3=-\left(b+c\right)^3\)
\(\Leftrightarrow a^3+d^3+3ad\left(a+d\right)=-\left[b^3+c^3+3bc\left(b+c\right)\right]\)
\(\Leftrightarrow a^3+d^3+3ad\left(a+d\right)=-b^3-c^3-3bc\left(b+c\right)\)
\(\Leftrightarrow a^3+d^3+b^3+c^3=-3ad\left(a+d\right)-3bc\left(b+c\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ad\left(a+d\right)+3bc\left(a+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=\left(a+d\right)\left(-3ad+3bc\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=\left(a+d\right)\cdot3\cdot\left(-ad+bc\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-\left(b+c\right)\cdot3\cdot\left[-\left(ad-bc\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\cdot\left(b+c\right)\cdot\left(ad-bc\right)\)(đpcm)
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)
thử bài bất :D
Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)
Hoàn toàn tương tự:
\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)
\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)
Cộng (*),(**),(***) vế theo vế ta được:
\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)
Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )
Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi a=b=c=1
1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D
a+b+c+d=0
nên a+b=-(c+d)
\(a^3+b^3+c^3+d^3\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+\left(c+d\right)^3-3cd\left(c+d\right)\)
\(=\left[-\left(c+d\right)\right]^3-3ab\cdot\left[-\left(c+d\right)\right]+\left(c+d\right)^3-3cd\left(c+d\right)\)
\(=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(=3\left(c+d\right)\left(ab-cd\right)\)
\(a+b+c+d=0\Rightarrow a+b=-\left(c+d\right)\)
\(\Rightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Rightarrow\left(a+b\right)^3+\left(c+d\right)^3=0\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)+c^3+d^3+3cd\left(c+d\right)=0\)
\(\Rightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
\(\Rightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\) (do \(a+b=-\left(c+d\right)\)
\(\Rightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\)