CMR nếu a+b+c+d=0 thì \(a^3+b^3+c^3+d^3=3\left(c+d\right).\left(ab-cd\right)\)
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29 tháng 9 2019
a ) \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Nếu : \(a+b+c=0\) thì đẳng thức trên đúng .
\(\Rightarrowđpcm\)
b ) \(a+b+c+d=0\)
\(\Rightarrow a+b=-\left(c+d\right)\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(cb-cd\right)\left(đpcm\right)\)
Chúc bạn học tốt !!!
29 tháng 9 2019
a ) a^3+b^3+c^3=3abca3+b3+c3=3abc
\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0⇔(a+b)3+c3−3ab(a+b)−3abc=0
\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0⇔(a+b+c)(a2+b2+c2−ab−bc−ac)=0
Nếu : a+b+c=0a+b+c=0 thì đẳng thức trên đúng .(đpcm)
b ) a+b+c+d=0a+b+c+d=0
\Rightarrow a+b=-\left(c+d\right)\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3⇒a+b=−(c+d)⇔(a+b)3=−(c+d)3
\Leftrightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)⇔a3+b3+c3+d3=−3ab(a+b)−3cd(c+d)
\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)⇔a3+b3+c3+d3=3ab(c+d)−3cd(c+d)
\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(cb-cd\right)\left(đpcm\right)⇔a3+b3+c3+d3=3(c+d)(cb−cd)(đpcm)
P
0
KB
15 tháng 9 2018
2 ) b )
\(a+b+c+d=0\)
\(\Leftrightarrow a+b=-\left(c+d\right)\)
\(\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+3a^2b+3b^2a=-c^3-3c^2d-3d^2c-d^3\)
\(\Leftrightarrow a^3+b^3+3a^2b+3b^2a+c^3+3c^2d+3d^2c+d^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3a^2b-3b^2a-3c^2d-3d^2c\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\) \(\left(đpcm\right)\)
LT
1
25 tháng 5 2022
a+b+c+d=0
nên a+b=-(c+d)
\(a^3+b^3+c^3+d^3\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+\left(c+d\right)^3-3cd\left(c+d\right)\)
\(=\left[-\left(c+d\right)\right]^3-3ab\cdot\left[-\left(c+d\right)\right]+\left(c+d\right)^3-3cd\left(c+d\right)\)
\(=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(=3\left(c+d\right)\left(ab-cd\right)\)
29 tháng 6 2018
\(a.a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Nếu : \(a+b+c=0\) thì đẳng thức trên đúng .
\(\Rightarrowđpcm\)
\(b.a+b+c+d=0\Rightarrow a+b=-\left(c+d\right)\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(ab-cd\right)\left(đpcm\right)\)
TT
1
5 tháng 7 2023
\(a+b+c+d=0\Rightarrow a+b=-\left(c+d\right)\)
\(\Rightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Rightarrow\left(a+b\right)^3+\left(c+d\right)^3=0\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)+c^3+d^3+3cd\left(c+d\right)=0\)
\(\Rightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
\(\Rightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\) (do \(a+b=-\left(c+d\right)\)
\(\Rightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\)
26 tháng 6 2018
\(a.a^3+b^3+c^3=3abc\)
⇔ \(a^3+b^3+c^3-3abc=0\)
⇔ \(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
⇔ \(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
⇔\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
⇔ \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
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)
NN
1
2 tháng 7 2017
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)