CM : a) Nếu a+b +c = 0 thì \(a^3+b^3+c^3=3abc\)
b) 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 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 .
\(\Rightarrowđpcm\)
\(b.a+b+c+d=0\Rightarrow a+b=-\left(c+d\right)\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(ab-cd\right)\left(đpcm\right)\)
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)
EC
0
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)
KB
18 tháng 9 2018
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 : \(\dfrac{\left(a^2+b^2+c^2\right)^2}{2}=\dfrac{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)}{2}\)
\(=\dfrac{a^4+b^4+c^4+a^4+b^4+c^4}{2}=\dfrac{2\left(a^4+b^4+c^4\right)}{2}\)
\(=a^4+b^4+c^4\left(đpcm\right)\)
KB
18 tháng 9 2018
b ) \(a+b+c+d=0\)
\(\Leftrightarrow a+b=-\left(c+d\right)\)
\(\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow\left(a+b\right)^3+\left(c+d\right)^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+d^3+3a^2b+3b^2a+3c^2d+3d^2c=0\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3a^2b-3b^2a-3c^2d-3d^2c\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(-a^2b-b^2a-c^2d-d^2c\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left[-ab\left(a+b\right)-cd\left(c+d\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left[ab\left(c+d\right)-cd\left(c+d\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\left(đpcm\right)\)
CV
3
28 tháng 11 2016
ta có
a+b+c+d=0
=> b+c=-(a+d) => (b+c)3=-(a+d)3
=> b3+c3+3bc(b+c)= -[a3+d3+3ad(a+d)]
=> a3+b3+c3+d3=-3ad(a+d)-3bc(b+c)= 3ad(b+c)-3bc(b+c)
=3(b+c)(ad-bc)
KB
15 tháng 9 2018
2 ) b )
\(a+b+c+d=0\)
\(\Leftrightarrow a+b=-\left(c+d\right)\)
\(\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+3a^2b+3b^2a=-c^3-3c^2d-3d^2c-d^3\)
\(\Leftrightarrow a^3+b^3+3a^2b+3b^2a+c^3+3c^2d+3d^2c+d^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3a^2b-3b^2a-3c^2d-3d^2c\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\) \(\left(đpcm\right)\)
13 tháng 7 2016
Ta có : \(a+b+c+d=0\Leftrightarrow a+d=-\left(b+c\right)\)
\(\Leftrightarrow\left(a+d\right)^3=-\left(b+c\right)^3\)
\(\Leftrightarrow a^3+d^3+3ad\left(a+d\right)=-\left[c^3+b^3+3bc\left(b+c\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ad\left(a+d\right)-3bc\left(b+c\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ad\left(b+c\right)-3bc\left(b+c\right)\) (vì a + d = - b - c )
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(b+c\right)\left(ad-bc\right)\)
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 !!!
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)