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8 tháng 9 2016

Ta có :

\(a+b+c+d=0\)

\(\Rightarrow b+c=-\left(a+d\right)\)

\(\Rightarrow\left(b+c\right)^2=\left(a+d\right)^2\)

\(\Rightarrow\left(b+c\right)^2-\left(a+d\right)^2=0\)

\(\Rightarrow b^2+c^2+2bc-a^2-d^2-2ad=0\)

Lại có :

\(a^3+b^3+c^3+d^3\)

\(=\left(a+d\right)\left(a^2+d^2-ad\right)+\left(b+c\right)\left(b^2+c^2-bc\right)\)

\(=\left(b+c\right)\left(b^2+c^2-bc\right)-\left(b+c\right)\left(a^2+d^2-ad\right)\)

\(=\left(b+c\right)\left[\left(b^2+c^2-bc\right)-\left(a^2+d^2-ad\right)\right]\)

\(=\left(b+c\right)\left[\left(b^2+c^2+2bc-a^2-d^2-2ad\right)+3ad-3bc\right]\)

\(=\left(b+c\right)\left[0+3\left(ad-bc\right)\right]\)

\(=3\left(b+c\right)\left(ad-bc\right)\)

Vậy ...

8 tháng 9 2016

  Ta có : a + b +c + d = 0

                  => a + d = - b - c

                 => (a + d) = -(b + c) 

                => (a + d)3 = -(b + c)3

a3 + 3a2d + 3ad2 + d3 = -(b3 + 3b2c + 3bc2 + c3)

a3 + 3a2d + 3ad2 + d3 = -b3 - 3b2c - 3bc2 - c3

       a3 + b3 + c3 + d3 = -3a2d - 3ad2 - 3b2c - 3bc2 

       a3 + b3 + c3 + d3 = -3ad(a + d) - 3bc(b + c)

       a3 + b3 + c3 + d3 = -3ad(-b - c) - 3bc(b + c)

       a3 + b3 + c3 + d3 = 3ad(b + c) - 3bc(b + c)

       a3 + b3 + c3 + d3 = 3(b + c)(ad - bc)

Ta có: a+b+c+d=0

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

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

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

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

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

\(\Leftrightarrow a^3+b^3+c^3+d^3=-3bc\left(b+c\right)-3ad\cdot\left[-\left(b+c\right)\right]\)

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

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

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 8 2017

4) Ta có : A=(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)

=> (a+d)2 - (b+c)2= (a-d)2 - (c-b)2

=> a2+ d2+ 2ad - b2- c2- 2bc=a2 + d2 - 2ad - c2-b2+2bc

Rút gọn ta được: 4ad = 4bc => ad = bc =>\(\dfrac{a}{c}=\dfrac{b}{d}\)

13 tháng 8 2017

1) a2+b2+c2+3=2(a+b+c) =>(a-1)2+(b-1)2+(c-1)2=0

=> a-1=b-1=c-1=0 => a=b=c=1 =>đpcm

a+b+c+d=0

nên a+b=-(c+d)

\(a^3+b^3+c^3+d^3\)

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

\(=\left[-\left(c+d\right)\right]^3-3ab\cdot\left[-\left(c+d\right)\right]+\left(c+d\right)^3-3cd\left(c+d\right)\)

\(=3ab\left(c+d\right)-3cd\left(c+d\right)\)

\(=3\left(c+d\right)\left(ab-cd\right)\)