Cho a+b+c+d=0. Chứng minh: \(a^3+b^3+c^3+d^3=3.\left(b+c\right).\left(ad-bc\right)\)
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Cho a+b+c+d=0. Chứng minh: \(a^3+b^3+c^3+d^3=3.\left(b+c\right).\left(ad-bc\right)\)
Cho a+b+c+d=0. Chứng minh: \(a^3+b^3+c^3+d^3=3.\left(b+c\right).\left(ad-bc\right)\)
Cho a + b + c + d = 0. Chứng minh rằng: \(a^3+b^3+c^3+d^3=3\left(b+c\right)\left(ad-bc\right)\)
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)
chứng minh rằng:
Nếu a+b+c+d=0 thì \(a^3+b^3+c^3+d^3=3\left(b+c\right)\left(ad-bc\right)\)
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 )
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(b+c\right)\left(ad-bc\right)\)
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Chứng minh rằng nếu a + b + c + d = 0 thì
a)\(a^3+b^3+c^3+d^3=3\left(b+d\right)\left(ac-bd\right)\)
b)\(\left(b+d\right)\left(ac-bd\right)=\left(b+c\right)\left(cd-bc\right)\)
Cho a+b+c+d=0.CMR: \(a^3+b^3+c^3+d^3=3\left(b+c\right).\left(ad-bc\right)\)
Cho : a + b + c + d = 0
Chứng minh rằng \(a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\)
\(a+b+c+d=0\Rightarrow a+b=-\left(c+d\right)\)
\(\Rightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Rightarrow\left(a+b\right)^3+\left(c+d\right)^3=0\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)+c^3+d^3+3cd\left(c+d\right)=0\)
\(\Rightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
\(\Rightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\) (do \(a+b=-\left(c+d\right)\)
\(\Rightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\)
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Cho a+b+c+d= 0
CMR : \(a^3+b^3+c^3+d^3=3\left(b+c\right)\left(ad-bc\right)\)
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 ...
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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)
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Cho a+b+c=0
a) Chứng minh rằng: \(a^3+a^2c-abc+b^2c+b^3=0\)
b) Áp dụng phân tích thành nhân tử đa thức: \(A=bc\left(a+d\right)\left(b-c\right)-ac\left(b+d\right)\left(a-c\right)+ab\left(c+d\right)\left(a-b\right)\)