Cho a+b+c+d=0, Chứng Minh Rằng : a^3+b^3+c^3+d^3=3.(b+c).(ad-bc)
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Cho a+b+c+d=0. Chứng minh rằng a^3+b^3+c^3+d^3=3(b+c)(ad-bc)
Hiuhiu mọi ngừi giúp mik vứii aaaT.T
a+b+c+d=0
=>a+d=-(b+c)
=>(a+d)^3=-(b+c)^3
=>\(a^3+d^3+3ad\left(a+d\right)=-b^3-c^3-3bc\left(b+c\right)\)
=>\(a^3+d^3+3ad\left(a+d\right)=-b^3-c^3+3bc\left(a+d\right)\)
=>\(a^3+d^3+b^3+c^3=3bc\left(a+d\right)-3ad\left(a+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(a+d\right)\left(bc-ad\right)\)
=>\(a^3+b^3+c^3+d^3=3\left(b+c\right)\left(ad-bc\right)\)
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Cho a + b + c + d = 0. Chứng minh 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)
Cho a+b+c+d=0. Chứng minh rằng :
a3+b3+c3+d3=3(b+c)(ad-bc)
ta có : a+b+c+d=0
=>a+b=-(c+d)
=> (a+b)3=-(c+d)3
=> a3+b3+3ab(a+b)=-c3-d3-3cd(c+d)
=> a3+b3+c3+d3=-3ab(a+b)-3cd(c+d)
=> a3+b3+c3+d3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d))
=> a3 +b3+c3+d3==3(c+d)(ab-cd)
(dpcm)
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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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Cho a+b+c+d=0. Chứng minh rằng :
a3+b3+c3+d3=3(b+c)(ad-bc)
a+b+c+d=0 => a+d= -b-c; (a+b)3=a3+b3+3ab(a+b) => a3+b3=(a+b)3-3ab(a+b)
a3+d3+b3+d3
=(a+d)3- 3ad(a+d)+ (b+c)3-3bc(b+c) (1)
Do a+d=-b-c nên pt (1) trở thành:
-(b+c)3-3ad(-b-c)+ (b+c)3-3bc(b+c)
=3ad(b+c)-3bc(b+c)
=3(b+c)(ad-bc) <đccm>
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: \(a^3+b^3+c^3+d^3=3.\left(b+c\right).\left(ad-bc\right)\)
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)
Cho a+b+c+d=0. Chứng minh: \(a^3+b^3+c^3+d^3=3.\left(b+c\right).\left(ad-bc\right)\)
a)Tìm số chính phương có 4 chữ số biết rằng 2 chữ số đầu giống nhau, 2 chữ số cuối giống nhau.
b)Cho a+b+c+d=0. Chứng minh rằng:
a3+b3+c3+d3=3(b+c)(ad-bc).