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Ta có : \(a+b+c+d=0\)
\(\Leftrightarrow a+b=-c-d\)
\(\Leftrightarrow\left(a+b\right)^3=\left(-c-d\right)^3\)
\(\Leftrightarrow a^3+b^3+3ab.\left(a+b\right)=-c^3-d^3+3cd.\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3cd.\left(c+d\right)-3ab.\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3.cd.\left(a+b\right)+3ab.\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3.\left(c+d\right)\left(cd+ab\right)\)
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=3cd.(c+d)−3ab.(a+b)
⇔a3+b3+c3+d3=3.cd.(a+b)+3ab.(c+d)
⇔a3+b3+c3+d3=3.(c+d)(cd+ab)
Bài 3:
a: \(\left(a-b\right)^2=\left(a+b\right)^2-4ab=7^2-4\cdot12=1\)
b: \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=7^3-3\cdot12\cdot7\)
\(=343-252=91\)
a )
`VP= (a+b)^3-3ab(a+b)`
`=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2`
`=a^3+b^3 =VT (đpcm)`
b)
b) Ta có
`VT=a3+b3+c3−3abc`
`=(a+b)3−3ab(a+b)+c3−3abc`
`=[(a+b)3+c3]−3ab(a+b+c)`
`=(a+b+c)[(a+b)2+c2−c(a+b)]−3ab(a+b+c)`
`=(a+b+c)(a2+b2+2ab+c2−ac−bc−3ab)`
`=(a+b+c)(a2+b2+c2−ab−bc−ca)=VP`
a) Ta có:
`VP= (a+b)^3-3ab(a+b)`
`=a^3 + b^3+3ab ( a + b )- 3ab ( a + b )`
`=a^3 + b^3=VT(dpcm)`
b) Ta có
`VT=a^3+b^3+c^3−3abc`
`=(a+b)^3−3ab(a+b)+c^3−3abc`
`=[(a+b)^3+c^3]−3ab(a+b+c)`
`=(a+b+c)[(a+b)^2+c^2−c(a+b)]−3ab(a+b+c)`
`=(a+b+c)(a^2+b^2+2ab+c^2−ac−bc−3ab)`
`=(a+b+c)(a^2+b^2+c^2−ab−bc−ca)=VP`
Ta có:
\(a^3+b^3+c^3+d^3\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+\left(c+d\right)^3-3cd\left(c+d\right)\)
\(=-\left(c+d\right)^3+3ab\left(c+d\right)+\left(c+d\right)^3-3cd\left(c+d\right)\) (vì \(a+b=-\left(c+d\right)\))
\(=3\left(c+d\right)\left(ab-cd\right)\)
Vậy đẳng thức được chứng minh.
a) B = x2 + 4y2 - 5x + 10y - 4xy + 17
= ( x2 - 4xy + 4y2 ) - ( 5x - 10y ) + 17
= ( x - 2y )2 - 5( x - 2y ) + 17
= 52 - 5.5 + 17
= 17
b) C = 2( a3 + b3 ) - 3( a2 + b2 )
= 2( a + b )( a2 - ab + b2 ) - 3( a2 + b2 )
= 2( a2 - ab + b2 ) - 3a2 - 3b2 ( gt a + b = 1 )
= 2a2 - 2ab + 2b2 - 3a2 - 3b2
= -a2 - 2ab - b2
= -( a2 + 2ab + b2 )
= -( a + b )2
= -1
c) a + b + c + d = 0
<=> a + b = -( c + d )
<=> ( a + b )3 = -( c + d )3
<=> a3 + 3a2b + 3ab2 + b3 = -( c3 + 3c2d + 3cd2 + d3 )
<=> a3 + 3a2b + 3ab2 + b3 = -c3 - 3c2d - 3cd2 - d3
<=> a3 + b3 + c3 + d3 = -3c2d - 3cd2 - 3a2b - 3ab2
<=> a3 + b3 + c3 + d3 = -3cd( c + d ) - 3ab( a + b )
<=> a3 + b3 + c3 + d3 = 3ab( c + d ) - 3cd( c + d ) < Do ( a + b ) = -( c + d ) >
<=> a3 + b3 + c3 + d3 = 3( ab - cd )( c + d )
<=> a3 + b3 + c3 + d3 - 3( ab - cd )( c + d ) = 0
Cảm ơn bạn TRẦN NHẬT QUỲNH nha'