cho a+b =1.tính M =a^3+b^3+3ab*(a^2+b^2)+6a^2b^2(a+b) 17/08/2021 Bởi Caroline cho a+b =1.tính M =a^3+b^3+3ab*(a^2+b^2)+6a^2b^2(a+b)
Ta có: M = a³ + b³ + 3ab(a² + b²) + 6a²b²(a + b) = (a+b)(a² – ab + b²) + 3ab[(a+b)² – 2ab] + 6a²b²(a +b ) = (a+b) [(a +b)² – 3ab] + 3ab[(a+b)² – 2ab] + 6a²b²(a +b ) _______thay a + b = 1 __________________: M = 1.(1 – 3ab) + 3ab(1 – 2ab) + 6a²b² M = 1 – 3ab + 3ab – 6a²b² + 6a² b² = 1 Bình luận
M= a³+b³+3ab.(a^2+b^2)+6a²b²(a+b) = (a+b)³ – 3ab( a+b) + 3ab[(a+b)² – 2ab] + 6a²b²(a+b) =(a+b) [(a +b)² – 3ab] + 3ab[(a+b)² – 2ab] + 6a²b²(a +b ) Thay a + b = 1 vào M, ta được: M = 1.(1 – 3ab) + 3ab(1 – 2ab) + 6a²b² = 1 – 3ab + 3ab – 6a²b² + 6a²b² = 1 Bình luận
Ta có:
M = a³ + b³ + 3ab(a² + b²) + 6a²b²(a + b)
= (a+b)(a² – ab + b²) + 3ab[(a+b)² – 2ab] + 6a²b²(a +b )
= (a+b) [(a +b)² – 3ab] + 3ab[(a+b)² – 2ab] + 6a²b²(a +b )
_______thay a + b = 1 __________________:
M = 1.(1 – 3ab) + 3ab(1 – 2ab) + 6a²b²
M = 1 – 3ab + 3ab – 6a²b² + 6a² b² = 1
M= a³+b³+3ab.(a^2+b^2)+6a²b²(a+b)
= (a+b)³ – 3ab( a+b) + 3ab[(a+b)² – 2ab] + 6a²b²(a+b)
=(a+b) [(a +b)² – 3ab] + 3ab[(a+b)² – 2ab] + 6a²b²(a +b )
Thay a + b = 1 vào M, ta được:
M = 1.(1 – 3ab) + 3ab(1 – 2ab) + 6a²b²
= 1 – 3ab + 3ab – 6a²b² + 6a²b²
= 1