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\(a+b+c+d=0\Rightarrow a+b=-\left(c+d\right)\)
\(\Rightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Rightarrow\left(a+b\right)^3+\left(c+d\right)^3=0\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)+c^3+d^3+3cd\left(c+d\right)=0\)
\(\Rightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
\(\Rightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\) (do \(a+b=-\left(c+d\right)\)
\(\Rightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu hỏi của ✰✰ βєsէ ℱƐƝƝIƘ ✰✰ - Toán lớp 8 - Học toán với OnlineMath
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Tìm trước khi hỏi : Câu hỏi của Phan Thị Hồng Nhung - Toán lớp 9 - Học toán với OnlineMath
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ \(a+b+c+d=0\) \(\Rightarrow\) \(a+b=-\left(c+d\right)\) \(\Rightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3-d^3-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(ab-cd\right)\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a+b+c+d=0
=>c+d=-a-b
\(a^3+b^3+c^3+d^3\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+\left(c+d\right)^3-3cd\left(c+d\right)\)
\(=\left(a+b\right)^3+\left(-a-b\right)^3+3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(=3ab\left(c+d\right)-3cd\left(c+d\right)\)
=3(c+d)(ab-cd)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo đề, a+b+c+d=0
\(\Rightarrow a+b=-\left(c+d\right)\)
Ta có: \(VT=\left(a+b\right)\left(a^2-ab+b^2\right)+\left(c+d\right)\left(c^2-cd+d^2\right)\)
\(\Leftrightarrow VT=\left(c+d)\left(c^2-cd+d^2-a^2+ab-b^2\right)\right)\)
Để có ĐPCM ta xét hiệu: \(c^2-cd+d^2-a^2+ab-b^2-3\left(ab+cd\right)=c^2-4cd+d^2-a^2-2ab-b^2=c^2-4cd+d^2-\left(a+b\right)^2=c^2-4cd+d^2-\left(c+d\right)^2=-6cd\)
S nó ko = 0 ta:::xem lại đề..Hay mk lm sai j đó
![](https://rs.olm.vn/images/avt/0.png?1311)
cau 1 ne:
a^2 + b^2 + c^2 + 3
theo bat dang thuc cosi ban se co
a^2 + a + 1 >= 3a
b^2 + b + 1 >= 3b
c^2 + c + 1 >= 3c
cong 3 ve bat dang thuc lai voi nhau ban se co
a^2 + b^2 + c^2 + (a + b + c) + 3>= 3(a + b + c)
=> a^2 + b^2 + c^2 + 3 >= 2(a + b + c)
dau = xay ra <=> a= b= c = 1
ma theo de bai ta lai co a^2 + b^2 + c^2 + 3 = 2(a + b + c)
=> a = b = c = 1 (dpcm)
b) (a - b)^2 + (b-c)^2 + (c - a)^2 = (a + b - 2c)^2 + (b + c - 2a)^2 + (c + a - 2b)^2
hay (a + b - 2b)^2 + (b + c - 2c)^2 + (c + a - 2a)^2 = (a + b - 2c)^2 + (b + c - 2a)^2 + (c + a - 2b)^2
dat. a + b = A
b + c = B
c + a = C
=> ban se co:
(A - 2b)^2 + (B - 2c)^2 + (C - 2a)^2 = (A - 2c)^2 + (B - 2a)^2 + (C - 2b)^2
tu day ban nhan pha ra roi rut gon 2 ve cho nhau ban se co
Ab + Bc + Ca = Ac + Ba + Cb
hay (a + b)b + (b + c)c + (c + a)a = (a + b)c + (b + c)a + (c + a)b
hay ab + b^2 + bc + c^2 + ac + a^2 = 2ab + 2bc + 2ac
hay a^2 + b^2 + c^2 - ab - bc - ac = 0
hay 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac = 0
hay (a-b)^2 + (b-c)^2 +(c - a)^2 = 0
dau = xay ra <=> a = b = c (dpcm)
c) a^3 + b^3 + c^3 + d^3 = (a + b)(a^2 -ab +b^2) + (c+d)(c^2 - cd + d^2) (**)
ban nhan thay a + b + c + d = 0
=> a + b = - c - d
thay vao pt (**) ban se co
-(c + d)(a^2 - ab + b^2) + (c + d)(c^2 - cd + d^2)
(c + d)(c^2 - cd + d^2 -a^2 + ab - b^2)
hay (c + d)(ab - cd + (c^2 + d^2 - a^2 - b^2)) (***)
ban co a + b = - c - d
hay (a + b)^2 = (c + d)^2
hay a^2 + b^2 + 2ab = c^2 + d^2 + 2cd
hay c^2 + d^2 - a^2 - b^2 = 2ab - 2cd
thay vao pt (***) ban se co
(c + d)(ab - cd + 2ab - 2cd)
hay (c +d)(3ab - 3cd) = 3(c+d)(ab - cd) (dpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(a+b+c+d=0\)
\(\Rightarrow c=-\left(a+b\right)\)
Do đó :
\(\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)\)
\(=-c^3-d^3-3cd\left(c+d\right)\)
\(\Rightarrow a^3+b^3+c^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
Vì \(a+b=-\left(c+d\right)\)
\(\Rightarrow3ab\left(c+d\right)-3cd\left(c+d\right)=3\left(c+d\right)\left(ab-cd\right)\)
a+b+c+d=0
=>a+b=-(c+d)
=> (a+b)^3=-(c+d)^3
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d)
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d)
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d))
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd) (đpcm)
hey you, còn câu b,c?