Tìm a,b,c \(\in\) Z để a3 +b3 + c3 =2019
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NA
1
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 8 2021
Đặt \(P=\dfrac{a^3}{a^2+b^2+ab}+\dfrac{b^3}{b^2+c^2+bc}+\dfrac{c^3}{c^2+a^2+ca}\)
Ta có: \(\dfrac{a^3}{a^2+b^2+ab}=a-\dfrac{ab\left(a+b\right)}{a^2+b^2+ab}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^3b^3}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)
Tương tự: \(\dfrac{b^3}{b^2+c^2+bc}\ge\dfrac{2b-c}{3}\) ; \(\dfrac{c^3}{c^2+a^2+ca}\ge\dfrac{2c-a}{3}\)
Cộng vế:
\(P\ge\dfrac{a+b+c}{3}=673\)
Dấu "=" xảy ra khi \(a=b=c=673\)
23 tháng 8 2021
\(A=x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)=x\left(y^2-z^2\right)+y\left(-y^2+z^2-x^2+y^2\right)+z\left(x^2-y^2\right)=\left(y^2-z^2\right)\left(x-y\right)+\left(x^2-y^2\right)\left(z-y\right)=\left(y-z\right)\left(y+z\right)\left(x-y\right)-\left(x-y\right)\left(x+y\right)\left(y-z\right)=\left(x-y\right)\left(y-z\right)\left(y+z-x-y\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
23 tháng 8 2021
\(B=a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c+abc+b^2c\right)\)
I
1
AH
Akai Haruma
Giáo viên
29 tháng 7 2021
Lời giải:
$a^3+b^3=2(c^3-8d^3)$
$a^3+b^3+c^3+d^3=c^3+d^3+2(c^3-8d^3)$
$=3c^3-15d^3=3(c^3-5d^3)\vdots 3$
Khi đó:
$(a+b+c+d)^3=(a+b)^3+(c+d)^3+3(a+b)(c+d)(a+b+c+d)$
$=a^3+b^3+c^3+d^3+3ab(a+b)+3cd(c+d)+3(a+b)(c+d)(a+b+c+d)\vdots 3$ do:
$a^3+b^3+c^3+d^3\vdots 3$
$3ab(a+b)\vdots 3$
$3cd(c+d)\vdots 3$
$3(a+b)(c+d)(a+b+c+d)\vdots 3$
Vậy:
$(a+b+c+d)^3\vdots 3$
$\Rightarrow a+b+c+d\vdots 3$
NH
1
5 tháng 10 2021
a(b3 - c3) + b(c3 - a3) + c(a3 - b3)
= a(b3 - c3 ) + b( c3 - b3 + b3 - a3) + c(a3 - b3)
= a(b3 - c3) + b(c3 - b3) + b(b3 - a3) + c(a3 - b3)
= a(b3 - c3) - b(b3 - c3) - [b(a3 - b3) - c(a3- b3)]
= (b3 - c3)(a - b) - (a3- b3)(b - c)
= (b - c)(b2 + bc + c2)(a - b) - (a - b)(a2 + ab + b2)(b - c)
= (b - c)(a - b)(b2 + bc + c2 - a2 + ab - b2)
= (b - c)(a - b) [ (c2 - a2) + (bc - ab) ]
= (b - c)(a - b) [ (c - a)(c + a) + b(c - a) ]
= (b - c)(a -b) [ (c - a)(c + a + b) ]
= (a- b)(b - c)(c - a)(a + b + c)
21 tháng 8 2021
a: Ta có: \(a+b+c=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
Ta có: a+b+c=0
\(\Leftrightarrow\left(a+b+c\right)^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
b: Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a+b+c=0\)
21 tháng 8 2021
a) \(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=0\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)(đúng do a+b+c = 0)
21 tháng 8 2021
a: Ta có: a+b+c=0
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
Ta có: a+b+c=0
\(\Leftrightarrow\left(a+b+c\right)^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
b: Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow a+b+c=0\)
MH
26 tháng 10 2021
a(b3 - c3) + b(c3 - a3) + c(a3 - b3)
= a(b3 - c3 ) + b( c3 - b3 + b3 - a3) + c(a3 - b3)
= a(b3 - c3) + b(c3 - b3) + b(b3 - a3) + c(a3 - b3)
\(=\left[a\left(b^3-c^3\right)-b\left(b^3-c^3\right)\right]-\left[b\left(a^3-b^3\right)-c\left(a^3-b^3\right)\right]\)
= (b3 - c3)(a - b) - (a3- b3)(b - c)
= (b - c)(b2 + bc + c2)(a - b) - (a - b)(a2 + ab + b2)(b - c)
= (b - c)(a - b)(b2 + bc + c2 - a2 + ab - b2)
= (b - c)(a - b) [ (c2 - a2) + (bc - ab) ]
= (b - c)(a - b) [ (c - a)(c + a) + b(c - a) ]
= (b - c)(a -b) [ (c - a)(c + a + b) ]
= (a- b)(b - c)(c - a)(a + b + c)
Y
22 tháng 8 2018
a) Áp dụng nhiều lần công thức \(\left(x+y\right)^3=x^3-y^3+3xy\left(x+y\right)\), ta có:
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(Đpcm\right)\)
b) Ta có:
\(a^3+b^3+c^3-3abc\)
\(=a^3+3ab\left(a+b\right)+b^2+c^3-3abc-3ab\left(a+b\right)\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)
Mình nghĩ bằng thế này mới đúng, bạn chắc ghi sai đề rồi 
22 tháng 8 2018
a) Ta có: (a + b + c)3 - a3 - b3 - c3 = [ (a + b + c)3 - a3 ] - ( b3 + c3)
= (a + b + c - a) ( a2 + b2 + c2 + 2ab + 2bc + 2ac + a2 + ab + ac + a2) - (b + c) ( b2 - bc + c3)
= (b + c) ( 3a2 + b2 + c2 + 3ab + 2bc + 3ac) - (b + c) ( b2 - bc + c3)
= ( b + c) ( 3a2 + b2 + c2 + 3ab + 2bc + 3ac - b2 + bc - c3)
= ( b + c) ( 3a2 + 3ab + 3bc + 3ac)
= 3 (b + c) [a (a + b) + c (a + b)]
= 3 (b + c) (a + b) (a + c) (đpcm)