Chứng ming hằng đẳng thức
(a+b+c)³=a³+b³+c³+3(a+b)*(b+c)*(c+a)
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NT
1
25 tháng 7 2017
Ta có \(VT=\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3=\left(a+b\right)^3+3\left(a+b\right)^2.c+3\left(a+b\right)c^2+c^3\)
\(=a^3+3a^2b+3ab^2+b^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left[\left(a+b\right)c+c^2+ab\right]\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left[a\left(b+c\right)\right]+c\left(b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)
Vậy \(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
6 tháng 7 2015
(a+b+c)^3=((a+b)+c)^3=(a+b)^3+c^3+3(a+b)c(a+b+c)
=a^3+b^3+3ab(a+b)+c^3+3(a+b)c(a+b+c)
=a^3+b^3+c^3+3(a+b)(ab+c(a+b+c))
=a^3+b^3+c^3+3(a+b)(ab+ac+bc+c^2)
=a^3+b^3+c^3+3(a+b)(a+c)(b+c)
25 tháng 6 2019
(a+b+c)3=[(a+b)+c]3=(a+b)3+c3+3(a+b)c(a+b+c)
=a3+b3+3ab(a+b)+c3+3(a+b)c(a+b+c)
=a3+b3+c3+3(a+b)[ab+c(a+b+c)]
=a3+b3+c3+3(a+b)(ab+ac+bc+c2)
==a3+b3+c3+3(a+b)[(ab+ac)+(bc+c2)]
=a3+b3+c3+3(a+b)(a+c)(b+c)
T
25 tháng 6 2019
#)Giải :
\(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+ca+c^2\right)\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right)c\left(a+b+c\right)\)
\(=\left(a+b^3\right)+c^3+3\left(a+b\right)c\left(a+b+c\right)\)
\(=\left(a+b+c\right)^3\)
\(\Rightarrowđpcm\)
PT
1
CM
5 tháng 9 2019
Biến đổi vế trái:
a + b + c 3 = a + + c 3 = a + b 3 +3 a + b 2 c+3(a+b) c 2 + c 3
= a 3 + 3 a 2 b + 3a b 2 + b 3 + 3( a 2 + 2ab + b 2 )c + 3a c 2 + 3b c 2 + c 3
= a 3 + 3 a 2 b + 3a b 2 + b 3 + 3 a 2 c + 6abc + 3 b 2 c + 3a c 2 + 3b c 2 + c3
= a 3 + b 3 + c 3 + 3 a 2 b + 3a b 2 + 3 a 2 c + 6abc + 3 b 2 c + 3a c 2 + 3b c 2
= a 3 + b 3 + c 3 + (3 a 2 b + 3a b 2 ) +( 3 a 2 c + 3abc)+ (3abc + 3 b 2 c)+(3a c 2 + 3b c 2 )
= a 3 + b 3 + c 3 + 3ab(a + b) + 3ac(a + b) + 3bc(a + b) + 3 c 2 (a + b)
= a 3 + b 3 + c 3 + 3(a + b)(ab + ac + bc + c 2 )
= a 3 + b 3 + c 3 + 3(a + b)[a(b + c) + c(b + c)]
= a 3 + b 3 + c 3 + 3(a + b)(b + c)(a + c) (đpcm)
TN
13 tháng 6 2016
(a+b+c)3=((a+b)+c)3=(a+b)3+c3+3(a+b)c(a+b+c)
=a3+b3+3ab(a+b)+c3+3(a+b)c(a+b+c)
=a3+b3+c3+3(a+b)(ab+c(a+b+c))
=a3+b3+c3+3(a+b)(ab+ac+bc+c2)
=a3+b3+c3+3(a+b)(a+c)(b+c)
14 tháng 6 2016
(a+b+c)3=((a+b)+c)3=(a+b)3+c3+3(a+b)c(a+b+c)
=a3+b3+3ab(a+b)+c3+3(a+b)c(a+b+c)
=a3+b3+c3+3(a+b)(ab+c(a+b+c))
=a3+b3+c3+3(a+b)(ab+ac+bc+c2)
=a3+b3+c3+3(a+b)(a+c)(b+c)
22 tháng 9 2018
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
=> ĐPCM
b) \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)^2+c^3-3abc\)
\(=\left[\left(a+b\right)^3+c^3\right]-\left(3a^2b+3abc+3ab^2\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2-3ab\right]\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
=> ĐPCM
P/s: Có sao sót xin bỏ qua
22 tháng 9 2018
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2\cdot c+3\left(a+b\right)c^2+c^3\)\(-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3a^2b+3ab^2+3\left(a^2+2ab+b^2\right)c\)\(+3ac^2+3bc^2-a^3-b^3-c^3\)
\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=\left(3abc+3a^2c+3b^2c+3bc^2\right)\)\(+\left(3a^2b+3a^2c+3ab^2+3abc\right)\)
\(=c\left(3ab+3ac+3b^2+3bc\right)\)\(+a\left(3ab+3ac+3b^2+3bc\right)\)
\(=\left(a+c\right)\left[\left(3ab+3b^2\right)+\left(3ac+3bc\right)\right]\)
\(=\left(a+c\right)\left[3b\left(a+b\right)+3c\left(a+b\right)\right]\)
\(=3\left(a+c\right)\left(a+b\right)\left(b+c\right)\)
b) \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)( do \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\))
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]\)\(-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ab-ac\right)\)\(-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Biến đổi vế trái ta có:
\(\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3=\left(a+b\right)^3+3c\left(a+b\right)\left(a+b+c\right)+c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+3c\left(a+b\right)\left(a+b+c\right)+c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)
=>đpcm