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\(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)
Lần lượt thay vào M, N, P ta có :
\(\Rightarrow\hept{\begin{cases}M=a\cdot\left(-c\right)\cdot\left(-b\right)=a\cdot b\cdot c\\N=b\cdot\left(-a\right)\cdot\left(-c\right)=a\cdot b\cdot c\\P=c\cdot\left(-b\right)\cdot\left(-a\right)=a\cdot b\cdot c\end{cases}}\)
\(\Rightarrow M=N=P\left(đpcm\right)\)
\(a+b+c=0\)
\(\Rightarrow a+b=-c;a+c=-b;b+c=-a\)
THAY \(a+b=-c;a+c=-b;b+c=-a\)VÀO M;N;P TA CÓ:
\(M=a.\left(-c\right).\left(-b\right)=a.b.c\)(1)
\(N=b.\left(-a\right).\left(-c\right)=a.b.c\)(2)
\(P=c.\left(-b\right).\left(-a\right)=a.b.c\)(3)
Từ (1) ; (2) ; (3) Ta có
\(M=N=P\left(=a.b.c\right)\)(đpcm)
Ta có :
a. ( a+b+c) = - 12 (1)
b.(a+b+c) = 18 (2)
c.(a+b+c) =30 (3)
=> (1) + (2) + (3) = a.(a+b+c) + b(a+b+c) + c(a+b+c)= -12+18+30
\(\Rightarrow\left(a+b+c\right)\left(a+b+c\right)=36\)
\(\Rightarrow\left(a+b+c\right)^2=6^2=\left(-6\right)^2\)
\(\Rightarrow a+b+c\in\left\{6;-6\right\}\)
Với \(a+b+c=6\)
Từ (1) => a = -12 : 6 = - 2
Từ (2) => b = 18 : 6 = 3
Từ (3) => c = 30 : 6 = 5
Với a + b + c = -6
Từ (1) => a = -12 : ( -6 ) = 2
Từ (2) => b = 18 : (-6) = -3
Từ (3) => c = 30: ( -6) = -5
4) Ta có : A=(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)
=> (a+d)2 - (b+c)2= (a-d)2 - (c-b)2
=> a2+ d2+ 2ad - b2- c2- 2bc=a2 + d2 - 2ad - c2-b2+2bc
Rút gọn ta được: 4ad = 4bc => ad = bc =>\(\dfrac{a}{c}=\dfrac{b}{d}\)
1) a2+b2+c2+3=2(a+b+c) =>(a-1)2+(b-1)2+(c-1)2=0
=> a-1=b-1=c-1=0 => a=b=c=1 =>đpcm
a) \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2b-a^2c+c^2a-c^2b+b^2\left(c-a\right)\)
\(=\left(a^2b-c^2b\right)-\left(a^2c-c^2a\right)-b^2\left(a-c\right)\)
\(=b\left(a^2-c^2\right)-ac\left(a-c\right)-b^2\left(a-c\right)\)
\(=b\left(a-c\right)\left(a+c\right)-ac\left(a-c\right)-b^2\left(a-c\right)\)
\(=\left(a-c\right)\left[b\left(a+c\right)-ac-b^2\right]\)
\(=\left(a-c\right)\left(ab+bc-ac-b^2\right)\)
\(=\left(a-c\right)\left[\left(ab-b^2\right)+\left(bc-ac\right)\right]\)
\(=\left(a-c\right)\left[b\left(a-b\right)+c\left(b-a\right)\right]\)
\(=\left(a-c\right)\left[b\left(a-b\right)-c\left(a-b\right)\right]\)
\(=\left(a-c\right)\left(a-b\right)\left(b-c\right)\)
b) \(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(=a^3b-a^3c+c^3a-c^3b+b^3\left(c-a\right)\)
\(=\left(a^3b-c^3b\right)-\left(a^3c-c^3a\right)-b^3\left(a-c\right)\)
\(=b\left(a^3-c^3\right)-ac\left(a^2-c^2\right)-b^3\left(a-c\right)\)
\(=b\left(a-c\right)\left(a^2+ac+c^2\right)-ac\left(a-c\right)\left(a+c\right)-b^3\left(a-c\right)\)
\(=\left(a-c\right)\left[b\left(a^2+ac+c^2\right)-ac\left(a+c\right)-b^3\right]\)
\(=\left(a-c\right)\left(ba^2+abc+bc^2-a^2c-ac^2-b^3\right)\)
\(=\left(a-c\right)\left[\left(ba^2-a^2c\right)+\left(abc-ac^2\right)+\left(bc^2-b^3\right)\right]\)
\(=\left(a-c\right)\left[a^2\left(b-c\right)+ac\left(b-c\right)+b\left(c^2-b^2\right)\right]\)
\(=\left(a-c\right)\left[a^2\left(b-c\right)+ac\left(b-c\right)-b\left(b^2-c^2\right)\right]\)
\(=\left(a-c\right)\left[a^2\left(b-c\right)+ac\left(b-c\right)-b\left(b-c\right)\left(b+c\right)\right]\)
\(=\left(a-c\right)\left(b-c\right)\left[a^2+ac-b\left(b+c\right)\right]\)
\(=\left(a-c\right)\left(b-c\right)\left(a^2+ac-b^2-bc\right)\)
\(=\left(a-c\right)\left(b-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]\)
\(=\left(a-c\right)\left(b-c\right)\left(a-b\right)\left(a+b+c\right)\)
a) Ta có:
\(\left(a+b\right)\left(a+c\right)+\left(c+a\right)\left(c+b\right)\)
\(=a^2+ac+ab+bc+c^2+bc+ac+ab\)
\(=a^2+c^2+2ac+2bc+2ab\)
Thay \(a^2+c^2=2b^2\) vào biểu thức ta được:
\(=2b^2+2ac+2bc+2ab\)
\(=2\left(b^2+ac+bc+ab\right)\)
\(=2\left[\left(b^2+bc\right)+\left(ac+ab\right)\right]\)
\(=2\left[b\left(b+c\right)+a\left(c+b\right)\right]\)
\(=2\left(b+a\right)\left(b+c\right)\)
\(\RightarrowĐpcm\)
Câu 1:
a) a(a+2b)3 - b(2a+b)3 = a( a3 + 6a2b + 12ab2 + 8b2) - b
= a( a3 + 6a2b + 12ab2 + 8b3) - b( 8a3 + 12a2b + 6ab2 + b3)
= a4 + 6a3b + 12a2b2 + 8ab3 - 8a3b -12a2b2 - 6ab3 - b4
= a4 - 2a3b + 2ab3 - b4
= (a - b )(a + b)(a2 +b2) - 2ab(a - b)(a + b)
= (a - b )(a + b)(a2 +b2 -2ab)
= (a - b )3(a + b)
M=a(a+b)(a+c)
N=b(b+c)(b+a)
P=c(c+a)(c+b)
Có a+b+c=0
=> a+b=-c
b+c=-a
a+c=-b
=> M=a(-c)(-b)
=abc
N=b(-a)(-c)
=bac
P=c(-b)(-a)
=cba
=> M=N=P(đpcm)