A=\(a^2\left(a+b\right)-b\left(a^2-b^2\right)+2013\)
vs a=1 b= -1
bạn nào làm đúng mik tich cho
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27 tháng 10 2018
\(a+b+c=9\)
\(\Leftrightarrow\)\(\left(a+b+c\right)^2=81\)
\(\Leftrightarrow\)\(a^2+b^2+c^2+2ab+2bc+2ca=81\)
\(\Leftrightarrow\)\(2\left(ab+bc+ca\right)=54\)
\(\Leftrightarrow\)\(ab+bc+ca=27\)
\(\Rightarrow\)\(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow\)\(2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)
\(\Rightarrow\)\(B=\left(a-4\right)^{2018}+\left(b-4\right)^{2019}+\left(c-4\right)^{2020}=4^{2018}-4^{2019}+4^{2020}\)
\(\Rightarrow\)\(B=13.4^{2018}\)
Vậy \(B=13.4^{2018}\)
Chúc bạn học tốt ~
27 tháng 10 2018
Phùng Minh Quân : sửa dòng thứ 4 từ dưới lên
Mà \(a+b+c=9\)
\(\Rightarrow a=b=c=3\)
\(B=\left(a-4\right)^{2018}+\left(b-4\right)^{2019}+\left(c-4\right)^{2020}\)
\(B=\left(3-4\right)^{2018}+\left(3-4\right)^{2019}+\left(3-4\right)^{2020}\)
\(B=\left(-1\right)^{2018}+\left(-1\right)^{2019}+\left(-1\right)^{2020}\)
\(B=1-1+1\)
\(B=1\)
5 tháng 2 2017
Ta có:
a2(b - c) + b2(c - a) + c2(a - b)
= (a - b)(c - a)(c - b)
Ta lại có:
a4(b2 - c2) + b4(c2 - a2) + c4(a2 - b2)
= (a - b)(c - a)(c - b)(a +b)(b + c)(c + a)
Từ đây ta có phân số ban đầu sẽ bằng
\(\frac{\left(a-b\right)\left(c-a\right)\left(c-b\right)}{\left(a-b\right)\left(c-a\right)\left(c-b\right)\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{1}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
BH
1 tháng 6 2018
a) a2(a-b)-b2(a-c)-c2(b-a)
=a2(a-b)-b2(a-c)+c2(a-b)
=(a-b)(a2-c2)-b2(a-c)
=(a-b)(a-c)(a+c)-b2(a-c)
=(a-c)[(a-b)(a+c)-b2]
b)a(b-c)3+b(c-a)3+c(a-b)3
=a(b-c)3-b[(a-b)+(b-c)]+c(a-b)3
=a(b-c)3-b[(a-b)3+3(a-b)2(b-c)+3(a-b)(b-c)2+(b-c)3]+c(a-b)3
=a(b-c)3-b(a-b)3+3b(a-b)2(b-c)+3b(a-b)(b-c)2+b(b-c)3+c(a-b)3
=(b-c)3(a-b)-(a-b)3(b-c)-3b(a-b)(b-c)(a-b+b-c)
=(b-c)3(a-b)-(a-b)3(b-c)-3b(a-b)(b-c)(a-c)
=(a-b)(b-c)[(b-c)2-(a-b)2-3b(a-c)]
=(a-b)(b-c)[(b-c-a+b)(b-c+a-b)-3b(a-c)]
=(a-b)(b-c)[(2b-a-c)(a-c)-3b(a-c)]
=(a-b)(b-c)(a-c)(2b-a-c-3b)
=-(a-b)(b-c)(a-c)(a+b+c)
=(a-b)(b-c)(c-a)(a+b+c)
c)abc-(ab+ac+bc)+(a+b+c)-1
=abc-ab-ac-bc+a+b+c-1
=abc-bc-ab+b-ac+c+a-1
=bc(a-1)-b(a-1)-c(a-1)+a-1
=(a-1)(bc-b-c+1)
=(a-1)[b(c-1)-(c-1)]
=(a-1)(c-1)(b-1)
=(a-1)(b-1)(c-1)
BA
17 tháng 11 2019
MTC: \(abc\left(a-b\right)\left(b-c\right)\left(a-c\right)\)nên
\(A=\frac{bc\left(b-c\right)\left(a-2\right)\left(a-1014\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{ac\left(a-c\right)\left(b-2\right)\left(b-1004\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)\left(c-2\right)\left(c-1004\right)}{abc\left(a-c\right)\left(a-b\right)\left(b-c\right)}\)
\(=\frac{2008b^2c+2008a^2c+2008a^2b-2008bc^2-2008a^2c-2008ab^2}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008\left[\left(c^2a-c^2b\right)+\left(a^2b-a^2c\right)+\left(b^2a-b^2c\right)\right]}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008}{abc}\) ( với \(abc\ne0\))
Y
20 tháng 5 2019
gt \(\Rightarrow\left\{{}\begin{matrix}b\left(a^2+2ac+c^2\right)+ac\left(a+c\right)+b^2\left(a+c\right)=0\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a+c\right)\left[b\left(a+c\right)+ac+b^2\right]=0\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\\a^{2013}+b^{2013}+c^{2013}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a+b=0\Rightarrow a^{2013}+b^{2013}=0\\b+c=0\Rightarrow b^{2013}+c^{2013}=0\\a+c=0\Rightarrow a^{2013}+c^{2013}=0\end{matrix}\right.\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)
\(\Rightarrow Q=1\)
17 tháng 4 2019
1a)
Đặt \(a^2+a+1=t\Rightarrow a^2+a+2=t+1\)
\(\Rightarrow A=t\left(t+1\right)-12=t^2+t-12=t^2-3t+4t-12=\left(t-3\right)\left(t+4\right)\)
\(=\left(a^2+a-2\right)\left(a^2+a+5\right)\)
Mà \(a>1\Rightarrow\hept{\begin{cases}a^2+a-2>0\\a^2+a+5>0\end{cases}}\forall a>1\)
Vậy A là hợp số
17 tháng 4 2019
1b)
Ta có :
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{1006}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{1006}+1\right)+1=....=\left(2^{1006}-1\right)\left(2^{1006}+1\right)+1\)
\(=2^{2012}-1+1=2^{2012}\)
\(A=a^2\left(a+b\right)-b\left(a-b\right)\left(a+b\right)+2013\)
\(A=\left[a^2-b\left(a-b\right)\right]\left(a+b\right)+2013\)
Thay a=1,b= -1 vào ta có:
\(A=\left[1^2+1\left(1+1\right)\right]\left(1-1\right)+2013\)
\(A=0+2013\)
\(A=2013\)
\(a=1;b=-1\)=> \(a+b=0\)
\(A=a^2\left(a+b\right)-b\left(a^2-b^2\right)+2013\)
\(=a^2\left(a+b\right)-b\left(a-b\right)\left(a+b\right)+2013\)
\(=2013\)