![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
bạn khai thác gt ta đc : (b+c)(a+b)(a+c)=0
b=-c
a=-b
a=-1
M=(a^3+b^3)(b^7+c^7)(a^2011+|c^2011)
vì
ta có 3 trường hợp
b=-c nên (b^7+c^7=0)
a=-b nên (a^3+b^3)=0
a=-1nên (a^2011+b^2011)=0
M=0
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(A=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(a^2+b^2\right)+6a^2b^2\)
\(=1^3-3ab+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\)
\(=1-3ab+3ab\left(1-2ab\right)+6a^2b^2=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)=\)\(0\)
\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
đến đây bạn tự thay vào tính P nhé P được \(2\) giá trị là \(-1\)hoặc\(8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
** \(a^3+b^3 +c^3 -3abc \)
\(=(a+b)^3+c^3 - 3ab(a+b) - 3abc \)
\(=(a+b+c)[(a+b)^2 - c(a+b)+ c^2] - 3ab(a+b+c) \)
\(=(a+b+c)(a^2 + 2ab+b^2-ca-bc+c^2) - 3ab(a+b+c) \)
\(=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \)
\(=a^2+b^2+c^2-ab-bc-ca\)
** \((a-b)^2 + (b-c)^2+(c- a)^2\)
\(=a^2+b^2+b^2+c^2+c^2+a^2 - 2(ab+bc+ca)\)
\(=2(a^2+b^2+c^2-ab-bc-ca)\)
\(\Rightarrow A=\dfrac{a^2+b^2+c^2-ab-bc-ca}{2\left(a^2+b^2+c^2-ab-bc-ca\right)}=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}=3\) (abc=1) (tự c/m)
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\)
=>ab+bc=-ca => (ab+bc)3=-c3a3
=>a3b3+b3c3+3a2b2.bc+3ab.b2c2=-c3a3
=>a3b3+b3c3+3ab2c(ab+bc)=-c3a3
=>a3b3+b3c3+3ab2c.(-ca)=-c3a3
=>a3b3+b3c3-3a2b2c2=-c3a3
=>a3b3+b3c3+c3a3=3a2b2c2 = 3 (do abc=1)
Vậy F=3.3=9
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu hỏi của Rarah Venislan - Toán lớp 8 - Học toán với OnlineMath
Câu hỏi của Nguyễn Lê Nhật Linh - Toán lớp 9 - Học toán với OnlineMath
\(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2+c^2-\left(a+b\right)c\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[a^2+b^2+2ab+c^2-ac-bc-3ab\right]=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)=0.2\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
TH1 : \(a+b+c=0\)
\(\Rightarrow A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
\(=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}\)
\(=\frac{\left(-c\right)}{b}.\frac{\left(-a\right)}{c}.\frac{\left(-b\right)}{a}=-1\)
TH2 : \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Mà \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\)
\(\Rightarrow a-b=b-c=c-a=0\)
\(\Rightarrow a=b=c\)
\(\Rightarrow A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Vậy ...