Cho 1/a+1/b+1/c=1/ a+b+c. Tính GTBT: A= (a+b+c)^3. (1/a+1/b+1/c)^3.
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)
\(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2+c^2-\left(a+b\right)c\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[a^2+b^2+2ab+c^2-ac-bc-3ab\right]=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)=0.2\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
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 ...
![](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)
Chỗ a+b+c=a*b*c* đó là sao bạn? Nếu như đó là a+b+c=abc thì mình giải theo cách này.
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
=>\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{bc}\right)=4-2.\frac{a+b+c}{abc}\)= 2 (vì a+b+c=abc)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](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)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a+b-2017c}{c}=\frac{b+c-2017a}{a}=\frac{c+a-2017b}{b}\)
\(=\frac{a+b-2017c+b+c-2017a+c+a-2017b}{a+b+c}=\frac{-2015\left(a+b+c\right)}{a+b+c}=-2015\)
Do đó :
\(\frac{a+b-2017c}{c}=-2015\)\(\Leftrightarrow\)\(a+b=2c\) \(\left(1\right)\)
\(\frac{b+c-2017a}{a}=-2015\)\(\Leftrightarrow\)\(b+c=2a\) \(\left(2\right)\)
\(\frac{c+a-2017b}{b}=-2015\)\(\Leftrightarrow\)\(c+a=2b\) \(\left(3\right)\)
Thay (1), (2) và (3) vào \(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}\) ta được :
\(B=\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=\frac{8abc}{abc}=8\)
Vậy \(B=8\)
Chúc bạn học tốt ~
A= (a+b+c)3.(\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) )3= (a+b+c)3.(\(\frac{1}{a+b+c}\))3=1
Lạ nhể. Lúc nãy lại kn nghĩ ra .