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\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\Rightarrow\frac{1}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{-a-b}{c\left(a+b+c\right)}\Rightarrow c\left(a+b\right)\left(a+b+c\right)=ab\left(-a-b\right)\)
\(\Rightarrow\left(a+b\right)\left(ca+cb+c^2\right)+ab\left(a+b\right)=0\Rightarrow\left(a+b\right)\left(ca+cb+c^2+ab\right)=0\)
\(\Rightarrow\left(a+b\right)\left(c+a\right)\left(b+c\right)=0\)
=> Trong 3 số a,b,c có 2 số đối nhau.Giả sử a = -b thì a9 + b9 = 0.
Tương tự giả sử b = -c hay a = -c thì b99 + c99 = 0 hay c999 + a999 = 0
Vậy biểu thức cần tính bằng 0.
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a^3+b^3\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left[\left(a+b\right)^3+c^3\right]-\left[3ab\left(a+b\right)+3abc\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Nếu \(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow a=b=c\)
Khi đó \(A=2^3=8\)
Nếu \(a+b+c=0\Rightarrow a+b=-c;b+c=-a;c+a=-b\)
Thay vào ta được:
\(A=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{-abc}{abc}=-1\)
Vậy A = 8 hoặc A = -1
Ta có:
\(a^3+b^3+c^3=3abc=>a^3+b^3+c^3-3abc=0\)
\(=>\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)
\(=>\left[\left(a+b\right)^3+c^3\right]-3a^2b-3ab^2-3abc=0\)
\(=>\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)=0\)
\(=>\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(=>\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)=0\)
\(=>\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Vì a3+b3+c3=3abc và a+b+c khác 0
=>\(a^2+b^2+c^2-ab-bc-ca=0\)
\(=>2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(=>\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(=>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Tổng 3 số không âm = 0 <=> chúng đều = 0
\(< =>\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}< =>a=b=c}\)
Vậy \(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}\)
\(\)
Ta có ; \(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ac\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\frac{a+b+c}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Vì \(a+b+c\ne0\) nên ta có \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
a) Thay a = b = c vào biểu thức được : \(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
b) Thay a = b = c vào P : \(P=\frac{2}{a}.\frac{2}{b}\frac{2}{c}=\frac{8}{abc}\)
a) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\Leftrightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)
\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
- TH1: Nếu a + b + c = 0 \(\Rightarrow P=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=\frac{-\left(abc\right)}{abc}=-1\)
- TH2 : Nếu \(a+b+c\ne0\) \(\Rightarrow a=b=c\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
b) Đề bài sai ^^
Ta có \(\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\) \(\frac{a+b}{ab}=\frac{1}{a+b+c}-\frac{1}{c}=\frac{c-\left(a+b+c\right)}{c\left(a+b+c\right)}=\frac{-a-b}{c\left(a+b+c\right)}\)
\(\Leftrightarrow\) \(c\left(a+b\right)\left(a+b+c\right)+ab\left(a+b\right)=0\)
\(\Leftrightarrow\) \(\left(a+b\right)\left(ca+cb+c^2+ab\right)=0\)
\(\Leftrightarrow\) \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\) \(\hept{\begin{cases}a+b=0\\b+c=0\\c+a\end{cases}}=0\Leftrightarrow\hept{\begin{cases}a=-b\\b=-c\\c=-a\end{cases}\Leftrightarrow\hept{\begin{cases}a^3=-b^3\\b^3=-c^3\\c^3=-a^3\end{cases}}\Leftrightarrow\hept{\begin{cases}a^3+b^3=0\\b^3+c^3=0\\c^3+a^3=0\end{cases}}}\)
(ko có kí hiệu hoặc cho 3 cái nên mk dùng kí hiệu và nhé)
Do đó \(A=\left(a^3+b^3\right)\left(b^3+c^3\right)\left(c^3+a^3\right)=0\)
em mới học lớp 5 nên ko giúp đc gì, mong chị tha lỗi. chúc chị học giỏi nha
Ta có: \(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Vì \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Mà \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0}\)
\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow a=b=c}\)
\(\Rightarrow A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{2a.2a.2a}{a.a.a}=\frac{8a^3}{a^3}=8\)
xét a + b + c = 0 khi đó a + b = -c ; b + c = -a ; a + c = -b
Ta có : \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{\left(-a\right)\left(-b\right)\left(-c\right)}{abc}=-1\)
xét a + b + c \(\ne\)0 . thì \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow a+b=2c;b+c=2a\)\(\Rightarrow a-c=2\left(c-a\right)\)\(\Rightarrow a=c\)( loại vì a khác c )
Vậy A = -1
Sửa đề: tính P=(1+a/b)(1+b/c)(1+c/a)
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\left(1\right)\\a^2+b^2+c^2-ab-bc-ca=0\left(2\right)\end{cases}}\)
- Xét (1) ta có: \(a+b+c=0\Leftrightarrow\hept{\begin{cases}-a=b+c\\-b=c+a\\-c=a+b\end{cases}}\)
=> \(P=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{\left(-c\right).\left(-a\right).\left(-b\right)}{bca}=-\frac{abc}{abc}=-1\)
- Xét (2) ta có: \(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Mà \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0}\)
\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow a=b=c}\)
=>\(P=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{2a}{a}\cdot\frac{2a}{a}\cdot\frac{2a}{a}=2.2.2=8\)
Vậy P=-1 hoặc P=8
Ta có; \(a^3+b^3+c^3=3abc\) hay \(a^3+b^3+c^3-3abc=0\)
Suy ra \(a+b+c=0\) hoặc a = b = c. (bạn tự chứng minh)
* Nếu a + b + c = 0 thì:
\(P=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=-1\)
*Nếu a = b = c thì \(P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Thay a3+b3=(a+b)3-3ab(a+b) vào giả thiết ta có:
(a+b)3-3ab(a+b)+c3-3abc=0
<=> [(a+b)+c].\(\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]\)-3ab(a+b+c)=0
<=> (a+b+c) (a2+b2+c2-ab-bc+c2-3ab)=0
<=> (a+b+c)(a2+b2+c2-ab-bc-ca)=0
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)
\(\Rightarrow A=\frac{b+a}{b}\cdot\frac{c+b}{c}\cdot\frac{a+c}{a}=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}\Rightarrow A=-1\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
<=> a=b=c
Khi đó \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)