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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(a+\frac{1}{a}=b+\frac{1}{b}=c+\frac{1}{c}\)
=> \(\hept{\begin{cases}a+\frac{1}{a}=b+\frac{1}{b}\\b+\frac{1}{b}=c+\frac{1}{c}\end{cases}}\Rightarrow\hept{\begin{cases}a-b=\frac{1}{b}-\frac{1}{a}\\b-c=\frac{1}{c}-\frac{1}{b}\end{cases}}\Rightarrow\hept{\begin{cases}a-b=\frac{a-b}{ab}\\b-c=\frac{b-c}{bc}\end{cases}}\)
=> \(\hept{\begin{cases}\frac{a-b}{ab}-\left(a-b\right)=0\\\frac{b-c}{bc}-\left(b-c\right)=0\end{cases}\Rightarrow\hept{\begin{cases}\left(a-b\right)\left(\frac{1}{ab}-1\right)=0\\\left(b-c\right)\left(\frac{1}{bc}-1\right)=0\end{cases}}}\Rightarrow\hept{\begin{cases}\frac{1}{ab}-1=0\\\frac{1}{bc}-1=0\end{cases}}\)(Vì a ;b;c đôi một khác nhau)
=> \(\frac{1}{ab}=\frac{1}{bc}=1\Rightarrow ab=bc=1\Rightarrow ab-bc=0\Rightarrow b\left(a-c\right)=0\Rightarrow b=0\)
Khi đó P = x.a.b.c = x.a.0.c = 0
Vậy P = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
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 ^^
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tham khảo: Câu hỏi của Nguyễn Thị Nhàn - Toán lớp 8 - Học toán với OnlineMath
Học tốt=)
tth : mẫu nó khác bạn nhé
- mẫu nó là 2bc 2ac 2ab
mẫu mk ko có nhân 2
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ gt => ab+bc+ca=0
\(a^2+2bc=a^2+bc+\left(-ab-ac\right)=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)
Tương tự \(\hept{\begin{cases}b^2+2ac=\left(b-a\right)\left(b-c\right)\\c^2+2ab=\left(c-a\right)\left(c-b\right)\end{cases}}\)
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}=\frac{b-c+c-a+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow ab+bc+ca=0\)
\(\Rightarrow bc=-ab-ca\)
Vậy thì \(a^2+2bc=a^2+bc-ab-ac=a\left(a-b\right)-c\left(a-b\right)=\left(a-c\right)\left(a-b\right)\)
Tương tự ta cũng có:
\(b^2+2ac=\left(b-c\right)\left(b-a\right)\)
\(c^2+2ab=\left(c-a\right)\left(c-b\right)\)
Vậy thì \(A=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\)
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{-b+c+a-c-a+b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(A=0.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow ab+bc+ac=0\Rightarrow\hept{\begin{cases}ab=-bc-ac\\bc=-ac-ab\\ac=-ab-bc\end{cases}}\)(*)
Thay (*) vào M ta được:
\(M=\frac{1}{a^2+bc-ab-ac}+\frac{1}{b^2+ac-ab-bc}+\frac{1}{c^2+ab-bc-ac}\)
\(=\frac{1}{a\left(a-b\right)-c\left(a-b\right)}+\frac{1}{a\left(c-b\right)-b\left(c-b\right)}+\frac{1}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(a-b\right)\left(c-b\right)}-\frac{1}{\left(c-b\right)\left(a-c\right)}\)
\(=\frac{c-b}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}-\frac{a-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}\)
\(=\frac{c-b+a-c-a+b}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=0\)
Vậy M = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
TH1: Nếu \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\a+c=-b\end{cases}}\)
Thay vào biểu thức M ta có: \(M=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\left(-a\right).\left(-b\right).\left(-c\right)}{abc}=-1\)
TH2: Nếu \(a+b+c\ne0\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}}\)
Thay vào biểu thức M ta có: \(M=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{8abc}{abc}=8\)
Vậy \(M=-1\)hoặc \(M=8\)
Ta có: \(\frac{ab+1}{b}=\frac{bc+1}{c}\Rightarrow\left(ab+1\right)c=\left(bc+1\right)b\Rightarrow abc+c=b^2c+b\Rightarrow bc\left(a-b\right)=b-c\).
Tương tự: \(ca\left(b-c\right)=c-a;ab\left(c-a\right)=a-b\).
Nhân vế với vế rồi rút gọn ta có: \(\left(abc\right)^2=1\Rightarrow abc=\pm1\).
Vậy abc = 1 hoặc abc = -1.