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Có bài y như bạn, tham khảo nha. 

~ Học tốt ~

Ta có \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)

=> \(\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)

=> \(\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)

Nếu a + b + c = 0

=> a + b = -c

b + c = -a

a + c = -b

Khi đó P = \(\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)=\frac{a+b}{a}.\frac{b+c}{b}.\frac{a+c}{c}=\frac{-c}{a}.\frac{-a}{b}.\frac{-b}{c}=\frac{-abc}{abc}=-1\)

Nếu a + b + c \(\ne\)0

=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)

=> a = b = c

Khi đó P \(\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2.2.2=8\)

Vậy khi a + b + c = 0 thì P = -1

khi a + b + c  \(\ne\)0 thì P = 8

\(\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)

\(\frac{a+b+c}{c}=\frac{b+c+a}{a}=\frac{c+a+b}{b}\)

=>a=b=c

=>A=(1+b/a)(1+a/c)(1+c/b) = (1+1)(1+1)(1+1) =2.2.2 =8

Giải đỡ em đi anh

\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)

\(\Rightarrow S=\left(\frac{a+b+c}{b+c}\right)+\left(\frac{a+b+c}{c+a}\right)+\left(\frac{a+b+c}{a+b}\right)-3\)

\(\Rightarrow S=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3=2016.\frac{1}{90}-3=\frac{97}{5}\)

Vậy....................

\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\Rightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)    (cộng 3 vế với 1)

TH1: \(a+b+c=0\)

Khi đó: \(M=\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{c+a}{a}\right)=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=-1\)

TH2: \(a=b=c\) (ko thỏa mãn a,b,c đôi 1 khác nhau)

Vây M = -1

Chúc bạn học tốt.

ta có: \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=\frac{2.\left(a+b+c\right)}{a+b+c}.\)

Nếu \(a+b+c\ne0\)thì \(\frac{2.\left(a+b+c\right)}{a+b+c}=2\)

=> a + b = 2c

b+c = 2a

=> a-c = 2.(c-a)

=> c=a ( trái với đề bài)

=> a + b +c = 0

\(\Rightarrow M=\left(1+\frac{a}{b}\right).\left(1+\frac{b}{c}\right).\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{c+b}{c}\cdot\frac{a+c}{a}=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{c}=-1\)

Ta có: a + b +  c = 0 => a + b = -c; b + c = -a; a + c = -b

a + b + c = 0 <=> a + b = -c

<=> (a + b)³ = (-c)³ 

<=> a³ + 3a²b + 3ab² + b³ = -c³

<=> a³ + b³ + c³ = -3ab(a + b)

<=> a³ + b³ + c³ = 3abc (vì a + b = -c)

Khi đó: Q = \(\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)\)

Q = \(1+\frac{a\left(a-b\right)}{c\left(b-c\right)}+\frac{b\left(a-b\right)}{c\left(c-a\right)}+\frac{c\left(b-c\right)}{a\left(a-b\right)}+1+\frac{b\left(b-c\right)}{a\left(c-a\right)}+\frac{c\left(c-a\right)}{b\left(a-b\right)}+\frac{a\left(c-a\right)}{b\left(b-c\right)}+1\)

Q = \(3+\left(\frac{a\left(a-b\right)}{c\left(b-c\right)}+\frac{a\left(c-a\right)}{b\left(b-c\right)}\right)+\left(\frac{b\left(a-b\right)}{c\left(c-a\right)}+\frac{b\left(b-c\right)}{a\left(c-a\right)}\right)+\left(\frac{c\left(b-c\right)}{a\left(a-b\right)}+\frac{c\left(c-a\right)}{b\left(a-b\right)}\right)\)

Q = \(3+\frac{ab\left(a-b\right)+ac\left(c-a\right)}{bc\left(b-c\right)}+\frac{ab\left(a-b\right)+bc\left(b-c\right)}{ac\left(c-a\right)}+\frac{bc\left(b-c\right)+ca\left(c-a\right)}{ab\left(a-b\right)}\)

Q = \(3+\frac{a\left(ab-b^2+c^2-ac\right)}{bc\left(b-c\right)}+\frac{b\left(a^2-ab+bc-c^2\right)}{ac\left(c-a\right)}+\frac{c\left(b^2-bc+ac-a^2\right)}{ab\left(a-b\right)}\)

Q = \(3+\frac{a\left[a\left(b-c\right)-\left(b-c\right)\left(b+c\right)\right]}{bc\left(b-c\right)}+\frac{b\left[b\left(c-a\right)-\left(c-a\right)\left(c+a\right)\right]}{ac\left(c-a\right)}+\frac{c\left[c\left(a-b\right)-\left(a-b\right)\left(a+b\right)\right]}{ab\left(a-b\right)}\)

Q = \(3+\frac{a\left[a-\left(b+c\right)\right]}{bc}+\frac{b\left(b-\left(c+a\right)\right)}{ac}+\frac{c\left[c-\left(a+b\right)\right]}{ab}\)

Q = \(3+\frac{a\left(a+a\right)}{bc}+\frac{b\left(b+b\right)}{ac}+\frac{c\left(c+c\right)}{ab}\)

Q = \(3+\frac{2a^2}{bc}+\frac{2b^2}{ac}+\frac{2c^2}{ab}\)

Q = \(3+\frac{2a^3+2b^3+2c^3}{abc}\)

Q = \(3+\frac{2\left(a^3+b^3+c^3\right)}{abc}\)

Q = \(3+\frac{2.3abc}{abc}=3+6=9\)

Bài làm:

Đặt \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)

\(\Leftrightarrow abc.M=ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)

\(\Leftrightarrow abc.M=ab\left(a-b\right)+b^2c-bc^2+c^2a-ca^2\)

\(\Leftrightarrow abc.M=ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)\)

\(\Leftrightarrow abc.M=\left(a-b\right)\left(ab+c^2-ac-bc\right)\)

\(\Leftrightarrow abc.M=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)

\(\Rightarrow M=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{abc}\)

Đặt \(N=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)

\(\Rightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right).N=c\left(b-c\right)\left(c-a\right)+a\left(a-b\right)\left(c-a\right)+b\left(a-b\right)\left(b-c\right)\)

Mà \(a+b+c=0\Rightarrow\hept{\begin{cases}a=-b-c\\b=-c-a\\c=-a-b\end{cases}}\)

Thay vào ta được:

\(N=\frac{c\left(b-c\right)\left(c-a\right)-\left(b+c\right)\left(a-b\right)\left(c-a\right)+b\left(a-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(N=\frac{c\left(c-a\right)\left(b-c-a+b\right)+b\left(a-b\right)\left(b-c-c+a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(N=\frac{c\left(c-a\right)\left(2b-c-a\right)+b\left(a-b\right)\left(a+b-2c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(N=\frac{c\left(c-a\right)\left(2b+b\right)+b\left(a-b\right)\left(-c-2c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(N=\frac{3bc\left(c-a\right)-3bc\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(N=\frac{3bc\left(b+c-2a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(N=\frac{-9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{9abc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

Mà \(Q=M.N=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{abc}.\frac{9abc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=9\)

Vậy Q = 9 

\(S+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\)

\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)

\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)

\(=2010.\frac{1}{3}=670\)

\(\Rightarrow S=670-3=667\)