Đặt \(\hept{\begin{cases}a-b=x\\b-c=y\\c-a=z\end{cases}}\)

Thế vào bài toán trở thành 

Cho: \(\frac{x+z}{xz}+\frac{x+y}{xy}+\frac{y+z}{yz}=2013\left(1\right)\)

Tính \(M=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

Từ (1) ta có

\(\left(1\right)\Leftrightarrow\frac{xy+yz+zx+yz+xy+zx}{xyz}=2013\)

\(\Leftrightarrow\frac{2\left(xy+yz+zx\right)}{xyz}=2013\)

\(\Leftrightarrow\frac{xy+yz+zx}{xyz}=\frac{2013}{2}\)

Ta lại có

\(M=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{xy+yz+zx}{xyz}=\frac{2013}{2}\)

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

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

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

\(=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=2013\)

\(\Rightarrow M=\frac{2013}{2}\)

đây là bài trong đề thi tớ mà, lúc đó là 5/12 sao bạn chép ra đây để hỏi?

1) \(M=a^2b^2c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)

Em chú ý bài toán sau nhé: Nếu a+b+c=0 <=> \(a^3+b^3+c^3=3abc\)

CM: có:a+b=-c <=> \(\left(a+b\right)^3=-c^3\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)

Chú ý: a+b=-c nên \(a^3+b^3+c^3=3abc\)

Do \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

Thay vào biểu thwusc M ta được M=3abc (ĐPCM)

2, em có thể tham khảo trong sách Nâng cao phát triển toán 8 nhé, anh nhớ không nhầm thì bài này trong đó

Nếu không thấy thì em có thể quy đồng lên mà rút gọn

vâng e cảm ơn anh 

TH1: a+b+c khác 0(a,b,c khác 0)

ta có:

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

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

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

\(\Rightarrow a=b=c\)

thay a=b=c vào M ta có:

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

\(\Rightarrow M=\left(1+1\right).\left(1+1\right).\left(1+1\right)=2^3=8\)

TH2: a+b+c=0(a,b,c khác 0)

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

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

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

thay các giá trị của a,b,c vào bt M ta có:

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

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

\(M=-1\)

vậy ...

p/s: bài này cô linh giải đúng rồi, nhưng ngắn quá => hơi khó hiểu

còn bn Darwin Watterson thiếu 1 trường hợp 

\(^{\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c+b+c-a+c+a-b}{a+b+c}=\frac{a+b+c}{a+b+c}=1}\)

Ta có: \(\frac{a+b-c}{c}=1\Rightarrow a+b-c=c\)

\(\Rightarrow a+b=2c\)

\(\frac{b+c-a}{a}=1\Rightarrow b+c-a=a\)

\(b+c=2a\)

\(\frac{c+a-b}{b}=1\Rightarrow c+a-b=b\)

\(\Rightarrow c+a=2b\)

\(M=\left(1+\frac{a}{b}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}\)( Quy đồng lên như bình thường )

Thay từ trên vào biểu thức, ta có: \(=\frac{2a.2b.2c}{abc}=\frac{8\left(abc\right)}{abc}=8\)

ĐK: a,b,c khác 0

+) a+b+c=0 => a+b=-c; a+c=-b; b+c=-a

=> M=-1

+)\(a+b+c\ne0\)

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

=> a+b=2c; b+c=2a; a+c=2b

=> \(M=\frac{a+b}{a}.\frac{b+c}{b}.\frac{c+a}{c}=8\)

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

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

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

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

\(=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=VP\left(đpcm\right)\)

Ta có :

\(VT=\frac{1}{2}\left[\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}\right]\)

\(=\frac{1}{2}\left[\frac{\left(b-c\right)^2}{\left(a-b\right)\left(a-c\right)}+\frac{\left(a-c\right)^2}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}+\frac{\left(a-b\right)^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)

\(=\frac{1}{2}\left[\frac{\left(b-c\right)^2+\left(a-c\right)^2+\left(a-b\right)^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)

\(=\frac{1}{2}\left[\frac{b^2-2bc+c^2+a^2-2ac+c^2+a^2-2ab+b^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)

\(=\frac{1}{2}\left[\frac{2a^2+2b^2+2c^2-2ab-2bc-2ac}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)

\(=\frac{a^2+b^2+c^2-ab-bc-ac}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)(1)

Lại có :

\(VP=\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\)

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

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

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

Từ (1) và (2) \(\RightarrowĐPCM\)

Xin phép thủ công :"))

\(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}=1008\)

\(\Leftrightarrow\frac{\left(b-c\right)\left(c-a\right)+\left(a-b\right)\left(c-a\right)+\left(a-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1008\)

\(\Leftrightarrow\frac{bc-c^2-ab+ac+ac-bc-a^2+ab+ab-b^2-ac+bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1008\)

\(\Leftrightarrow-\frac{a^2+b^2+c^2-ab-ac-bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1008\)

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

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

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

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

\(=2.1008=2016\)

a) 9x² - 36

=(3x)²-6²

=(3x-6)(3x+6)

=4(x-3)(x+3)

b) 2x³y-4x²y²+2xy³

=2xy(x²-2xy+y²)

=2xy(x-y)²

c) ab - b²-a+b

=ab-a-b²+b

=(ab-a)-(b²-b)

=a(b-1)-b(b-1)

=(b-1)(a-b)

P/s đùng để ý đến câu trả lời của mình

dễ!Ta có:

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

Chứng minh tương tự,Ta được:

\(\hept{\begin{cases}\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{a-b}+\frac{1}{b-c}\\\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}+\frac{1}{b-c}\end{cases}}\)

\(\Rightarrow\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-a\right)\left(b-c\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=2013\)\(\Rightarrow\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}=\frac{2013}{2}\)

Xong!

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

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

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

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

\(\Rightarrow a=b=c\) 

\(\Rightarrow\frac{a}{b}=1;\frac{b}{c}=1;\frac{c}{a}=1\)

\(\Rightarrow M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

ok , cảm ơn bạn !!!

Bài toán rất hay và bổ ích !!!

Đây nhé 

Đặt b + c = x ; c + a = y ;  a + b = z 

\(\Rightarrow\hept{\begin{cases}x+y=2c+b+a=2c+z\\y+z=2a+b+c=2a+x\\x+z=2b+a+c=2b+y\end{cases}}\)

\(\Rightarrow\frac{x+y-z}{2}=c;\frac{y+z-x}{2}=a;\frac{x+z-y}{2}=b\)

Thay vào PT đã cho ở đề bài , ta có : 

\(\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\)

\(=\frac{1}{2}\left(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}-3\right)\)

\(\ge\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)

( cái này cô - si cho x/y + /x ; x/z + z/x ; y/z + z/y) 

e cũng có 1 vài cách chứng minh khá là cổ điển ạ !

Sử dụng BĐT AM-GM ta có :

\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2}{b+c}.\frac{b+c}{4}}=2.\frac{a}{2}=a\)

Bằng cách chứng minh tương tự :

\(\frac{b^2}{a+c}+\frac{a+c}{4}\ge b;\frac{c^2}{a+b}+\frac{a+b}{4}\ge c\)

Cộng theo vế các bđt cùng chiều ta được :

\(\frac{a^2}{c+b}+\frac{b^2}{a+c}+\frac{c^2}{a+b}+\frac{2\left(a+b+c\right)}{4}\ge a+b+c\)

\(< =>\frac{a^2}{b+c}+\frac{a}{2}+\frac{b^2}{a+c}+\frac{b}{2}+\frac{c^2}{a+b}+\frac{c}{2}\ge a+b+c\)

\(< =>\frac{a^2}{b+c}+a+\frac{b^2}{a+c}+b+\frac{c^2}{a+b}+c\ge\frac{3}{2}\left(a+b+c\right)\)

\(< =>\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{a+c}+\frac{c\left(a+b+c\right)}{b+a}\ge\frac{3}{2}\left(a+b+c\right)\)

\(< =>\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\left(Q.E.D\right)\)