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

<=>\(\frac{\left(b-a\right)-\left(c-a\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(c-b\right)-\left(a-b\right)}{\left(b-c\right)\left(b-a\right)}+\frac{\left(a-c\right)-\left(b-c\right)}{\left(c-a\right)\left(c-b\right)}=2013\)

<=>\(\frac{1}{c-a}+\frac{1}{a-b}+\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{b-c}+\frac{1}{c-a}=2013\)

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

<=>\(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}=\frac{2013}{2}=1006,5\)

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

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

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

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

Khi a + b = 0

=> (a + b)(b + c)(c + a) = 0 (2)

Nếu a + b \(\ne0\)

=> ab = -(a + b + c).c

=> ab + (a + b + c).c = 0

=> ab + ac + bc + c² = 0

=> (a + c)(b + c) = 0

=> (a + b)(b + c)(a + c) = 0 (1)

Từ (2)(1) => (a + b)(b + c)(a + c) = 0 \(\forall a;b;c\)

=> a = -b hoặc b = -c hoặc = c = -a

Nếu a = -b => a¹¹ = -b¹¹ => a¹¹ + b¹¹ = 0

=> P = 0 (3)

Nếu b = -c => b⁹ = - c⁹ => b⁹ + c⁹ = 0

=>P = 0 (4)

Nếu c = -a => c²⁰⁰¹ = -a²⁰⁰¹ => c²⁰⁰¹ + a²⁰⁰¹ = 0

=> P = 0 (5)

Từ (3);(4);(5) => P = 0 trong cả 3 trường hợp 

Vạy P = 0

Xyz là ad ak?

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

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

\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

b) Đề bài sai ^^

thi hsg co cao khong

dang no giong bai bdt vap LHP chuyen nam 2017-2018

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?

\(a^2\left(b+c\right)+b^2\left(c+a\right)+c^2\left(a+b\right)+2abc=0\)

=>\(\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

=>a=-b hoặc a=-c hoặc b=-c (1)

=>a=1 hoăc b=1 hoặc c=1 (2)

từ 1 và 2 => Q=1