$\frac{b+c}{bc}=\frac{2}{a}$

$2bc=a\left(b+c\right)$

$bc+bc=ab+ac$

$bc-ab=ac-bc$

$b\left(c-a\right)=c\left(a-b\right)$

$\Rightarrow \frac{b}{c}=\frac{a-b}{c-a}$ ( đpcm )

\(P=\dfrac{ab}{\left(a-c\right)\left(b-c\right)}+\dfrac{bc}{\left(b-a\right)\left(c-a\right)}+\dfrac{ca}{\left(c-b\right)\left(a-b\right)}\)

\(=\dfrac{ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

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

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

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

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

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3abc-3a^2b-3ab^2=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3abc\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)=0\)

\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

Vì a;b;c đôi 1 khác nhau nên \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ne0\)

\(\Rightarrow a+b+c=0\) (đpcm)

chuyển vế -> phân tích a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca) -> cm a²+b²+c²-ab-bc-ca >= 0

ta có: a²+b²+c²-ab-bc-ca >= 0 <=> 2a²+2b²+2c²-2ab-2bc-2ca >= 0 <=> (a²-2ab+b²)+(b²-2bc+c²)+(c²-2ca+a²) >=0

<=>(a-b)²+(b-c)²+(c-a)² >=0

dấu "=" xảy ra khi a=b=c mà a,b,c đôi một khác nhau => a²+b²+c²-ab-bc-ca khác 0 <=> a+b+c=0

cảm ơn Đức Hùng

\(P=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^2}{\left(b-c\right)\left(b-a\right)}+\dfrac{c^2}{\left(c-b\right)\left(c-a\right)}\)

\(=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}+\dfrac{-b^2}{\left(b-c\right)\left(a-b\right)}+\dfrac{c^2}{\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

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

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

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

\(=1\)

\(b^2\left(a+c\right)=a^2\left(b+c\right)=2013\)nên \(a^2b+a^2c-b^2a-b^2c=0\Leftrightarrow ab\left(a-b\right)+c\left(a-b\right)\left(a+b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(ab+ca+bc\right)=0\Leftrightarrow ab+bc+ca=0\) vì \(a\ne b\ne c\ne0\)

\(\Leftrightarrow\left(ab+bc+ca\right)b=0\Leftrightarrow b^2\left(a+c\right)=-abc\)nên \(-abc=2013\)

\(\Leftrightarrow\left(ab+bc+ca\right)c=0\Leftrightarrow c^2\left(a+b\right)=-abc=2013\)

Có gì sai sót xin lượng thứ nha

Câu hỏi của Chu Hoàng THủy Tiên - Toán lớp 7 - Học toán với OnlineMath