Ta có: \(A=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}-\dfrac{b^2}{\left(b-a\right)\left(c-b\right)}-\dfrac{c^2}{\left(c-a\right)\left(b-c\right)}\)

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

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

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

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

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

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

\(M=\dfrac{\left(a+b\right)\left(b+c\right)}{\left(a-b\right)\left(b-c\right)}.\dfrac{\left(b+c\right)\left(c+a\right)}{\left(b-c\right)\left(c-a\right)}.\dfrac{\left(c+a\right)\left(a+b\right)}{\left(c-a\right)\left(a-b\right)}\) (loạn mắt quá !!! )

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

Thôi để bữa sau a làm (đau mắt quá :)))

Bài này mình làm một lần ở trường rồi nhưng không có điện thoại chụp được:((

Ta có: \(\dfrac{a^3}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^3}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^3}{\left(c-a\right)\left(c-b\right)}=\dfrac{a^3\left(c-b\right)+b^3\left(a-c\right)-c^3\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}\)\(=\dfrac{a^3\left(c-b\right)+b^3a-b^3c-c^3a+c^3b}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=\dfrac{a^3\left(c-b\right)-a\left(c^3-b^3\right)+bc\left(c^2-b^2\right)}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=\dfrac{a^3\left(c-b\right)-a\left(c-b\right)\left(a^2+bc+b^2\right)+bc\left(c-b\right)\left(c+b\right)}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}\)\(=\dfrac{\left(c-b\right)\left(a^3-ac^2-abc-ab^2+bc^2+b^2c\right)}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}=\dfrac{\left(c-b\right)\left[a\left(a^2-b^2\right)-c^2\left(a-b\right)-bc\left(a-b\right)\right]}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}\)\(=\dfrac{\left(c-b\right)\left[a\left(a-b\right)\left(a+b\right)-c\left(a-b\right)-bc\left(a-b\right)\right]}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=\dfrac{\left(c-b\right)\left(a-b\right)\left(a^2+ab-c-bc\right)}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}\)

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

Vì a, b, c là các số nguyên

=> a+b+c là các số nguyên

=> Đpcm.

Đấy mình làm chi tiết tiền tiệt lắm luôn, không hiểu thì mình chịu rồi, trời lạnh mà đánh máy nhiều thế này buốt tay lắm luôn:vv

1)\(\dfrac{c-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}+\dfrac{a-c}{\left(b-a\right)\left(b-c\right)\left(a-c\right)}+\dfrac{b-a}{\left(b-a\right)\left(c-b\right)\left(c-a\right)}=\dfrac{c-b+a-c+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

Lời giải:
Ta có:

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

\(\Leftrightarrow \frac{-(b-c)^2}{(a-b)(b-c)(c-a)}+\frac{-(c-a)^2}{(a-b)(b-c)(c-a)}+\frac{-(a-b)^2}{(a-b)(b-c)(c-a)}=2013\)

\(\Leftrightarrow \frac{-[(a-b)^2+(b-c)^2+(c-a)^2]}{(a-b)(b-c)(c-a)}=2013\)

\(\Rightarrow \frac{2(a^2+b^2+c^2-ab-bc-ac)}{(a-b)(b-c)(c-a)}=-2013(*)\)

Lại có:

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

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

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

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

\(=\frac{-1}{2}.-2013=\frac{2013}{2}\) (theo $(*)$)

Áp dụng t/c dtsbn ta có:

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

\(\dfrac{2b+c-a}{a}=2\Rightarrow2b+c-a=2a\Rightarrow2b=3a-c\)\(\dfrac{2c-b+a}{b}=2\Rightarrow2c-b+a=2b\Rightarrow2c=3b-a\)

\(\dfrac{2a+b-c}{c}=2\Rightarrow2a+b-c=2c\Rightarrow2a=3c-b\)

\(P=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{2a.2b.2c}=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{8abc}\)

\(B=\dfrac{\left(4a^2-1\right)\left(b-c\right)-\left(4b^2-1\right)\left(a-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\dfrac{4c^2-1}{\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{4a^2b-4a^2c-b+c-4ab^2+4b^2c+a-c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\dfrac{4ac^2-4bc^2-a+b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)

\(=\dfrac{4a^2b-4a^2c+a-b-4ab^2+4b^2c+4ac^2-4bc^2-a+b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)

\(=\dfrac{4a^2b-4ab^2-4a^2c+4ac^2-4bc^2+4b^2c}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)

\(=\dfrac{4a^2\left(b-c\right)+4bc\left(b-c\right)-4a\left(b^2-c^2\right)}{\left(b-c\right)\left(a-c\right)\left(a-b\right)}\)

\(=\dfrac{4a^2+4bc-4a\left(b+c\right)}{\left(a-c\right)\left(a-b\right)}\)

\(=\dfrac{4a^2-4ab+4bc-4ac}{\left(a-c\right)\left(a-b\right)}\)

\(=\dfrac{4a\left(a-b\right)-4c\left(a-b\right)}{\left(a-c\right)\left(a-b\right)}=4\)