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

\(\Leftrightarrow\dfrac{a\left(y+z\right)}{abc}=\dfrac{b\left(z+x\right)}{abc}=\dfrac{c\left(x+y\right)}{abc}\)

\(\Leftrightarrow\dfrac{\left(x+y\right)-\left(z+x\right)}{ab-ac}=\dfrac{y-z}{a\left(b-c\right)}\)

\(\Leftrightarrow\dfrac{\left(y+z\right)-\left(x+y\right)}{bc-ab}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{\left(z+x\right)-\left(y+z\right)}{ac-bc}=\dfrac{x-y}{c\left(a-b\right)}\)

\(\Rightarrow\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\left(đpcm\right)\)

`VT = (b-c)/((a-b)(a-c)) + (c-a)/((b-c)(b-a)) +(a-b)/((c-a)(c-b)) = 2/(a-b) + 2/(b-c) + 2/(c-a)`

`=-((a-b-a+c)/((a-b)(a-c))+(b-c-b+a)/((b-c)(b-a))+(c-a-c+b)/((c-a)(c-b)))`

`=-((a-b)/((a-b)(a-c))-(a-c)/((a-b)(a-c))+(b-c)/((b-c)(b-a))-(b-a)/((b-c)(b-a))+(c-a)/((c-a)(c-b))-(c-b)/((c-a)(c-b)))`

`= 1/(c-a)+1/(a-b)+1/(a-b)+1/(b-c)+1/(b-c)+1/(c-a)`

`=2/(a-b)+2/(b-c)+2/(c-a)=VP(đpcm)`

Biến đổi tương đương thôi em, dễ mà =)

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

\(ĐKXĐ:\)a,b,c đôi một khác nhau.

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

\(\Leftrightarrow x\left(\dfrac{1}{\left(a-b\right).\left(a-c\right)}+\dfrac{1}{\left(b-a\right).\left(b-c\right)}+\dfrac{1}{\left(c-a\right).\left(c-b\right)}\right)=2\)

\(\Leftrightarrow x\left(\dfrac{-\left(b-c\right)-\left(c-a\right)-\left(a-b\right)}{\left(a-b\right).\left(b-c\right).\left(c-a\right)}\right)=2\)

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

\(\Leftrightarrow x\left(\dfrac{0}{\left(a-b\right).\left(b-c\right).\left(c-a\right)}\right)=2\)

\(\Rightarrow0x=2\) (vô lý)

-Vậy khi a,b,c đôi một khác nhau thì phương trình vô nghiệm.

sky oi say oh yeah