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Ta có:
\(\dfrac{b-c}{1\left(a-b\right)\left(a-c\right)}+\dfrac{c-a}{\left(b-c\right)\left(b-a\right)}+\dfrac{a-b}{\left(c-a\right)\left(c-b\right)}\)
\(=\dfrac{c-b}{1\left(a-b\right)\left(c-a\right)}+\dfrac{a-c}{\left(b-c\right)\left(a-b\right)}+\dfrac{b-a}{\left(c-a\right)\left(b-c\right)}\)
Quy đồng rút gọn ta được
\(=\dfrac{2\left(ab+bc+ca-a^2-b^2-c^2\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{2\left[\left(a-b\right)\left(b-c\right)+\left(b-c\right)\left(c-a\right)+\left(c-a\right)\left(a-b\right)\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=2\left(\dfrac{1}{a-b}+\dfrac{1}{b-c}+\dfrac{1}{c-a}\right)\)
PS: Hôm qua đi chơi nên nay mới giải nhé.
\(ĐKXĐ:\)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.
\(\text{ }\dfrac{\left(x-a\right)\left(x-c\right)}{\left(b-a\right)\left(b-c\right)}+\dfrac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-a\right)}{\left(b-a\right)\left(b-c\right)}.\left(x-c\right)+\dfrac{\left(x-b\right)}{\left(a-b\right)\left(a-c\right)}.\left(x-c\right)=1\)
\(\Leftrightarrow\left(x-c\right)\left(\dfrac{\left(x-a\right)}{\left(b-a\right)\left(b-c\right)}+\dfrac{\left(x-b\right)}{\left(a-b\right)\left(a-c\right)}\right)=1\)
\(\Leftrightarrow\left(x-c\right)\dfrac{\left(a-x\right)\left(a-c\right)+\left(x-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
\(\Leftrightarrow\left(x-c\right)\left[\left(a^2-b^2\right)-x\left(a-b\right)-c\left(a-b\right)\right]=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
\(\Leftrightarrow\left(x-c\right)\left(a-b\right)\left(a+b-x-c\right)=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
\(\Leftrightarrow\left(x-c\right)\left(a+b-x-c\right)-\left(b-c\right)\left(a-c\right)=0\)
\(\Leftrightarrow ax+bx-x^2-xc-ac-bc+xc+c^2-ab+bc+ac-c^2=0\)
\(\Leftrightarrow x^2-ax-bx+ab=0\)
\(\Leftrightarrow x\left(x-a\right)+b\left(x-a\right)=0\)
\(\Leftrightarrow\left(x-a\right)\left(x-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-a=0\\x-b=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=a\\x=b\end{matrix}\right.\)
Vậy\(S=\left\{a,b\right\}\)
a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
b) \(\dfrac{\left(a^2-\left(b+c\right)^2\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
\(=\dfrac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(\left(a-c\right)^2-b^2\right)}\)
\(=\dfrac{\left(a-c-b\right)\left(a-c+b\right)}{\left(a-c-b\right)\left(a-c+b\right)}=1\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^3-x\left(x+1\right)\left(x-1\right)+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{x^3-3x^2+3x-1-x^3+x+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{4x-1}{x^3\left(x-1\right)^2}\)
d) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)
\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)
\(=\dfrac{x}{x+y}\)
