Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐK \(x\ne\left\{1;2;3;4\right\}\)
Ta có \(\frac{x^2-2x+2}{x-1}+\frac{x^2-8x+20}{x-4}=\frac{x^2-4x+6}{x-2}+\frac{x^2-6x+12}{x-3}\)
\(\Leftrightarrow\frac{\left(x-1\right)^2+1}{x-1}+\frac{\left(x-4\right)^2+4}{x-4}=\frac{\left(x-2\right)^2+2}{x-2}+\frac{\left(x-3\right)^2+3}{x-3}\)
\(\Leftrightarrow x-1+\frac{1}{x-1}+x-4+\frac{4}{x-4}=x-2+\frac{2}{x-2}+x-3+\frac{3}{x-3}\)
\(\Leftrightarrow\frac{1}{x-1}+\frac{4}{x-4}=\frac{2}{x-2}+\frac{3}{x-3}\)
\(\Leftrightarrow\frac{5x-8}{x^2-5x+4}=\frac{5x-12}{x^2-5x+6}\)\(\Leftrightarrow\left(5x-8\right)\left(x^2-5x+6\right)=\left(5x-12\right)\left(x^2-5x+4\right)\)
\(\Leftrightarrow5x^3-25x^2+30x-8x^2+40x-48=5x^3-25x^2+20x-12x^2+60x-48\)
\(\Leftrightarrow4x^2-10x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{2}\end{cases}\left(tm\right)}\)
Vậy x=0 hoặc x=5/2
ĐKXĐ: \(x\ne-1,-2,-3,-4\)
\(\Leftrightarrow\frac{\left(x+1\right)^2+1}{x+1}+\frac{\left(x+4\right)^2+4}{x+4}=\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow x+1+\frac{1}{x+1}+x+4+\frac{4}{x+4}=x+2+\frac{2}{x+2}+x+3+\frac{3}{x+3}\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{1}{x+4}=\frac{1}{x+2}+\frac{1}{x+3}\)
\(\Leftrightarrow\frac{x}{x+1}+\frac{x}{x+4}=\frac{x}{x+2}+\frac{x}{x+3}\)
\(\Leftrightarrow x\left(\frac{1}{x+1}+\frac{1}{x+4}-\frac{1}{x+2}-\frac{1}{x+3}\right)=0\)
\(\Leftrightarrow x\left(\frac{1}{x^2+3x+2}-\frac{1}{x^2+7x+12}\right)=0\)
\(\Leftrightarrow-x\left(\frac{4x+10}{\left(x^2+3x+2\right)\left(x^2+7x+12\right)}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{5}{2}\end{cases}}\)Thỏa mãn ĐKXĐ
Ta có Pt
<=>\(\frac{\left(x+1\right)^2+1}{x+1}+\frac{\left(x+4\right)^2+4}{x+4}=\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+3\right)^2+3}{x+3}\)
<=>\(x+1+\frac{1}{x+1}+x+4+\frac{4}{x+4}=x+2+\frac{2}{x+2}+x+3+\frac{3}{x+3}\)
<=>\(\frac{1}{x+1}+\frac{4}{x+4}=\frac{2}{x+2}+\frac{3}{x+3}\)
<=>\(1-\frac{1}{x+1}+1-\frac{4}{x+4}=1-\frac{2}{x+2}+1-\frac{3}{x+3}\)
<=>\(\frac{x}{x+1}+\frac{x}{x+4}=\frac{x}{x+2}+\frac{x}{x+3}\Leftrightarrow x\left(\frac{1}{x+1}+\frac{1}{x+4}-\frac{1}{x+2}-\frac{1}{x+3}\right)=0\)
<=>\(\orbr{\begin{cases}x=0\\\frac{1}{x+1}+\frac{1}{x+4}-\frac{1}{x+2}-\frac{1}{x+3}=0\left(1\right)\end{cases}}\)
Giải pt (1) , ta có
\(\frac{x+2-x-1}{\left(x+1\right)\left(x+2\right)}-\frac{x+4-x-3}{\left(x+3\right)\left(x+4\right)}=0\)
<=>\(\frac{1}{x^2+3x+2}-\frac{1}{x^2+7x+12}=0\Leftrightarrow x^2+3x+2=x^2+7x+12\)
<=>\(4x+10=0\Leftrightarrow x=-\frac{5}{2}\)
nhớ đối chiếu đk nhé !
^_^
ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x-2\ne0\\x-3\ne0\\x-4\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne3\\x\ne4\end{matrix}\right.\)
Ta có : \(\frac{x^2-2x+2}{x-1}+\frac{x^2-8x+20}{x-4}=\frac{x^2-4x+6}{x-2}+\frac{x^2-6x+12}{x-3}\)
=> \(\frac{x^2-2x+1+1}{x-1}+\frac{x^2-8x+16+4}{x-4}=\frac{x^2-4x+4+2}{x-2}+\frac{x^2-6x+9+3}{x-3}\)
=> \(\frac{\left(x-1\right)^2+1}{x-1}+\frac{\left(x-4\right)^2+4}{x-4}=\frac{\left(x-2\right)^2+2}{x-2}+\frac{\left(x-3\right)^2+3}{x-3}\)
=> \(x-1+\frac{1}{x-1}+x-4+\frac{4}{x-4}=x-2+\frac{2}{x-2}+x-3+\frac{3}{x-3}\)
=> \(\frac{1}{x-1}+\frac{4}{x-4}=\frac{2}{x-2}+\frac{3}{x-3}\)
=> \(\frac{x-4}{\left(x-1\right)\left(x-4\right)}+\frac{4\left(x-1\right)}{\left(x-4\right)\left(x-1\right)}=\frac{2\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{3\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\)
=> \(\frac{\left(x-4+4\left(x-1\right)\right)}{\left(x-1\right)\left(x-4\right)}=\frac{2\left(x-3\right)+3\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
=> \(\frac{x-4+4x-4}{\left(x-1\right)\left(x-4\right)}=\frac{2x-6+3x-6}{\left(x-2\right)\left(x-3\right)}\)
=> \(\frac{5x-8}{x^2-5x+4}=\frac{5x-12}{x^2-5x+6}\)
=> \(5x^3-25x^2+30x-8x^2+40x-48=5x^3-25x^2+20x-12x^2+60x-48\)
=> \(5x^3-25x^2+30x-8x^2+40x-48-5x^3+25x^2-20x+12x^2-60x+48=0\)
=> \(4x^2-10x=0\)
=> \(2x\left(2x-5\right)=0\)
=> \(\left[{}\begin{matrix}2x=0\\2x-5=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\x=\frac{5}{2}\end{matrix}\right.\) ( TM )
Vậy phương trình trên có tập nghiệm là \(S=\left\{0,\frac{5}{2}\right\}\)