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26 tháng 7 2017
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
30 tháng 9 2019
\(\frac{2x-1}{3x^2+7x+2}+\frac{3}{9x^2+15x+4}-\frac{2x+7}{3x^2-5x-12}=\frac{5}{x+2}\)
\(\Leftrightarrow\frac{2x-1}{\left(3x+1\right)\left(x+2\right)}+\frac{3}{\left(3x+1\right)\left(3x+4\right)}-\frac{2x+7}{\left(4x+3\right)\left(x-3\right)}=\frac{5}{\left(x+2\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{3x+1}+\frac{1}{3x+1}-\frac{1}{3x+4}+\frac{1}{3x+4}-\frac{1}{x-3}=\frac{5}{x+2}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x-3}=\frac{5}{x+2}\)
\(\Leftrightarrow\frac{x-3-x-2}{\left(x+2\right)\left(x-3\right)}=\frac{5\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow5x-3=-5\)
\(\Leftrightarrow x=-\frac{2}{5}\)
Chúc bạn học tốt !!!
2 tháng 1 2018
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
VT
2 tháng 1 2018
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 !
^_^
1 tháng 3 2020
\(x-\frac{\frac{x}{2}-\frac{3+x}{4}}{2}=3-\frac{\left(1-\frac{6-x}{3}\right).\frac{1}{2}}{2}\)
\(\Leftrightarrow2x-\frac{x}{2}+\frac{3+x}{4}=6-\frac{1}{2}+\frac{6-x}{6}\)
\(\Leftrightarrow24x-6x+9+3x=72-6+12-2x\)
\(\Leftrightarrow23x=69\)
\(\Leftrightarrow x=3\)
Vậy nghiệm của pt x=3
`Answer:`
\(\frac{x+2}{x-2}-\frac{3}{x+2}=\frac{12}{x^2-4}+2\left(ĐKXĐ:x\ne\pm2\right)\)
\(\Leftrightarrow\frac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{12}{\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(\Rightarrow\left(x+2\right)^2-3\left(x-2\right)=12+2\left(x+2\right)\left(x-2\right)\)
\(\Leftrightarrow x^2+4x+4-3x+6=12+2x^2-8\)
\(\Leftrightarrow x^2-2x^2+4x-3x=12-8-6-4\)
\(\Leftrightarrow-x^2+x=-6\)
\(\Leftrightarrow-x^2+x+6=0\)
\(\Leftrightarrow-x^2+3x-2x+6=0\)
\(\Leftrightarrow\left(-x^2+3x\right)-\left(2x-6\right)=0\)
\(\Leftrightarrow-x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(-x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x-2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\text{(Loại)}\\x=3\end{cases}}}\)