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\(A=\left(\dfrac{1+2x}{4+2x}-\dfrac{x}{3x-6}+\dfrac{2x^2}{12-3x^2}\right).\dfrac{24-12x}{6+13x}\)\(=\left(\dfrac{1+2x}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^2}{3\left(x-2\right)\left(x+2\right)}\right).\dfrac{-12\left(x-2\right)}{13x+6}\)\(=\left(\dfrac{3\left(1+2x\right)\left(x-2\right)}{6\left(x+2\right)\left(x-2\right)}-\dfrac{2x\left(x+2\right)}{6\left(x+2\right)\left(x-2\right)}-\dfrac{4x^2}{6x\left(x+2\right)\left(x-2\right)}\right).\dfrac{-2\left(x-2\right)}{13x+6}\)\(=\dfrac{6x^2-9x-6-2x^2-4x-4x^2}{6\left(x+2\right)\left(x-2\right)}.\dfrac{-12\left(x-2\right)}{13x+6}\)\(=\dfrac{-\left(13x+6\right)}{6\left(x+2\right)\left(x-2\right)}.\dfrac{-12\left(x-2\right)}{13x+6}\)
\(=\dfrac{2}{x+2}\)
Giải:
a) \(8\left(3x-2\right)-13x=5\left(12-3x\right)+7x\)
\(\Leftrightarrow24x-16-13x=60-15x+7x\)
\(\Leftrightarrow24x-13x+15x-7x=60+16\)
\(\Leftrightarrow19x=76\)
\(\Leftrightarrow x=\dfrac{76}{19}=4\)
Vậy ...
b) \(\dfrac{5x}{x+2}-\dfrac{3}{x-2}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\) (1)
ĐKXĐ: \(x\ne\pm2\)
\(\left(1\right)\Leftrightarrow\dfrac{5x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow5x\left(x-2\right)-3\left(x+2\right)+3x^2+6=0\)
\(\Leftrightarrow5x^2-10x-3x-6+3x^2+6=0\)
\(\Leftrightarrow8x^2-13x=0\)
\(\Leftrightarrow x\left(8x-13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\8x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=\dfrac{13}{8}\left(TM\right)\end{matrix}\right.\)
Vậy ...
c) \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\) (2)
ĐKXĐ: \(x\ne-1;x\ne3\)
\(\left(2\right)\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{4x}{2\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)
\(\Leftrightarrow x\left(x+1+x-3\right)=4x\)
\(\Leftrightarrow x\left(2x-2\right)=4x\)
\(\Leftrightarrow2x-2=4\)
\(\Leftrightarrow x=3\)
Vậy ...
1) ĐKXĐ của \(x\):
\(\left\{{}\begin{matrix}2x-6\ne0\\2x^2+6x\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-3\right)\ne0\\2x\left(x+3\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne0;x\ne-3\end{matrix}\right.\)
ĐKXĐ: \(x\ne0;x\ne\pm3\)
Ta có: \(\dfrac{3}{2x-6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x-3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3}{2\left(x-3\right)}+\dfrac{x-6}{2x\left(x-3\right)}\)
\(=\dfrac{3.2+x-6}{2x\left(x-3\right)}\)
\(=\dfrac{6+x-6}{2x\left(x-3\right)}\)
\(=\dfrac{x}{2x\left(x-3\right)}\)
\(=\dfrac{1}{2\left(x-3\right)}\)
2) ĐKXĐ của câu này bạn làm tương tự câu trên nhé, ở đây ngoặc không đủ
ĐKXĐ: \(x\ne0;x\ne\pm2;x\ne3\)
Ta có: \(A=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\)
\(A=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2-x}{2+x}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(A=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{\left(2-x\right)\left(2+x\right)}-\dfrac{2-x}{2+x}\right).\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(A=\dfrac{\left(2+x\right)\left(2+x\right)-4x^2-\left(2-x\right)\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(A=\dfrac{4+4x+x^2-4x^2-\left(4-4x+x^2\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(A=\dfrac{-4x^2+8x}{\left(2-x\right)\left(2+x\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(A=\dfrac{-4x\left(x-2\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(A=\dfrac{-4x^2\left(x-2\right)}{\left(2+x\right)\left(x-3\right)}\)