a: =>x-2+2=x2+2x
\(\Leftrightarrow x^2+x=0\)
=>x(x+1)=0
=>x=0(loại) hoặc x=-1(nhận)
b: Bạn xem lại đề
c: \(\Leftrightarrow\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow2\left(x^2+10x+25\right)-\left(x-5\right)^2=x^2+25x\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\)
=>5x+25=0
hay x=-5(loại)
a,ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)
\(\dfrac{1}{x}+\dfrac{2}{x^2-2x}=\dfrac{x+2}{x-2}\\ \Leftrightarrow\dfrac{x-2+2-x\left(x-2\right)}{x\left(x-2\right)}=0\\ \Rightarrow x-x^2+2x=0\\ \Leftrightarrow-x^2+3x=0\\ \Leftrightarrow-x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\left(tm\right)\end{matrix}\right.\)
b, ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-9\\x\ne\pm\dfrac{3}{4}\end{matrix}\right.\)
\(\dfrac{x+5}{x+9}+\dfrac{3x-7}{9-12x}=\dfrac{4x^2+x-7}{16x^2-9}\\ \Leftrightarrow\dfrac{3\left(x+5\right)\left(4x-3\right)\left(4x+3\right)}{3\left(x+9\right)\left(4x-3\right)\left(4x+3\right)}-\dfrac{\left(x+9\right)\left(3x-7\right)\left(4x+3\right)}{3\left(4x-3\right)\left(x+9\right)\left(4x+3\right)}-\dfrac{3\left(4x^2+x-7\right)\left(x+9\right)}{3\left(4x-3\right)\left(4x+3\right)\left(x+9\right)}=0\)
\(\Rightarrow48x^3+240x^2-27x-135-12x^3-89x^2+192x+189-12x^3-111x^2-6x+189=0\)
\(\Leftrightarrow24x^3+40x^2+159x+243=0\)
nghiệm xấu lắm bạn
c,ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne\pm5\end{matrix}\right.\)
\(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\\ \Leftrightarrow\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}-\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\dfrac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}-\dfrac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}=0\\ \Leftrightarrow\dfrac{2\left(x^2+10x+25\right)-\left(x^2-10x+25\right)-\left(x^2+25x\right)}{2x\left(x+5\right)\left(x-5\right)}=0\)
\(\Rightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\\ \Leftrightarrow5x+25=0\\ \Leftrightarrow x=-5\left(ktm\right)\)
a,ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)
\(\dfrac{1}{x}+\dfrac{2}{x^2-2x}=\dfrac{x+2}{x-2}\\ \Leftrightarrow\dfrac{x-2+2-x\left(x+2\right)}{x\left(x-2\right)}=0\\ \Rightarrow x-x^2-2x=0\\ \Leftrightarrow-x^2-x=0\\ \Leftrightarrow-x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
b, ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-9\\x\ne\pm\dfrac{3}{4}\end{matrix}\right.\)
\(\dfrac{x+5}{x+9}+\dfrac{3x-7}{9-12x}=\dfrac{4x^2+x-7}{16x^2-9}\\ \Leftrightarrow\dfrac{3\left(x+5\right)\left(4x-3\right)\left(4x+3\right)}{3\left(x+9\right)\left(4x-3\right)\left(4x+3\right)}-\dfrac{\left(x+9\right)\left(3x-7\right)\left(4x+3\right)}{3\left(4x-3\right)\left(x+9\right)\left(4x+3\right)}-\dfrac{3\left(4x^2+x-7\right)\left(x+9\right)}{3\left(4x-3\right)\left(4x+3\right)\left(x+9\right)}=0\)
\(\Rightarrow48x^3+240x^2-27x-135-12x^3-89x^2+192x+189-12x^3-111x^2-6x+189=0\)
\(\Leftrightarrow24x^3+40x^2+159x+243=0\)
nghiệm xấu lắm bạn
c,ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne\pm5\end{matrix}\right.\)
\(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\\ \Leftrightarrow\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}-\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\dfrac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}-\dfrac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}=0\\ \Leftrightarrow\dfrac{2\left(x^2+10x+25\right)-\left(x^2-10x+25\right)-\left(x^2+25x\right)}{2x\left(x+5\right)\left(x-5\right)}=0\)
\(\Rightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\\ \Leftrightarrow5x+25=0\\ \Leftrightarrow x=-5\left(ktm\right)\)
