bài 2 giải các phương trình saub,\(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\) m,\(\dfrac{3x-1}{x+1}=\dfrac{2x+1}{x-1}\)d,\(\dfrac{3x-14}{x+5}=\dfrac{2}{3}\) p,\(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\)f,\(\dfrac{6}{x}-1=\dfrac{2x-3}{3}\) r,\(\dfrac{1}{x+3}+\dfrac{1}{x-1}=\dfrac{10}{\left(x+3\right)\left(x-1\right)}\)h,\(\dfrac{1}{x-2}+3=\dfrac{x-3}{2-x}\) ...
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bài 2 giải các phương trình sau
b,\(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\) m,\(\dfrac{3x-1}{x+1}=\dfrac{2x+1}{x-1}\)
d,\(\dfrac{3x-14}{x+5}=\dfrac{2}{3}\) p,\(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\)
f,\(\dfrac{6}{x}-1=\dfrac{2x-3}{3}\) r,\(\dfrac{1}{x+3}+\dfrac{1}{x-1}=\dfrac{10}{\left(x+3\right)\left(x-1\right)}\)
h,\(\dfrac{1}{x-2}+3=\dfrac{x-3}{2-x}\) t,\(\dfrac{3x}{x-2}-\dfrac{x}{x-5}=\dfrac{3x}{\left(x-2\right)\left(5-x\right)}\)
j,\(\dfrac{5}{3x+2}=2x-1\) u,\(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=\dfrac{2\left(x^2+x-1\right)}{x\left(x+1\right)}\)
w,\(\dfrac{5x}{2x+2}+1=-\dfrac{6}{x+1}\) s, \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{2x}{\left(x-1\right)\left(x-3\right)}\)
ơ,\(\dfrac{1}{x-1}+\dfrac{2}{x+1}=\dfrac{x}{x^2-1}\) v,\(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)
z,\(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\) ư,\(\dfrac{x+2}{x-2}-\dfrac{-2}{x^2-2x}=\dfrac{1}{x}\)
o,\(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\) ô,\(1-\dfrac{1}{1-x}=\dfrac{x^2}{x^2-1}\) zz,\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)
e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)
\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)
\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)
\(\Leftrightarrow x=-1\left(TM\right)\)