Bài 11:
a) \(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\) (ĐKXĐ: \(x\ne-1;x\ne2\))
\(\Leftrightarrow\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}-\dfrac{5x+5}{\left(x+1\right)\left(x-2\right)}=\dfrac{-15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow x-2-5x-5=-15\)
\(\Leftrightarrow x=2\) (KTMĐK)
Vậy: \(S=\varnothing\)
b) \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\) (ĐKXĐ: \(x\ne2;x\ne-2\))
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{-5x+2}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2-2x-x+2-x^2-2x=-5x+2\)
\(\Leftrightarrow-5x=-5x\) (luôn đúng)
Vậy phương trình có vô số nghiệm.
\(b)\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0.\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0.\\ \Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0.\\ \Leftrightarrow\left(x-2\right)\left(3x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(e)2x^3+6x^2=x^2+3x.\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0.\\ \Leftrightarrow\left(2x^2-x\right)\left(x+3\right)=0.\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0.\\x=\dfrac{1}{2}.\\x=-3.\end{matrix}\right.\)
\(c)\left(2x+5\right)^2=\left(x+2\right)^2.\\ \Leftrightarrow\left(2x+5+x+2\right)\left(2x+5-x-2\right)=0.\\ \Leftrightarrow\left(3x+7\right)\left(x+3\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7}{3}.\\x=-3.\end{matrix}\right.\)
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