a.
\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)
\(\Leftrightarrow\left(4x^2+8x+3\right)\left(x^2+2x+1\right)=18\)
Đặt \(t=x^2+2x+1=\left(x+1\right)^2\left(t\ge0\right)\)
\(\Rightarrow\left(4t-1\right)\cdot t=18\)
\(\Leftrightarrow\left(2t\right)^2-2\cdot2t\cdot\dfrac{1}{4}+\dfrac{1}{16}=\dfrac{289}{16}\)
\(\Leftrightarrow\left(2t-\dfrac{1}{4}\right)^2=\dfrac{289}{16}\Leftrightarrow\left(t-\dfrac{1}{8}\right)^2=\dfrac{289}{64}\)
\(\Leftrightarrow\left[{}\begin{matrix}t-\dfrac{1}{8}=\dfrac{17}{8}\\t-\dfrac{1}{8}=-\dfrac{17}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{9}{4}\\t=-2\left(loai\right)\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^2=\dfrac{9}{4}\Leftrightarrow\left[{}\begin{matrix}x+1=\dfrac{3}{2}\\x+1=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{5}{2};\dfrac{1}{2}\right\}\)
b.
Ta có:
- \(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
- \(x^2+11x+24=x^2+3x+8x+24=x\left(x+3\right)+8\left(x+3\right)=\left(x+3\right)\left(x+8\right)\)
- \(x^2+18x+80=x^2+8x+10x+80=x\left(x+8\right)+10\left(x+8\right)=\left(x+8\right)\left(x+10\right)\)
Thay vào phương trình, ta được:
\(\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{5}{\left(x+3\right)\left(x+8\right)}+\dfrac{2}{\left(x+8\right)\left(x+10\right)}=\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+10}=\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+10}=\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{x+10-\left(x+1\right)}{\left(x+1\right)\left(x+10\right)}=\dfrac{9}{25}\Leftrightarrow\dfrac{9}{\left(x+1\right)\left(x+10\right)}=\dfrac{9}{25}\)
\(\Leftrightarrow\left(x+1\right)\left(x+10\right)=25\)
\(\Leftrightarrow x^2+11x+\dfrac{121}{4}=\dfrac{181}{4}\)
\(\Leftrightarrow\left(x+\dfrac{11}{2}\right)^2=\dfrac{181}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{11}{2}=\dfrac{\sqrt{181}}{2}\\x+\dfrac{11}{2}=-\dfrac{\sqrt{181}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{181}}{2}\\x=\dfrac{-11-\sqrt{181}}{2}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{-11+\sqrt{181}}{2};\dfrac{-11-\sqrt{181}}{2}\right\}\)
