Câu 4:
\(x^2+\dfrac{25x^2}{\left(x+5\right)^2}=11\left(ĐKXĐ:x\ne-5\right)\)
\(\Leftrightarrow\left(x-\dfrac{5x}{x+5}\right)^2+2.x.\dfrac{5x}{x+5}=11\)
\(\Leftrightarrow\left(\dfrac{x^2+5x-5x}{x+5}\right)^2+\dfrac{10x^2}{x+5}=11\)
\(\Leftrightarrow\left(\dfrac{x^2}{x+5}\right)^2+\dfrac{10x^2}{x+5}=11\) (*)
-Đặt \(a=\dfrac{x^2}{x+5}\)
(*) \(\Leftrightarrow a^2+10a=11\)
\(\Leftrightarrow a^2+10a-11=0\)
\(\Leftrightarrow a^2-a+11a-11=0\)
\(\Leftrightarrow a\left(a-1\right)+11\left(a-1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(a+11\right)=0\)
\(\Leftrightarrow a-1=0\) hay \(a+11=0\)
\(\Leftrightarrow\dfrac{x^2}{x+5}-1=0\) hay \(\dfrac{x^2}{x+5}+11=0\)
\(\Leftrightarrow\dfrac{x^2-x-5}{x+5}=0\) hay \(\dfrac{x^2+11x+55}{x+5}=0\)
\(\Rightarrow x^2-x-5=0\) hay \(x^2+11x+55=0\)
\(\Leftrightarrow x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{21}{4}=0\) hay \(x^2+2.\dfrac{11}{2}x+\dfrac{121}{4}+\dfrac{99}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2-\dfrac{21}{4}=0\) hay \(\left(x+\dfrac{11}{2}\right)^2+\dfrac{99}{4}=0\) (vô nghiệm vì \(\left(x+\dfrac{11}{2}\right)^2+\dfrac{99}{4}\ge\dfrac{99}{4}\forall x\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}-\dfrac{\sqrt{21}}{2}\right)\left(x-\dfrac{1}{2}+\dfrac{\sqrt{21}}{2}\right)=0\)
\(\Leftrightarrow x-\dfrac{1}{2}-\dfrac{\sqrt{21}}{2}=0\) hay \(x-\dfrac{1}{2}+\dfrac{\sqrt{21}}{2}=0\)
\(\Leftrightarrow x=\dfrac{1\pm\sqrt{21}}{2}\left(tmđk\right)\)
-Vậy \(S=\left\{\dfrac{1\pm\sqrt{21}}{2}\right\}\)
