Đáp án + giải thích các bước giải:
`x^2-4x-3=\sqrt{x+5} (-5<=x<=2-\sqrt{7}` hoặc `x>=2+\sqrt{7})`
`->(x^2-4x-3)^3=x+5`
`->x^4-8x^3+10x^2+24x+9=x+5`
`->x^4-8x^3+10x^2+23x+4=0`
`->x^4+x^3-9x^3-9x^2+19x^2+19x+4x+4=0`
`->x^3(x+1)-9x^2(x+1)+19x(x+1)+4(x+1)=0`
`->(x^3-9x^2+19x+4)(x+1)=0`
`->(x^3-4x^2-5x^2+20x-x+4)(x+1)=0`
`->[x^2(x-4)-5x(x-4)-(x-4)](x+1)=0`
`->(x^2-5x-1)(x-4)(x+1)=0`
`->`\(\left[ \begin{array}{l}x^2-5x-1=0\\x-4=0\\x+1=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x^2-2.x.\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{29}{4}=0\\x=4(KTM)\\x=-1\end{array} \right.\)
`->`\(\left[ \begin{array}{l}(x-\dfrac{5}{2})^2=\dfrac{29}{4}\\x=-1\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x-\dfrac{5}{2}=\dfrac{\sqrt{29}}{2}\\x-\dfrac{5}{2}=\dfrac{-\sqrt{29}}{2}\\x=-1\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=\dfrac{\sqrt{29}+5}{2}(TM)\\x=\dfrac{-\sqrt{29}+5}{2}(KTM)\\x=-1(TM)\end{array} \right.\)
Vậy `S={(\sqrt{29}+5)/2 ;-1}`
