Đáp án:
`a) \ \ S={(3+sqrt(13))/2;(3-sqrt(13))/2}`
`b) \ \ S={(sqrt(29)-5)/2;(-sqrt(29)-5)/2}`
Giải thích các bước giải:
`a) \ \ x^2-3x-1=0`
`<=> x^2 - 2 . x . 3/2 + (3/2)^2 - 1 - (3/2)^2=0`
`<=> (x-3/2)^2-13/4=0`
`<=> (x-3/2)^2=13/4`
`<=>` \(\left[ \begin{array}{l}x-\dfrac{3}{2}=\dfrac{\sqrt{13}}{2}\\x-\dfrac{3}{2}=-\dfrac{\sqrt{13}}{2}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt{13}+3}{2}\\x=\dfrac{-\sqrt{13}+3}{2}\end{array} \right.\)
Vậy `S = {(3+sqrt(13))/2;(3-sqrt(13))/2}`
`b) \ \ x^2+5x-1=0`
`<=> x^2+2.x . 5/2 +(5/2)^2-1-(5/2)^2=0`
`<=> (x+5/2)^2-29/4=0`
`<=> (x+5/2)^2=29/4`
`<=>` \(\left[ \begin{array}{l}x+\dfrac{5}{2}=\dfrac{\sqrt{29}}{2}\\x+\dfrac{5}{2}=-\dfrac{\sqrt{29}}{2}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt{29}-5}{2}\\x=\dfrac{-\sqrt{29}-5}{2}\end{array} \right.\)
Vậy `S={(sqrt(29)-5)/2;(-sqrt(29)-5)/2}`