\(\begin{array}{l} pt \Leftrightarrow 2\sqrt {{x^2} - \frac{1}{4} + \sqrt {{{\left( {x + \frac{1}{2}} \right)}^2}} } = 2{x^3} + {x^2} + 2x + 1\\ \Leftrightarrow 2\sqrt {{x^2} - \frac{1}{4} + x + \frac{1}{2}} = 2{x^3} + {x^2} + 2x + 1\\ \Leftrightarrow 2\sqrt {{{\left( {x + \frac{1}{2}} \right)}^2}} = 2{x^3} + {x^2} + 2x + 1\\ \Leftrightarrow 2\left( {x + \frac{1}{2}} \right) = 2{x^3} + {x^2} + 2x + 1\\ \Leftrightarrow 2x + 1 = 2{x^3} + {x^2} + 2x + 1\\ \Leftrightarrow 2{x^3} + {x^2} = 0\\ \Leftrightarrow {x^2}\left( {2x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = - \frac{1}{2} \end{array} \right. \end{array}\)

Đáp án:

Giải thích các bước giải: b. Áp dụng hệ thức lượng trong tam giác AHB vuông tại H có HE vuông góc AB: \(\begin{array}{l} H{A^2} = AE.AB\\ Mat\;khac:Theo\;dinh\;ly\;Pytago\;trong\;tam\;giac\;AHC\;vuong\;tai\;H\\ A{C^2} = A{H^2} + H{C^2}\\ \Rightarrow H{A^2} = A{C^2} - H{C^2} \Rightarrow AE.AB = A{C^2} - H{C^2} \end{array}\) \(\begin{array}{l} c.\;AF = FH.\tan \widehat {FHA}\\ Ma:\widehat {FHA} = \widehat C(cung\;phu\;\widehat {CHF})\\ \Rightarrow AF = FH.\tan \widehat C\\ Mat\;khac:\;Tu\;giac\;AFHE\;co\;4\;goc\;vuong \Rightarrow AFHE\;la\;hcn \Rightarrow FH = AE\\ \Rightarrow AF = AE.\tan \widehat C\\ d.Ap\;dung\;he\;thuc\;luong\;ta\;co:\\ A{B^2} = BH.BC\\ A{C^2} = CH.BC\\ \left. \begin{array}{l} \Rightarrow \frac{{A{B^2}}}{{A{C^2}}} = \frac{{BH}}{{CH}} \Rightarrow \frac{{A{B^4}}}{{A{C^4}}} = \frac{{B{H^2}}}{{C{H^2}}}\\ \left. \begin{array}{l} B{H^2} = AB.BE\\ C{H^2} = AC.CF \end{array} \right\} \Rightarrow \frac{{B{H^2}}}{{C{H^2}}} = \frac{{AB}}{{AC}}.\frac{{BE}}{{CF}} \end{array} \right\}\frac{{A{B^4}}}{{A{C^4}}} = \frac{{AB}}{{AC}}.\frac{{BE}}{{CF}} \Rightarrow \frac{{A{B^3}}}{{A{C^3}}} = \frac{{BE}}{{CF}} \end{array}\)