Đáp án:

Giải thích các bước giải: \(\begin{array}{l} a)\,{\left( {x - 1} \right)^2} + \left( {x - 2} \right)\sqrt {{x^2} + 1} = 0\\ \Leftrightarrow {x^2} - 2x + 1 + \left( {x - 2} \right)\sqrt {{x^2} + 1} = 0\\ \Leftrightarrow \left( {{x^2} + 1} \right) - 2\left( {x - 2} \right) + \left( {x - 2} \right)\sqrt {{x^2} + 1} - 4 = 0\\ \Leftrightarrow \left( {\sqrt {{x^2} + 1} - 2} \right)\left( {\sqrt {{x^2} + 1} + 2} \right) + \left( {x - 2} \right)\left( {\sqrt {{x^2} + 1} - 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sqrt {{x^2} + 1} - 2 = 0\\ \sqrt {{x^2} + 1} + 2 + x - 2 = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \pm \sqrt 3 \\ \sqrt {{x^2} + 1} = - x\left( {x \le 0} \right)\,\left( 1 \right) \end{array} \right.\\ \left( 1 \right) \Leftrightarrow {x^2} + 1 = {x^2}\left( {VN} \right) \end{array}\) Vậy $x = \pm \sqrt 3$

a) Pt tuong duong vs

$$x^2 + 1 -(2x-4) + (x-2)\sqrt{x^2+1} = 4$$

Dat $a = \sqrt{x^2+1}, b = x-2$ ta co

$$a^2 + ba-2b-4=0$$

Giai ptrinh bac 2 an a va coi b la tham so. Khi do $\Delta = b^2 +4(2b+4) = b+4)^2$.

Vay $a = \dfrac{-b + (b+4)}{2} = 2$ hoac $a = \dfrac{-b - (b+4)}{2} = -b-2$.

TH1: $a= 2$ thi $x^2+1=4$ hay x = $\pm \sqrt{3}$.

TH2: $a = -b-2$.

DK: $b \leq -2$

Ta co

$x^2 + 1 = x^2 -4x + 4$ hay $x=3/4$ (ko thoa man)

Vay $x = \pm \sqrt{3}$.

b) Pt tuong duong vs

$$x^2 + 4 -(4x-16) + (x-4)\sqrt{x^2+4} = 16$$

Dat $a = \sqrt{x^2+4}, b = x-4$ ta co

$$a^2 + ba-4b-16=0$$

Giai ptrinh bac 2 an a va coi b la tham so. Khi do $\Delta = b^2 +4(4b+16) = b+8)^2$.

Vay $a = \dfrac{-b + (b+8)}{2} = 4$ hoac $a = \dfrac{-b - (b+8)}{2} = -b-4$.

TH1: $a= 4$ thi $x^2+4=16$ hay x = $\pm 2\sqrt{3}$.

TH2: $a = -b-4$.

DK: $b \leq -4$

Ta co

$x^2 + 4 = x^2 -8x + 16$ hay $x=3/2$ (ko thoa man)

Vay $x = \pm 2\sqrt{3}$.

Cac cau c), d) lam tuong tu nhe.