Giải phương trình:
a) (x – 1)² + (x – 2) √(x² + 1) = 0
b) (x – 2)² + (x – 4) √(x² + 4) = 0
c) (2x – 1)² + (x – 4) √(4x² + 1) = 0
d) x² – 2x + 2 + (x – 2) √(x² + 2) = 0
Giải phương trình:
a) (x – 1)² + (x – 2) √(x² + 1) = 0
b) (x – 2)² + (x – 4) √(x² + 4) = 0
c) (2x – 1)² + (x – 4) √(4x² + 1) = 0
d) x² – 2x + 2 + (x – 2) √(x² + 2) = 0
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.
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 $x = \pm \sqrt 3$