Bài 40 trang 60 SBT toán 10 Cánh diều: a) (sqrt { - 4x + 4}  = sqrt { - {x^2} + 1} )...

Giải các phương trình sau:

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

b) $\sqrt {3{x^2} - 6x + 1}  = \sqrt {{x^2} - 3}$

c) $\sqrt {2x - 1}  = 3x - 4$

d) $\sqrt { - 2{x^2} + x + 7}  = x - 3$

+ $\sqrt {f\left( x \right)}  = \sqrt {g\left( x \right)}  \Leftrightarrow \left\{ \begin{array}{l}f\left( x \right) \ge 0\\f\left( x \right) = g\left( x \right)\end{array} \right.$

+ $\sqrt {f\left( x \right)}  = g\left( x \right) \Leftrightarrow \left\{ \begin{array}{l}g\left( x \right) \ge 0\\f\left( x \right) = {\left[ {g\left( x \right)} \right]^2}\end{array} \right.$

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

 $\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l} - 4x + 4 \ge 0\\ - 4x + 4 =  - {x^2} + 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \le 1\\{x^2} - 4x + 3 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \le 1\\\left[ \begin{array}{l}x = 1\\x = 3\;(L)\end{array} \right.\end{array} \right.\quad  \Leftrightarrow x = 1\end{array}$

Vậy $S = \left\{ 1 \right\}$

b) $\sqrt {3{x^2} - 6x + 1}  = \sqrt {{x^2} - 3}$

 $\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 3 \ge 0\\3{x^2} - 6x + 1 = {x^2} - 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 3 \ge 0\\2{x^2} - 6x + 4 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 3 \ge 0\\\left[ \begin{array}{l}x = 1\;(L)\\x = 2\end{array} \right.\end{array} \right.\quad  \Leftrightarrow x = 2\end{array}$

Vậy $S = \left\{ 2 \right\}$

c) $\sqrt {2x - 1}  = 3x - 4$

 $\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}3x - 4 \ge 0\\2x - 1 = {\left( {3x - 4} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge \frac{4}{3}\\2x - 1 = 9{x^2} - 24x + 16\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ge \frac{4}{3}\\9{x^2} - 26x + 17 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge \frac{4}{3}\\\left[ \begin{array}{l}x = 1\;(L)\\x = \frac{{17}}{9}\end{array} \right.\end{array} \right. \Leftrightarrow x = \frac{{17}}{9}\end{array}$

Vậy $S = \left\{ {\frac{{17}}{9}} \right\}$

d) $\sqrt { - 2{x^2} + x + 7}  = x - 3$

$\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}x - 3 \ge 0\\ - 2{x^2} + x + 7 = {\left( {x - 3} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 3\\ - 2{x^2} + x + 7 = {x^2} - 6x + 9\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ge 3\\3{x^2} - 7x + 2 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 3\\\left[ \begin{array}{l}x = 2\;(L)\\x = \frac{1}{3}\;(L)\end{array} \right.\end{array} \right.\end{array}$

Vậy $S = \emptyset$