\[\begin{array}{l} a)\,\,\sqrt {{x^2} - 6x + 9} = x \Leftrightarrow \sqrt {{{\left( {x - 3} \right)}^2}} = x\\ \Leftrightarrow \left| {x - 3} \right| = x \Leftrightarrow \left[ \begin{array}{l} x - 3 = x\\ x - 3 = - x \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} - 3 = 0\,\,\left( {vo\,\,ly} \right)\\ 2x = 3 \end{array} \right. \Leftrightarrow x = \frac{3}{2}.\\ b)\,\,x + \sqrt {4{x^2} - 4x + 1} = 2 \Leftrightarrow \sqrt {{{\left( {2x - 1} \right)}^2}} = 2 - x\\ \Leftrightarrow \left| {2x - 1} \right| = 2 - x \Leftrightarrow \left[ \begin{array}{l} 2x - 1 = 2 - x\\ 2x - 1 = x - 2 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} 3x = 3\\ x = - 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = - 1 \end{array} \right..\\ c)\,\,\,\sqrt {x + 2\sqrt x + 1} - \sqrt {x - 2\sqrt x + 1} = 2\,\,\,\left( {DK:\,\,\,x \ge 0} \right)\\ \Leftrightarrow \sqrt {{{\left( {\sqrt x + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt x + 1} \right)}^2}} = 1\\ \Leftrightarrow \left| {\sqrt x + 1} \right| - \left| {\sqrt x - 1} \right| = 1 \Leftrightarrow \sqrt x + 1 - \left| {\sqrt x - 1} \right| = 1\\ \Leftrightarrow \sqrt x = \left| {\sqrt x - 1} \right| \Leftrightarrow \left[ \begin{array}{l} \sqrt x = \sqrt x - 1\\ \sqrt x = 1 - \sqrt x \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} 0 = - 1\,\,\,\left( {VN} \right)\\ 2\sqrt x = 1 \end{array} \right. \Leftrightarrow \sqrt x = \frac{1}{2} \Leftrightarrow x = \frac{1}{4}\,\,\,\left( {tm} \right).\\ d)\,\,\sqrt {x + 4\sqrt x + 4} + \sqrt {x - 4\sqrt x + 4} = 4\,\,\,\,\left( {DK:\,\,\,x \ge 0} \right)\\ \Leftrightarrow \sqrt {{{\left( {\sqrt x + 2} \right)}^2}} + \sqrt {{{\left( {\sqrt x - 2} \right)}^2}} = 4\\ \Leftrightarrow \left| {\sqrt x + 2} \right| + \left| {\sqrt x - 2} \right| = 4\\ \Leftrightarrow \sqrt x + 2 + \left| {\sqrt x - 2} \right| = 4\\ \Leftrightarrow \left| {\sqrt x - 2} \right| = 4 - \sqrt x - 2\\ \Leftrightarrow \left| {\sqrt x - 2} \right| = 2 - \sqrt x \\ \Leftrightarrow \left[ \begin{array}{l} \sqrt x - 2 = 2 - \sqrt x \\ \sqrt x - 2 = \sqrt x - 2 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 2\sqrt x = 4\\ \,\,\forall x \ge 0 \end{array} \right.\\ \Leftrightarrow \sqrt x = 2 \Leftrightarrow x = 4.\\ Vay\,\,\,pt\,\,\,co\,\,nghiem\,\,voi\,\,moi\,\,x \ge 0. \end{array}\]

Đáp án:

a. x=3/2 b.x= $ \pm $1 c.x=1 d.$0 \le x \le 4$

Giải thích các bước giải:

a. $\sqrt {{x^2} - 6x + 9} = x$ ĐKXĐ: $x \ge 0$ pttt: $\sqrt {{{\left( {x - 3} \right)}^2}} = x$ $\left| {x - 3} \right| = x$ x-3=x hoặc x-3= -x 0=3(vô lí) x=3/2 ( thỏa mãn ĐKXĐ ) Vậy x=3/2 b. $\eqalign{ & x + \sqrt {4{x^2} - 4x + 1} = 2 \cr & \sqrt {{{(2x - 1)}^2}} = 2 - x \cr} $ $\left| {2x - 1} \right| = 2 - x$ ( với $x \le 2$ ) khi 2x-1 $ \ge $0 $\eqalign{ & 2x - 1 = 2 - x \cr & x = 1 \cr} $ khi 2x-1<0 $\eqalign{ & 2x - 1 = x - 2 \cr & x = - 1 \cr} $ (thỏa mãn $x \le 2$) Vậy x= $ \pm $1 c. ĐKXĐ x$ \ge $0 $\eqalign{ & \sqrt {x + 2\sqrt x + 1} - \sqrt {x - 2\sqrt x + 1} = 2 \cr & \sqrt {{{(\sqrt x + 1)}^2}} - \sqrt {{{(\sqrt x - 1)}^2}} = 2 \cr & \left| {\sqrt x + 1} \right| - \left| {\sqrt x - 1} \right| = 2 \cr & \sqrt x + 1 - \left| {\sqrt x - 1} \right| = 2 \cr & khi\sqrt x - 1 \ge 0 \cr & \sqrt x + 1 - \sqrt x - 1 = 2(L) \cr & khi\sqrt x - 1 < 0 \Leftrightarrow 0 \le x < 1 \cr & \sqrt x + 1 - 1 + \sqrt x = 2 \cr & 2\sqrt x = 2 \cr & x = 1 \cr} $ d.ĐKXĐ x$ \ge 0$ $\eqalign{ & \sqrt {x + 4\sqrt x + 4} + \sqrt {x - 4\sqrt x + 4} = 4 \cr & \sqrt {{{(\sqrt x + 2)}^2}} + \sqrt {{{(\sqrt x - 2)}^2}} = 4 \cr & \left| {\sqrt x + 2} \right| + \left| {\sqrt x - 2} \right| = 4 \cr & \sqrt x + 2 + \left| {\sqrt x - 2} \right| = 4 \cr & khi\sqrt x - 2 \ge 0 \Leftrightarrow x \ge 4 \cr & \sqrt x + 2 + \sqrt x - 2 = 4 \cr & 2\sqrt x = 4 \cr & x = 4(thoaman) \cr & khi\sqrt x - 2 < 0 \Leftrightarrow 0 \le x < 4 \cr & \sqrt x + 2 + 2 - \sqrt x = 4 \cr & 4 = 4 \cr} $ (luôn đúng) vậy $0 \le x \le 4$