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\(---\begin{gathered} a)\sqrt {1 - 6x + 9{x^2}} = 5 \hfill \\ \Leftrightarrow \sqrt {{{\left( {1 - 3x} \right)}^2}} = 5 \hfill \\ \Leftrightarrow \left| {1 - 3x} \right| = 5 \hfill \\ T{H_1}:1 - 3x \geqslant 0 \Rightarrow x \leqslant \frac{1}{3} \hfill \\ 1 - 3x = 5 \hfill \\ \Leftrightarrow - 3x = 5 - 1 \hfill \\ \Leftrightarrow - 3x = 4 \hfill \\ \Leftrightarrow x = - \frac{4}{3}\left( {TM} \right) \hfill \\ T{H_2}:1 - 3x < 0 \Rightarrow x > \frac{1}{3} \hfill \\ - \left( {1 - 3x} \right) = 5 \hfill \\ \Leftrightarrow - 1 + 3x = 5 \hfill \\ \Leftrightarrow 3x = 5 + 1 \hfill \\ \Leftrightarrow 3x = 6 \hfill \\ \Leftrightarrow x = \frac{6}{3} \hfill \\ \Leftrightarrow x = 2\left( {TM} \right) \hfill \\ b)\sqrt {{x^2} - 4x + 4} = 7 \hfill \\ \Leftrightarrow \sqrt {{{\left( {x - 2} \right)}^2}} = 7 \hfill \\ \Leftrightarrow \left| {x - 2} \right| = 7 \hfill \\ T{H_1}:x - 2 \geqslant 0 \Rightarrow x \geqslant 2 \hfill \\ x - 2 = 7 \hfill \\ \Leftrightarrow x = 7 + 2 \hfill \\ \Leftrightarrow x = 9\left( {TM} \right) \hfill \\ T{H_2}:x - 2 < 0 \Rightarrow x < 2 \hfill \\ - \left( {x - 2} \right) = 7 \hfill \\ \Leftrightarrow - x + 2 = 7 \hfill \\ \Leftrightarrow - x = 7 - 2 \hfill \\ \Leftrightarrow - x = 5 \hfill \\ \Leftrightarrow x = - 5\left( {TM} \right) \hfill \\ c)\sqrt {25 - 10x + {x^2}} = 7 - 2x \hfill \\ \Leftrightarrow \sqrt {{{\left( {5 - x} \right)}^2}} = 7 - 2x \hfill \\ \Leftrightarrow \left| {5 - x} \right| = 7 - 2x \hfill \\ \Leftrightarrow \left| {5 - x} \right| + 2x = 7 \hfill \\ T{H_1}:5 - x \geqslant 0 \Rightarrow x \leqslant 5 \hfill \\ 5 - x + 2x = 7 \hfill \\ \Leftrightarrow 5 + x = 7 \hfill \\ \Leftrightarrow x = 7 - 5 \hfill \\ \Leftrightarrow x = 2\left( {TM} \right) \hfill \\ T{H_2}:5 - x < 0 \Rightarrow x > 5 \hfill \\ - \left( {5 - x} \right) + 2x = 7 \hfill \\ \Leftrightarrow - 5 + x + 2x = 7 \hfill \\ \Leftrightarrow 3x = 7 + 5 \hfill \\ \Leftrightarrow 3x = 12 \hfill \\ \Leftrightarrow x = 4\left( {KTM} \right) \hfill \\ d)\sqrt {{x^2} + 6x + 9} = 3x - 1 \hfill \\ \Leftrightarrow \sqrt {{{\left( {x + 3} \right)}^2}} = 3x - 1 \hfill \\ \Leftrightarrow \left| {x + 3} \right| = 3x - 1 \hfill \\ \Leftrightarrow \left| {x + 3} \right| - 3x = - 1 \hfill \\ T{H_1}:x + 3 \geqslant 0 \Rightarrow x \geqslant - 3 \hfill \\ x + 3 - 3x = - 1 \hfill \\ \Leftrightarrow - 2x = - 1 - 3 \hfill \\ \Leftrightarrow - 2x = - 4 \hfill \\ \Leftrightarrow x = \frac{{ - 4}}{{ - 2}} \hfill \\ \Leftrightarrow x = 2\left( {TM} \right) \hfill \\ T{H_2}:x + 3 < 0 \Rightarrow x < - 3 \hfill \\ - \left( {x + 3} \right) - 3x = - 1 \hfill \\ \Leftrightarrow - x - 3 - 3x = - 1 \hfill \\ \Leftrightarrow - 4x = - 1 + 3 \hfill \\ \Leftrightarrow - 4x = 2 \hfill \\ \Leftrightarrow x = \frac{2}{{ - 4}} \hfill \\ \Leftrightarrow x = - \frac{1}{2}\left( {KTM} \right) \hfill \\ \end{gathered} \)
Những câu này bạn chỉ cần quy vế trước về hđthức xong bỏ căn đi ra giá trị tuyệt đối r tính bình thường thoi ạ