Đáp án:
Bạn tham khảo nhé:
Giải thích các bước giải:
\(\begin{array}{l}a)\,\,\,y = \sqrt {{x^2} + 4x + 4} + \sqrt {{x^2} - 2x + 1} = \sqrt {{{\left( {x + 2} \right)}^2}} + \sqrt {{{\left( {x - 1} \right)}^2}} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left| {x + 2} \right| + \left| {x - 1} \right|.\end{array}\)
\(\begin{array}{l}TH1:\,\,\,x < - 2 \Rightarrow \left\{ \begin{array}{l}\left| {x + 2} \right| = - x - 2\\\left| {x - 1} \right| = 1 - x\end{array} \right.\\ \Rightarrow y = - x - 2 + 1 - x = - 2x - 1\\TH2:\,\,\, - 2 \le x < 1 \Rightarrow \left\{ \begin{array}{l}\left| {x + 2} \right| = x + 2\\\left| {x - 1} \right| = 1 - x\end{array} \right.\\ \Rightarrow y = x + 2 + 1 - x = 3\\TH3:\,\,x \ge 1 \Rightarrow \left\{ \begin{array}{l}\left| {x + 2} \right| = x + 2\\\left| {x - 1} \right| = x - 1\end{array} \right.\\ \Rightarrow y = x + 2 + x - 1 = 2x + 1.\end{array}\)
Đồ thị hàm số như hình vẽ.
\(\begin{array}{l}b)\,\,\,y = \left| x \right| + \left| {x - 2} \right|\\TH1:\,\,\,x < 0 \Rightarrow \left\{ \begin{array}{l}\left| x \right| = - x\\\left| {x - 2} \right| = 2 - x\end{array} \right.\\ \Rightarrow y = - x + 2 - x = - 2x + 2\\TH2:\,\,0 \le x < 2 \Rightarrow \left\{ \begin{array}{l}\left| x \right| = x\\\left| {x - 2} \right| = 2 - x\end{array} \right.\\ \Rightarrow y = x + 2 - x = 2\\TH3:\,\,\,x \ge 2 \Rightarrow \left\{ \begin{array}{l}\left| x \right| = x\\\left| {x - 2} \right| = x - 2\end{array} \right.\\ \Rightarrow y = x + x - 2 = 2x - 2.\end{array}\)
Đồ thị hàm số như hình vẽ.
