1)|2-x|+2|x+1|=4
2)|x+3|-|x+4|=2
3)3|x-1|+|2-x|=5
4)|x+1|+|x-1|=2
5)|3-x|+4-x|=1
6)|10-x|+|9-x|=2
MỌI NGƯỜI GIÚP EM NHA!!! EM CẢM ƠN NHÌU
1)|2-x|+2|x+1|=4
2)|x+3|-|x+4|=2
3)3|x-1|+|2-x|=5
4)|x+1|+|x-1|=2
5)|3-x|+4-x|=1
6)|10-x|+|9-x|=2
MỌI NGƯỜI GIÚP EM NHA!!! EM CẢM ƠN NHÌU
Đáp án:
4) \(\left[ \begin{array}{l}
x = 1\\
x = – 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\left| {2 – x} \right| + 2\left| {x + 1} \right| = 4\\
\to \left[ \begin{array}{l}
2 – x + 2x + 2 = 4\left( {DK:2 \ge x \ge – 1} \right)\\
– 2 + x + 2x + 2 = 4\left( {DK:x > 2} \right)\\
2 – x – 2x – 2 = 4\left( {DK:x < – 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\left( {TM} \right)\\
3x = 4\\
– 3x = 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = \dfrac{4}{3}\left( l \right)\\
x = – \dfrac{4}{3}\left( {TM} \right)
\end{array} \right.\\
2)\left| {x + 2} \right| = 2 + \left| {x + 4} \right|\\
\to \left[ \begin{array}{l}
x + 2 = 2 + x + 4\left( {DK:x \ge – 2} \right)\\
– x – 2 = 2 + x + 4\left( {DK: – 2 > x \ge – 4} \right)\\
– x – 2 = 2 – x – 4\left( {DK:x < – 4} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
0 = 4\left( l \right)\\
2x = – 8\\
– 4 = – 4\left( {ld} \right)
\end{array} \right.\\
\to x = – 4\\
3)3\left| {x – 1} \right| + 2\left| {2 – x} \right| = 5\\
\to \left[ \begin{array}{l}
3x – 3 + 4 – 2x = 5\left( {DK:2 \ge x \ge 1} \right)\\
– 3x + 3 + 4 – 2x = 5\left( {DK:x < 1} \right)\\
3x – 3 – 4 + 2x = 5\left( {DK:x > 2} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
– 5x = – 2\\
5x = 12
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\left( l \right)\\
x = \dfrac{2}{5}\left( {TM} \right)\\
x = \dfrac{{12}}{5}\left( {TM} \right)
\end{array} \right.\\
4)\left| {x + 1} \right| + \left| {x – 1} \right| = 2\\
\to \left[ \begin{array}{l}
x + 1 + x – 1 = 2\left( {DK:x \ge 1} \right)\\
x + 1 – x + 1 = 2\left( {DK:1 > x > – 1} \right)\\
– x – 1 – x + 1 = 2\left( {DK:x \le – 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 2\\
2 = 2\left( {ld} \right)\\
– 2x = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
– 2x = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = – 1
\end{array} \right.\left( {TM} \right)
\end{array}\)