A) |2x-1|=5 B) |7x+2|=-3 C) |x-5|=2x+1 D) |x+2|+7=4x E) |x-2|=|4x+7| 13/08/2021 Bởi aihong A) |2x-1|=5 B) |7x+2|=-3 C) |x-5|=2x+1 D) |x+2|+7=4x E) |x-2|=|4x+7|
Đáp án: d) x=3 Giải thích các bước giải: \(\begin{array}{l}a)\left| {2x – 1} \right| = 5\\ \to \left[ \begin{array}{l}2x – 1 = 5\\2x – 1 = – 5\end{array} \right.\\ \to \left[ \begin{array}{l}x = 3\\x = – 2\end{array} \right.\\b)Do:\left| {7x + 2} \right| \ge 0\forall x\\\left| {7x + 2} \right| = – 3\left( {voly} \right)\\ \to x \in \emptyset \\c)\left| {x – 5} \right| = 2x + 1\\ \to \left[ \begin{array}{l}x – 5 = 2x + 1\left( {DK:x \ge 5} \right)\\x – 5 = – 2x – 1\left( {DK:x < 5} \right)\end{array} \right.\\ \to \left[ \begin{array}{l}x = – 6\left( l \right)\\3x = 4\end{array} \right.\\ \to x = \dfrac{4}{3}\\d)\left| {x + 2} \right| = 4x – 7\\ \to \left[ \begin{array}{l}x + 2 = 4x – 7\left( {DK:x \ge – 2} \right)\\x + 2 = – 4x + 7\left( {DK:x < – 2} \right)\end{array} \right.\\ \to \left[ \begin{array}{l}3x = 9\\5x = 5\end{array} \right.\\ \to \left[ \begin{array}{l}x = 3\\x = 1\left( l \right)\end{array} \right.\\e)\left| {x – 2} \right| = \left| {4x + 7} \right|\\ \to \left[ \begin{array}{l}x – 2 = 4x + 7\\x – 2 = – 4x – 7\end{array} \right.\\ \to \left[ \begin{array}{l}3x = – 9\\5x = – 5\end{array} \right.\\ \to \left[ \begin{array}{l}x = – 3\\x = – 1\end{array} \right.\end{array}\) Bình luận
Câu A bn viết kl giúp mk nha
Còn câu E mk nhác làm
Xin ctlhn và đánh giá 5 sao nha
Đáp án:
d) x=3
Giải thích các bước giải:
\(\begin{array}{l}
a)\left| {2x – 1} \right| = 5\\
\to \left[ \begin{array}{l}
2x – 1 = 5\\
2x – 1 = – 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = – 2
\end{array} \right.\\
b)Do:\left| {7x + 2} \right| \ge 0\forall x\\
\left| {7x + 2} \right| = – 3\left( {voly} \right)\\
\to x \in \emptyset \\
c)\left| {x – 5} \right| = 2x + 1\\
\to \left[ \begin{array}{l}
x – 5 = 2x + 1\left( {DK:x \ge 5} \right)\\
x – 5 = – 2x – 1\left( {DK:x < 5} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – 6\left( l \right)\\
3x = 4
\end{array} \right.\\
\to x = \dfrac{4}{3}\\
d)\left| {x + 2} \right| = 4x – 7\\
\to \left[ \begin{array}{l}
x + 2 = 4x – 7\left( {DK:x \ge – 2} \right)\\
x + 2 = – 4x + 7\left( {DK:x < – 2} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = 9\\
5x = 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = 1\left( l \right)
\end{array} \right.\\
e)\left| {x – 2} \right| = \left| {4x + 7} \right|\\
\to \left[ \begin{array}{l}
x – 2 = 4x + 7\\
x – 2 = – 4x – 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = – 9\\
5x = – 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – 3\\
x = – 1
\end{array} \right.
\end{array}\)