giải pt: 1) /3-x/ + x^2 -x(x+4)=0 2)/5x/-3x -2=0 3) /-2x/+x-5x-3=0 29/10/2021 Bởi Sadie giải pt: 1) /3-x/ + x^2 -x(x+4)=0 2)/5x/-3x -2=0 3) /-2x/+x-5x-3=0
Đáp án: 3) \(\left[ \begin{array}{l}x = \frac{3}{2}\\x = – \frac{1}{2}\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}1)\left| {3 – x} \right| = – {x^2} + x\left( {x + 4} \right)\\ \to \left| {3 – x} \right| = 4x\\ \to \left[ \begin{array}{l}3 – x = 4x\\3 – x = – 4x\end{array} \right.\\ \to \left[ \begin{array}{l}5x = 3\\ – 3x = 3\end{array} \right.\\ \to \left[ \begin{array}{l}x = \frac{3}{5}\\x = – 1\end{array} \right.\\2)\left| {5x} \right| = 3x + 2\\ \to \left[ \begin{array}{l}5x = 3x + 2\\5x = – 3x – 2\end{array} \right.\\ \to \left[ \begin{array}{l}2x = 2\\8x = – 2\end{array} \right.\\ \to \left[ \begin{array}{l}x = 1\\x = – \frac{1}{4}\end{array} \right.\\3)\left| { – 2x} \right| – 4x – 3 = 0\\ \to \left| {2x} \right| = 4x + 3\\ \to \left[ \begin{array}{l}2x = 4x + 3\\2x = – 4x – 3\end{array} \right.\\ \to \left[ \begin{array}{l}2x = 3\\6x = – 3\end{array} \right.\\ \to \left[ \begin{array}{l}x = \frac{3}{2}\\x = – \frac{1}{2}\end{array} \right.\end{array}\) Bình luận
Đáp án:
3) \(\left[ \begin{array}{l}
x = \frac{3}{2}\\
x = – \frac{1}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\left| {3 – x} \right| = – {x^2} + x\left( {x + 4} \right)\\
\to \left| {3 – x} \right| = 4x\\
\to \left[ \begin{array}{l}
3 – x = 4x\\
3 – x = – 4x
\end{array} \right.\\
\to \left[ \begin{array}{l}
5x = 3\\
– 3x = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \frac{3}{5}\\
x = – 1
\end{array} \right.\\
2)\left| {5x} \right| = 3x + 2\\
\to \left[ \begin{array}{l}
5x = 3x + 2\\
5x = – 3x – 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 2\\
8x = – 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = – \frac{1}{4}
\end{array} \right.\\
3)\left| { – 2x} \right| – 4x – 3 = 0\\
\to \left| {2x} \right| = 4x + 3\\
\to \left[ \begin{array}{l}
2x = 4x + 3\\
2x = – 4x – 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 3\\
6x = – 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \frac{3}{2}\\
x = – \frac{1}{2}
\end{array} \right.
\end{array}\)