Lớp 7: Tìm x: a) | 2x-1| = | 3x-4| b) |x-5| + 2=3x c) |x+2| + |x-3| = 4 06/11/2021 Bởi Quinn Lớp 7: Tìm x: a) | 2x-1| = | 3x-4| b) |x-5| + 2=3x c) |x+2| + |x-3| = 4
Đáp án: Giải thích các bước giải: a)$|2x-1|=|3x-4|$ $(2x-1)^2=(3x-4)^2$\(\left[ \begin{array}{l}2x-1-3x+4=0\\5x-5=0\end{array} \right.\) \(\left[ \begin{array}{l}x=3\\x=1\end{array} \right.\) b)$|x-5|+2=3x$ $|x-5|=3x-2$ \(\left[ \begin{array}{l}x-5=3x-2(x\geq \dfrac{2}{3})\\x-5=-3x+2(x<\dfrac{2}{3})\end{array} \right.\) \(\left[ \begin{array}{l}-2x=3(x\geq \dfrac{2}{3})\\4x=7(x<\dfrac{2}{3})\end{array} \right.\) \(\left[ \begin{array}{l}x=\dfrac{-3}{2}(loại)(x\geq \dfrac{2}{3})\\x=\dfrac{7}{4}(loại)(x<\dfrac{2}{3})\end{array} \right.\) Vậy pt vô nghiệm c) Giải theo bảng Bình luận
Đáp án: a) x=3 Giải thích các bước giải: \(\begin{array}{l}a)\left| {2x – 1} \right| = \left| {3x – 4} \right|\\ \to \left[ \begin{array}{l}2x – 1 = 3x – 4\left( {DK:x \ge \dfrac{1}{2}} \right)\\2x – 1 = – 3x + 4\left( {DK:x < \dfrac{1}{2}} \right)\end{array} \right.\\ \to \left[ \begin{array}{l}x = 3\left( {TM} \right)\\5x = 5\end{array} \right.\\ \to \left[ \begin{array}{l}x = 3\\x = 1\left( l \right)\end{array} \right.\\b)\left| {x – 5} \right| = 3x – 2\\ \to \left[ \begin{array}{l}x – 5 = 3x – 2\left( {DK:x \ge 5} \right)\\x – 5 = – 3x + 2\left( {DK:x < 5} \right)\end{array} \right.\\ \to \left[ \begin{array}{l}2x = – 3\\4x = 7\end{array} \right.\\ \to \left[ \begin{array}{l}x = – \dfrac{3}{2}\left( l \right)\\x = \dfrac{7}{4}\left( {TM} \right)\end{array} \right.\\c)\left| {x + 2} \right| + \left| {x – 3} \right| = 4\\ \to \left[ \begin{array}{l}x + 2 + x – 3 = 4\left( {DK:x \ge 3} \right)\\x + 2 – x + 3 = 4\left( {DK:3 > x \ge – 2} \right)\\ – x – 2 – x + 3 = 4\left( {DK:x < – 2} \right)\end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{5}{2}\left( l \right)\\5 = 4\left( l \right)\\x = – \dfrac{3}{2}\left( {TM} \right)\end{array} \right.\end{array}\) Bình luận
Đáp án:
Giải thích các bước giải:
a)$|2x-1|=|3x-4|$
$(2x-1)^2=(3x-4)^2$
\(\left[ \begin{array}{l}2x-1-3x+4=0\\5x-5=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=3\\x=1\end{array} \right.\)
b)$|x-5|+2=3x$
$|x-5|=3x-2$
\(\left[ \begin{array}{l}x-5=3x-2(x\geq \dfrac{2}{3})\\x-5=-3x+2(x<\dfrac{2}{3})\end{array} \right.\)
\(\left[ \begin{array}{l}-2x=3(x\geq \dfrac{2}{3})\\4x=7(x<\dfrac{2}{3})\end{array} \right.\)
\(\left[ \begin{array}{l}x=\dfrac{-3}{2}(loại)(x\geq \dfrac{2}{3})\\x=\dfrac{7}{4}(loại)(x<\dfrac{2}{3})\end{array} \right.\)
Vậy pt vô nghiệm
c) Giải theo bảng
Đáp án:
a) x=3
Giải thích các bước giải:
\(\begin{array}{l}
a)\left| {2x – 1} \right| = \left| {3x – 4} \right|\\
\to \left[ \begin{array}{l}
2x – 1 = 3x – 4\left( {DK:x \ge \dfrac{1}{2}} \right)\\
2x – 1 = – 3x + 4\left( {DK:x < \dfrac{1}{2}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\left( {TM} \right)\\
5x = 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = 1\left( l \right)
\end{array} \right.\\
b)\left| {x – 5} \right| = 3x – 2\\
\to \left[ \begin{array}{l}
x – 5 = 3x – 2\left( {DK:x \ge 5} \right)\\
x – 5 = – 3x + 2\left( {DK:x < 5} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = – 3\\
4x = 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – \dfrac{3}{2}\left( l \right)\\
x = \dfrac{7}{4}\left( {TM} \right)
\end{array} \right.\\
c)\left| {x + 2} \right| + \left| {x – 3} \right| = 4\\
\to \left[ \begin{array}{l}
x + 2 + x – 3 = 4\left( {DK:x \ge 3} \right)\\
x + 2 – x + 3 = 4\left( {DK:3 > x \ge – 2} \right)\\
– x – 2 – x + 3 = 4\left( {DK:x < – 2} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{5}{2}\left( l \right)\\
5 = 4\left( l \right)\\
x = – \dfrac{3}{2}\left( {TM} \right)
\end{array} \right.
\end{array}\)