Tìm x a, | 1$\frac{2}{3}$ -2x| <5 b, | 3-5x|>8 c, | $\frac{3}{4}$ -2x|+x <13 d, | -3$\frac{3}{4}$ -3x| -2x>15 27/07/2021 Bởi Raelynn Tìm x a, | 1$\frac{2}{3}$ -2x| <5 b, | 3-5x|>8 c, | $\frac{3}{4}$ -2x|+x <13 d, | -3$\frac{3}{4}$ -3x| -2x>15
Đáp án: d. \(\left[ \begin{array}{l}x < – \dfrac{{15}}{4}\\x > \dfrac{{45}}{4}\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}a.\left| {\dfrac{5}{3} – 2x} \right| < 5\\ \to \left[ \begin{array}{l}\dfrac{5}{3} – 2x < 5\\\dfrac{5}{3} – 2x > – 5\end{array} \right.\\ \to \left[ \begin{array}{l}2x > – \dfrac{{10}}{3}\\2x < \dfrac{{20}}{3}\end{array} \right.\\ \to \left[ \begin{array}{l}x > – \dfrac{5}{3}\\x < \dfrac{{10}}{3}\end{array} \right.\\b.\left| {3 – 5x} \right| > 8\\ \to \left[ \begin{array}{l}3 – 5x > 8\\3 – 5x < – 8\end{array} \right.\\ \to \left[ \begin{array}{l}5x < – 5\\5x > 11\end{array} \right.\\ \to \left[ \begin{array}{l}x < – 1\\x > \dfrac{{11}}{5}\end{array} \right.\\c.\left| {\dfrac{3}{4} – 2x} \right| < 13 – x\\ \to \left[ \begin{array}{l}\dfrac{3}{4} – 2x < 13 – x\\\dfrac{3}{4} – 2x > – 13 + x\end{array} \right.\\ \to \left[ \begin{array}{l}x > – \dfrac{{49}}{4}\\3x < \dfrac{{55}}{4}\end{array} \right.\\ \to \left[ \begin{array}{l}x > – \dfrac{{49}}{4}\\x < \dfrac{{55}}{{12}}\end{array} \right.\\d.\left| { – \dfrac{{15}}{4} – 3x} \right| > 15 + 2x\\ \to \left[ \begin{array}{l} – \dfrac{{15}}{4} – 3x > 15 + 2x\\ – \dfrac{{15}}{4} – 3x < – 15 – 2x\end{array} \right.\\ \to \left[ \begin{array}{l}5x < – \dfrac{{75}}{4}\\x > \dfrac{{45}}{4}\end{array} \right.\\ \to \left[ \begin{array}{l}x < – \dfrac{{15}}{4}\\x > \dfrac{{45}}{4}\end{array} \right.\end{array}\) Bình luận
Đáp án:
d. \(\left[ \begin{array}{l}
x < – \dfrac{{15}}{4}\\
x > \dfrac{{45}}{4}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left| {\dfrac{5}{3} – 2x} \right| < 5\\
\to \left[ \begin{array}{l}
\dfrac{5}{3} – 2x < 5\\
\dfrac{5}{3} – 2x > – 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x > – \dfrac{{10}}{3}\\
2x < \dfrac{{20}}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > – \dfrac{5}{3}\\
x < \dfrac{{10}}{3}
\end{array} \right.\\
b.\left| {3 – 5x} \right| > 8\\
\to \left[ \begin{array}{l}
3 – 5x > 8\\
3 – 5x < – 8
\end{array} \right.\\
\to \left[ \begin{array}{l}
5x < – 5\\
5x > 11
\end{array} \right.\\
\to \left[ \begin{array}{l}
x < – 1\\
x > \dfrac{{11}}{5}
\end{array} \right.\\
c.\left| {\dfrac{3}{4} – 2x} \right| < 13 – x\\
\to \left[ \begin{array}{l}
\dfrac{3}{4} – 2x < 13 – x\\
\dfrac{3}{4} – 2x > – 13 + x
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > – \dfrac{{49}}{4}\\
3x < \dfrac{{55}}{4}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > – \dfrac{{49}}{4}\\
x < \dfrac{{55}}{{12}}
\end{array} \right.\\
d.\left| { – \dfrac{{15}}{4} – 3x} \right| > 15 + 2x\\
\to \left[ \begin{array}{l}
– \dfrac{{15}}{4} – 3x > 15 + 2x\\
– \dfrac{{15}}{4} – 3x < – 15 – 2x
\end{array} \right.\\
\to \left[ \begin{array}{l}
5x < – \dfrac{{75}}{4}\\
x > \dfrac{{45}}{4}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x < – \dfrac{{15}}{4}\\
x > \dfrac{{45}}{4}
\end{array} \right.
\end{array}\)