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Toru
5 tháng 10 2023 lúc 17:58

1) \(\left|\dfrac{1}{2}x-\dfrac{1}{6}\right|=\dfrac{1}{3}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{6}=\dfrac{1}{3}\\\dfrac{1}{2}x-\dfrac{1}{6}=-\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{1}{2}\\\dfrac{1}{2}x=-\dfrac{1}{6}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

\(---\)

2) \(\left|\dfrac{1}{2}x+\dfrac{3}{5}\right|=\dfrac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+\dfrac{3}{5}=\dfrac{1}{2}\\\dfrac{1}{2}x+\dfrac{3}{5}=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-\dfrac{1}{10}\\\dfrac{1}{2}x=-\dfrac{11}{10}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{11}{5}\end{matrix}\right.\)

\(---\)

3) \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\left|\dfrac{-3}{4}\right|\)

\(\Rightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{4}\\\dfrac{3}{4}x-\dfrac{3}{4}=-\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{3}{2}\\\dfrac{3}{4}x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

\(---\)

4) \(14-\left|\dfrac{3x}{2}-1\right|=9\)

\(\Rightarrow\left|\dfrac{3x}{2}-1\right|=5\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3x}{2}-1=5\\\dfrac{3x}{2}-1=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3x}{2}=6\\\dfrac{3x}{2}=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=12\\3x=-8\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{8}{3}\end{matrix}\right.\)

\(---\)

5) \(17-\left|\dfrac{2}{3}-4x\right|=9\)

\(\Rightarrow\left|\dfrac{2}{3}-4x\right|=8\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}-4x=8\\\dfrac{2}{3}-4x=-8\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x=-\dfrac{22}{3}\\4x=\dfrac{26}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{6}\\x=\dfrac{13}{6}\end{matrix}\right.\)

\(---\)

6) \(5-\left|2x-3\right|=\dfrac{1}{2}\)

\(\Rightarrow\left|2x-3\right|=\dfrac{9}{2}\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{9}{2}\\2x-3=-\dfrac{9}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{15}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

#\(Toru\)

Toru
5 tháng 10 2023 lúc 18:31

11) \(\sqrt{\dfrac{3}{4}x}-\dfrac{1}{2}=\dfrac{3}{7}\left(x\ge0\right)\)

\(\Rightarrow\sqrt{\dfrac{3}{4}x}=\dfrac{13}{14}\)

\(\Rightarrow\dfrac{3}{4}x=\dfrac{169}{196}\)

\(\Rightarrow x=\dfrac{169}{147}\left(tm\right)\)

\(---\)

12) \(\dfrac{2}{3}+\sqrt{\dfrac{1}{3}:x}=\dfrac{3}{5}\left(x>0\right)\)

\(\Rightarrow\sqrt{\dfrac{1}{3}:x}=-\dfrac{1}{15}\)

Mặt khác: \(\sqrt{\dfrac{1}{3}:x}>0\forall x>0\)

\(\Rightarrow\) Không tìm được giá trị nào của \(x\) thoả mãn yêu cầu đề bài.

#\(Toru\)


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