Question

[Lumina]

Giúp mình câu 6 và 7 với ạ

Answer 1

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6.

a) \(\left|x-\frac{1}{2}\right|\le\frac{1}{3}\)\(\Rightarrow\left|x-\frac{1}{2}\right|\le\left|\frac{1}{3}\right|\)

\(\Rightarrow\frac{-1}{3}\le x-\frac{1}{2}\le\frac{1}{3}\)

\(\Rightarrow\frac{1}{6}\le x\le\frac{5}{6}\)

b) \(\left|2x-\frac{1}{2}\right|>\left|-1,5\right|\)

\(\Rightarrow\orbr{\begin{cases}-1,5< 2x-\frac{1}{2}< 1,5\\2x-\frac{1}{2}< -1,5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\frac{-3}{2}< 2x-\frac{1}{2}< \frac{3}{2}\\2x-\frac{1}{2}< \frac{-3}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\frac{-3}{2}< 2x-\frac{1}{2}< \frac{3}{2}\\2x-\frac{1}{2}< \frac{-3}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-1< 2x< 2\\2x< -1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\frac{-1}{2}< x< 1\\x< \frac{-1}{2}\end{cases}}\)

Answer 2

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7.

a) \(\left|-2x+1,5\right|=\frac{1}{4}\)

\(\Rightarrow\orbr{\begin{cases}-2x+\frac{3}{2}=\frac{1}{4}\\-2x+\frac{3}{2}=\frac{-1}{4}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-2x=\frac{7}{4}\\-2x=\frac{5}{4}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{-7}{8}\\x=\frac{-5}{8}\end{cases}}\)

b) \(\frac{3}{2}-\left|1\frac{1}{4}+3x\right|=\frac{1}{4}\)

\(\Rightarrow\left|\frac{5}{4}+3x\right|=\frac{3}{2}-\frac{1}{4}=\frac{5}{4}\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{4}+3x=\frac{5}{4}\\\frac{5}{4}+3x=\frac{-5}{4}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3x=0\\3x=\frac{-5}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-5}{6}\end{cases}}\)