`B1`

`a,|2/3-4x|-3/4=0`

`⇔|2/3-4x|=3/4`

`⇔\(\left[ \begin{array}{l}2/3-4x=3/4\\2/3-4x=-3/4\end{array} \right.\) `

`⇔\(\left[ \begin{array}{l}x=-1/48\\x=17/48\end{array} \right.\) `

Vậy `x=-1/48;x=17/48`

`b,(3x+1/2)(4-5x)=0`

`⇔\(\left[ \begin{array}{l}3x+1/2=0\\4-5x=0\end{array} \right.\) `

`⇔\(\left[ \begin{array}{l}x=-1/6\\x=4/5\end{array} \right.\) `

Vậy `x=-1/6;x=4/5`

`c,|3x-4|=7-x`.Vì `7-x≥0`

`⇒3x+4≥0`

`⇔x≥-4/3`

Vậy `x≥-4/3`

`B2`

Có `3|2x-1|≥0 ∀ x⇒3|2x-1|-5≥-5 ∀x`

`⇒A≥-5⇒MinA=-5`

Dấu ''='' xảy ra⇔`|2x-1|=0⇔2x-1=0⇔x=1/2`

Vậy `GTNN` của `A=-5⇔x=1/2`

Đáp án:

B1:

c. \(x = \dfrac{{11}}{4}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\left| {\dfrac{2}{3} - 4x} \right| = \dfrac{3}{4}\\
 \to \left[ \begin{array}{l}
\dfrac{2}{3} - 4x = \dfrac{3}{4}\\
\dfrac{2}{3} - 4x =  - \dfrac{3}{4}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
4x =  - \dfrac{1}{{12}}\\
4x = \dfrac{{17}}{{12}}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - \dfrac{1}{{48}}\\
x = \dfrac{{17}}{{48}}
\end{array} \right.\\
b.\left[ \begin{array}{l}
3x + \dfrac{1}{2} = 0\\
4 - 5x = 0
\end{array} \right. \to \left[ \begin{array}{l}
x =  - \dfrac{1}{6}\\
x = \dfrac{4}{5}
\end{array} \right.\\
c.\left| {3x - 4} \right| = 7 - x\\
 \to 3x - 4 = 7 - x\left( {Do:7 - x \ge 0 \to 7 \ge x} \right)\\
 \to 4x = 11\\
 \to x = \dfrac{{11}}{4}\left( {TM} \right)\\
B2:\\
Do:\left| {2x - 1} \right| \ge 0\forall x \in R\\
 \to 3\left| {2x - 1} \right| \ge 0\\
 \to 3\left| {2x - 1} \right| - 5 \ge  - 5\\
 \to Min =  - 5\\
 \Leftrightarrow 2x - 1 = 0\\
 \Leftrightarrow x = \dfrac{1}{2}
\end{array}\)