Toán Lớp 7: Tìm x ∈ Q :
c) l x – 1,5 l + l 2,5 – x l = 0
d, l x – $\frac{4}{5}$ l = $\frac{3}{4}$
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Toán Lớp 7: Tìm x ∈ Q :
c) l x – 1,5 l + l 2,5 – x l = 0
d, l x – $\frac{4}{5}$ l = $\frac{3}{4}$
Giúp em với ạ! Hứa 5 sao và CTLHN !
Comments ( 1 )
c)\left| {x – 1,5} \right| + \left| {2,5 – x} \right| = 0\\
Do:\left\{ \begin{array}{l}
\left| {x – 1,5} \right| \ge 0\\
\left| {2,5 – x} \right| \ge 0
\end{array} \right.\\
\Leftrightarrow \left| {x – 1,5} \right| + \left| {2,5 – x} \right| \ge 0\\
Khi:\left| {x – 1,5} \right| + \left| {2,5 – x} \right| = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
x – 1,5 = 0\\
2,5 – x = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1,5\\
x = 2,5
\end{array} \right.\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
d)\left| {x – \dfrac{4}{5}} \right| = \dfrac{3}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
x – \dfrac{4}{5} = \dfrac{3}{4}\\
x – \dfrac{4}{5} = – \dfrac{3}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{4} + \dfrac{4}{5} = \dfrac{{31}}{{20}}\\
x = \dfrac{4}{5} – \dfrac{3}{4} = \dfrac{1}{{20}}
\end{array} \right.\\
Vậy\,x = \dfrac{{31}}{{20}};x = \dfrac{1}{{20}}
\end{array}$