Bài 1: a) /2x – 5/=4 b)1/3 – /5/4 – 2x/=1/4 c)3/4 – /2x + 1/=7/8 d)/x – 1,5/= 2 e)/x + 1/ /y – 3/=0 23/07/2021 Bởi Arya Bài 1: a) /2x – 5/=4 b)1/3 – /5/4 – 2x/=1/4 c)3/4 – /2x + 1/=7/8 d)/x – 1,5/= 2 e)/x + 1/ /y – 3/=0
a, $/2x – 5/=4 $ \(⇔\left[ \begin{array}{l}2x – 5=4 \\2x-5=-4\end{array} \right.\) \(⇔\left[ \begin{array}{l}2x=9\\2x=1\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=\dfrac{9}{2} \\x=\dfrac{1}{2} \end{array} \right.\) Vậy $x=\dfrac{9}{2}$ $hoặc$ $x=\dfrac{1}{2}$ b, $\dfrac{1}{3}-$ $/\dfrac{5}{4}-2x/=$ $\dfrac{1}{4}$ $⇔/\dfrac{5}{4}-2x/=$ $\dfrac{1}{12}$ \(⇔\left[ \begin{array}{l}\dfrac{5}{4}-2x=\dfrac{1}{12} \\\dfrac{5}{4} -2x=\dfrac{-1}{12} \end{array} \right.\) \(⇔\left[ \begin{array}{l} x=\dfrac{7}{12} \\x=\dfrac{2}{3}\end{array} \right.\) Vậy $x=\dfrac{7}{12}$ hoặc $x=\dfrac{2}{3}$ c, $\dfrac{3}{4}-/2x+1/=$ $\dfrac{7}{8}$ $⇔/2x+1/=\dfrac{-1}{8}$ Vì $/2x+1/$ $\geq0$ ⇒ Phương trình vô nghiệm. d, $/x – 1,5/= 2$ \(⇔\left[ \begin{array}{l}x= \dfrac{7}{2} \\x=\dfrac{-1}{2} \end{array} \right.\) Vậy $x=\dfrac{7}{2}$ hoặc $x=\dfrac{-1}{2}$ e, $⇒/x+1/=0$ và $y-3=0$ $⇒x=-1$ và $y=3$ Bình luận
a, $/2x – 5/=4 $
\(⇔\left[ \begin{array}{l}2x – 5=4 \\2x-5=-4\end{array} \right.\)
\(⇔\left[ \begin{array}{l}2x=9\\2x=1\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=\dfrac{9}{2} \\x=\dfrac{1}{2} \end{array} \right.\)
Vậy $x=\dfrac{9}{2}$ $hoặc$ $x=\dfrac{1}{2}$
b, $\dfrac{1}{3}-$ $/\dfrac{5}{4}-2x/=$ $\dfrac{1}{4}$
$⇔/\dfrac{5}{4}-2x/=$ $\dfrac{1}{12}$
\(⇔\left[ \begin{array}{l}\dfrac{5}{4}-2x=\dfrac{1}{12} \\\dfrac{5}{4} -2x=\dfrac{-1}{12} \end{array} \right.\)
\(⇔\left[ \begin{array}{l} x=\dfrac{7}{12} \\x=\dfrac{2}{3}\end{array} \right.\)
Vậy $x=\dfrac{7}{12}$ hoặc $x=\dfrac{2}{3}$
c, $\dfrac{3}{4}-/2x+1/=$ $\dfrac{7}{8}$
$⇔/2x+1/=\dfrac{-1}{8}$
Vì $/2x+1/$ $\geq0$
⇒ Phương trình vô nghiệm.
d, $/x – 1,5/= 2$
\(⇔\left[ \begin{array}{l}x= \dfrac{7}{2} \\x=\dfrac{-1}{2} \end{array} \right.\)
Vậy $x=\dfrac{7}{2}$ hoặc $x=\dfrac{-1}{2}$
e, $⇒/x+1/=0$ và $y-3=0$
$⇒x=-1$ và $y=3$