Toán Lớp 7: Tìm x biết
A) |x +¼|- ¾ =5% b) 2-|3/2 x- ¼|=|-5/4|
C) 3/2+4/5|x-¾|=7/4 d) 4,5 – ¾|½ x + 5/3| = 5/6
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 7: Tìm x biết
A) |x +¼|- ¾ =5% b) 2-|3/2 x- ¼|=|-5/4|
C) 3/2+4/5|x-¾|=7/4 d) 4,5 – ¾|½ x + 5/3| = 5/6
Comments ( 1 )
a)\left| {x + \dfrac{1}{4}} \right| – \dfrac{3}{4} = 5\% \\
\Leftrightarrow \left| {x + \dfrac{1}{4}} \right| = \dfrac{3}{4} + \dfrac{5}{{100}}\\
\Leftrightarrow \left| {x + \dfrac{1}{4}} \right| = \dfrac{3}{4} + \dfrac{1}{{20}} = \dfrac{4}{5}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{1}{4} = \dfrac{4}{5}\\
x + \dfrac{1}{4} = – \dfrac{4}{5}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{4}{5} – \dfrac{1}{4} = \dfrac{9}{{20}}\\
x = – \dfrac{4}{5} – \dfrac{1}{4} = \dfrac{{ – 21}}{{20}}
\end{array} \right.\\
Vậy\,x = \dfrac{9}{{20}};x = – \dfrac{{21}}{{20}}\\
b)2 – \left| {\dfrac{3}{2}x – \dfrac{1}{4}} \right| = \left| {\dfrac{{ – 5}}{4}} \right|\\
\Leftrightarrow \left| {\dfrac{3}{2}x – \dfrac{1}{4}} \right| = 2 – \dfrac{5}{4}\\
\Leftrightarrow \left| {\dfrac{3}{2}x – \dfrac{1}{4}} \right| = \dfrac{3}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{3}{2}x – \dfrac{1}{4} = \dfrac{3}{4}\\
\dfrac{3}{2}x – \dfrac{1}{4} = – \dfrac{3}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{3}{2}x = \dfrac{3}{4} + \dfrac{1}{4} = 1\\
\dfrac{3}{2}x = – \dfrac{3}{4} + \dfrac{1}{4} = – \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{2}{3}\\
x = – \dfrac{1}{2}.\dfrac{2}{3} = \dfrac{{ – 1}}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{2}{3};x = – \dfrac{1}{3}\\
c)\dfrac{3}{2} + \dfrac{4}{5}\left| {x – \dfrac{3}{4}} \right| = \dfrac{7}{4}\\
\Leftrightarrow \dfrac{4}{5}\left| {x – \dfrac{3}{4}} \right| = \dfrac{7}{4} – \dfrac{3}{2}\\
\Leftrightarrow \dfrac{4}{5}\left| {x – \dfrac{3}{4}} \right| = \dfrac{1}{4}\\
\Leftrightarrow \left| {x – \dfrac{3}{4}} \right| = \dfrac{1}{4}.\dfrac{5}{4}\\
\Leftrightarrow \left| {x – \dfrac{3}{4}} \right| = \dfrac{5}{{16}}\\
\Leftrightarrow \left[ \begin{array}{l}
x – \dfrac{3}{4} = \dfrac{5}{{16}}\\
x – \dfrac{3}{4} = – \dfrac{5}{{16}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{{16}} + \dfrac{3}{4} = \dfrac{{17}}{{16}}\\
x = \dfrac{3}{4} – \dfrac{5}{{16}} = \dfrac{7}{{16}}
\end{array} \right.\\
Vậy\,x = \dfrac{{17}}{{16}};x = \dfrac{7}{{16}}\\
d)4,5 – \dfrac{3}{4}\left| {\dfrac{1}{2}x + \dfrac{5}{3}} \right| = \dfrac{5}{6}\\
\Leftrightarrow \dfrac{3}{4}\left| {\dfrac{1}{2}x + \dfrac{5}{3}} \right| = 4,5 – \dfrac{5}{6}\\
\Leftrightarrow \dfrac{3}{4}\left| {\dfrac{1}{2}x + \dfrac{5}{3}} \right| = \dfrac{9}{2} – \dfrac{5}{6} = \dfrac{{11}}{3}\\
\Leftrightarrow \left| {\dfrac{1}{2}x + \dfrac{5}{3}} \right| = \dfrac{{11}}{3}.\dfrac{4}{3}\\
\Leftrightarrow \left| {\dfrac{1}{2}x + \dfrac{5}{3}} \right| = \dfrac{{44}}{9}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{2}x + \dfrac{5}{3} = \dfrac{{44}}{9}\\
\dfrac{1}{2}x + \dfrac{5}{3} = – \dfrac{{44}}{9}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{2}x = \dfrac{{44}}{9} – \dfrac{5}{3} = \dfrac{{29}}{9}\\
\dfrac{1}{2}x = – \dfrac{{44}}{9} – \dfrac{5}{3} = \dfrac{{ – 59}}{9}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ – 58}}{9}\\
x = \dfrac{{ – 118}}{9}
\end{array} \right.\\
Vậy\,x = – \dfrac{{58}}{9};x = – \dfrac{{118}}{9}
\end{array}$