Tìm x:
a) |x+2| – |3x-5| = 0
b) x:3/4 + 1/4 = -2/3
c) (2x + 3/5) ^2 – 9/25 = 0
d) 3.(3x – 1/2) ^3 + 1/9 = 0
Tìm x: a) |x+2| – |3x-5| = 0 b) x:3/4 + 1/4 = -2/3 c) (2x + 3/5) ^2 – 9/25 = 0 d) 3.(3x – 1/2) ^3 + 1/9 = 0
By aikhanh
By aikhanh
Tìm x:
a) |x+2| – |3x-5| = 0
b) x:3/4 + 1/4 = -2/3
c) (2x + 3/5) ^2 – 9/25 = 0
d) 3.(3x – 1/2) ^3 + 1/9 = 0
Đáp án:a)|x+2|=|3x-5|
=》[x+2=3x-5
[x+2=-3x-5
TH1
x+2=3x-5
x-3x=-5-2
-2x=-7
x=3.5
TH2
x+2=-3x-5
x+3x=-5-2
4x=-7
x=-7/4
Giải thích các bước giải:-7/4 là -7 phần 4
Đáp án:
d. \(x = \dfrac{1}{{18}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left| {x + 2} \right| = \left| {3x – 5} \right|\\
\to \left[ \begin{array}{l}
x + 2 = 3x – 5\\
x + 2 = – 3x + 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 7\\
4x = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{7}{2}\\
x = \dfrac{3}{4}
\end{array} \right.\\
b.x:\dfrac{3}{4} + \dfrac{1}{4} = – \dfrac{2}{3}\\
\to \dfrac{{4x}}{3} = – \dfrac{2}{3} – \dfrac{1}{4}\\
\to \dfrac{{4x}}{3} = – \dfrac{{11}}{{12}}\\
\to x = – \dfrac{{11}}{{12}}:\dfrac{4}{3}\\
\to x = – \dfrac{{11}}{{16}}\\
c.{\left( {2x + \dfrac{3}{5}} \right)^2} = \dfrac{9}{{25}}\\
\to \left| {2x + \dfrac{3}{5}} \right| = \dfrac{3}{5}\\
\to \left[ \begin{array}{l}
2x + \dfrac{3}{5} = \dfrac{3}{5}\\
2x + \dfrac{3}{5} = – \dfrac{3}{5}
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 0\\
2x = – \dfrac{6}{5}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = – \dfrac{3}{5}
\end{array} \right.\\
d.3{\left( {3x – \dfrac{1}{2}} \right)^3} = – \dfrac{1}{9}\\
\to {\left( {3x – \dfrac{1}{2}} \right)^3} = – \dfrac{1}{{27}}\\
\to 3x – \dfrac{1}{2} = – \dfrac{1}{3}\\
\to 3x = \dfrac{1}{6}\\
\to x = \dfrac{1}{{18}}
\end{array}\)