a: \(\dfrac{x}{10.8}=\dfrac{3}{12}=\dfrac{1}{4}\)

nên x=2.7

c: \(\dfrac{x}{0.3}=\dfrac{7.5}{x}\)

nên \(x^2=2.25\)

=>x=1,5 hoặc x=-1,5

d: \(\dfrac{x-2}{40.8}=\dfrac{21}{28}=\dfrac{3}{4}\)

nên x-2=30,6

=>x=32,6

a: \(\left|-x+\frac25\right|+\frac12=3,5\)

=>\(\left|x-\frac25\right|=3,5-0,5=3\)

=>\(\left[\begin{array}{l}x-\frac25=3\\ x-\frac25=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3+\frac25=\frac{17}{5}\\ x=-3+\frac25=-\frac{13}{5}\end{array}\right.\)

b: \(\frac{21}{5}+3:\left|\frac{x}{4}-\frac23\right|=6\)

=>\(3:\left|\frac14x-\frac23\right|=6-\frac{21}{5}=\frac95\)

=>\(\left|\frac14x-\frac23\right|=3:\frac95=3\cdot\frac59=\frac53\)

=>\(\left[\begin{array}{l}\frac14x-\frac23=\frac53\\ \frac14x-\frac23=-\frac53\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac14x=\frac23+\frac53=\frac73\\ \frac14x=-\frac53+\frac23=-\frac33=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac73\cdot4=\frac{28}{3}\\ x=-1\cdot4=-4\end{array}\right.\)

c: 7,5-3|5-2x|=-4,5

=>3|5-2x|=7,5+4,5=12

=>3|2x-5|=12

=>|2x-5|=4

=>\(\left[\begin{array}{l}2x-5=4\\ 2x-5=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=9\\ 2x=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac92\\ x=\frac12\end{array}\right.\)

d: \(\frac13-\left|\frac54-2x\right|=\frac14\)

=>\(\left|\frac54-2x\right|=\frac13-\frac14=\frac{1}{12}\)

=>\(\left|2x-\frac54\right|=\frac{1}{12}\)

=>\(\left[\begin{array}{l}2x-\frac54=\frac{1}{12}\\ 2x-\frac54=-\frac{1}{12}\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac54+\frac{1}{12}=\frac{16}{12}=\frac43\\ 2x=\frac54-\frac{1}{12}=\frac{15}{12}-\frac{1}{12}=\frac{14}{12}=\frac76\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac23\\ x=\frac{7}{12}\end{array}\right.\)

e: \(\frac{21}{5}+3:\left|\frac{x}{4}-\frac23\right|=6\)

=>\(3:\left|\frac14x-\frac23\right|=6-\frac{21}{5}=\frac95\)

=>\(\left|\frac14x-\frac23\right|=3:\frac95=3\cdot\frac59=\frac53\)

=>\(\left[\begin{array}{l}\frac14x-\frac23=\frac53\\ \frac14x-\frac23=-\frac53\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac14x=\frac23+\frac53=\frac73\\ \frac14x=-\frac53+\frac23=-\frac33=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac73\cdot4=\frac{28}{3}\\ x=-1\cdot4=-4\end{array}\right.\)

a: Ta có: \(\left|\dfrac{2}{5}-x\right|+\dfrac{1}{2}=3.5\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=3\\x-\dfrac{2}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{13}{5}\end{matrix}\right.\)

b: Ta có: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)

\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)

\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)

em muốn hỏi là tại sao 3,5 bên trên xuống dưới lại là 3 và -x +2/5 của em xuống dưới lại chuyển thành x-2/5 ạ mong anh giải đáp

\(48x:\frac{21}{4}=7.5:\frac{25}{8}\)
\(48x:\frac{21}{4}=\frac{12}{5}\)

\(48x=\frac{12}{5}\cdot\frac{21}{4}\)

\(48x=\frac{63}{5}\)

\(x=\frac{63}{5}:48\)

\(x=\frac{21}{80}\)

\(P=\dfrac{\left(1+2+3+...+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63\cdot1,2-21\cdot3,6\right)}{1-2+3-4+5-6+...+99-100}\)

đề là vậy nhé mn

để ý chút thấy liền ah : 63.1,2-21.3,6=63.1,2-21.3.1,2= 63.1,2- 63.1,2=0

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Ta có P = \(\dfrac{\left(1+2+3+...+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+5-...+99-100}\)= \(\dfrac{\left(1+2+3+...+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)0}{1-2+3-4+5-...+99-100}\)= \(\dfrac{0}{1-2+3-4+5-6+...+99-100}=0\)