a, 2x+ 1/4 = 3/2
b, (x-5) - 1/3 = 2/5
c, (4,5-2x):3/4=4/3
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b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)
=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)
=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)
=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)
d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)
=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)
=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)
=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)
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
a) Ta có: \(\frac{4}{5}+x=\frac{2}{3}\)
hay \(x=\frac{2}{3}-\frac{4}{5}=\frac{10}{15}-\frac{12}{15}=\frac{-2}{15}\)
Vậy: \(x=\frac{-2}{15}\)
b) Ta có: \(\frac{3}{4}-x=\frac{1}{3}\)
hay \(x=\frac{3}{4}-\frac{1}{3}=\frac{9}{12}-\frac{4}{12}=\frac{5}{12}\)
Vậy: \(x=\frac{5}{12}\)
c) Ta có: \(x-\frac{5}{9}=\frac{-2}{3}\)
hay \(x=\frac{-2}{3}+\frac{5}{9}=\frac{-6}{9}+\frac{5}{9}=-\frac{1}{9}\)
Vậy: \(x=\frac{-1}{9}\)
d) Ta có: \(\frac{\left(4,5-2x\right)}{\frac{3}{4}}=1\frac{1}{3}\)
\(\Leftrightarrow\frac{9}{2}-2x=\frac{4}{3}\cdot\frac{3}{4}=1\)
\(\Leftrightarrow2x=\frac{9}{2}-1=\frac{7}{2}\)
hay \(x=\frac{7}{4}\)
Vậy: \(x=\frac{7}{4}\)
b: \(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
a, 2x+ 1/4 = 3/2
2x = 3/2-1/4
2x = 5/4
x=5/4:2
x= 5/8
b, (x-5) - 1/3 = 2/5
x-5= 2/5+1/3
x-5=11/15
x=11/15+5
x=86/15
c, (4,5-2x):3/4=4/3
(4,5-2x)=4/3. 3/4
4,5-2x=1
2x= 5,5
x=5,5:2
x=2,75
a) 2x+1/4=3/2
=> 2x =3/2-1/4
=> 2x =5/4
=> x =5/4:2
=> x =5/8
b) (x-5)-1/3=2/5
=> x-5 =2/5+1/3
=> x-5 =11/15
=> x =11/15+5
=> x =86/15
c) (4,5-2x):3/4=4/3
=> 4,5-2x =4/3x3/4
=> 4,5-2x =1
=> 2x =4,5-1
=> 2x =3,5
=> x =3,5:2
=> x =1,75