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bạn đã kiểm tra kĩ chưa vậy?mình đọc đề câu B mà loạn não luôn á;-;
a: =>|x+3/4|=2+1/5=11/5
=>x+3/4=11/5 hoặc x+3/4=-11/5
=>x=29/20 hoặc x=-59/20
b: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
c: =>|2x-1/3|=1/6
=>2x-1/3=1/6 hoặc 2x-1/3=-1/6
=>2x=1/2 hoặc 2x=1/6
=>x=1/4 hoặc x=1/12
e: =>x+2/3=0 hoặc -2x-3/5=0
=>x=-2/3 hoặc x=-3/10
1) -3-(-x) =5
=> -3 + x = 5
=> x = 5 - (-3)
=> x = 8
2)-9-(x-3)=12
=> x- 3 = (-9) - 12
=> x-3 =-21
=> x = (-21)+3
=> x = -18
3)-10-(x-5)=-5
x-5 = (-10) -(-5)
x-5 = -5
x = -5 +5
x = 0
4)210-(10-x) =110
10-x = 210 -110
10-x = 100
x = 10 -100
x = -90
5)4x-(2x-5)=21
=> 4x-2x + 5 =21
=> x(4-2) + 5 = 21
=> x.2 =21 - 5
=> x.2 = 16
=> x = 16:2
=> x = 8
6)14x-5=8x+10
7)15+5x =3x+30
8)2x-5=15-3x
9)2(3x+5)+3(x+1)=x+220
10)4.(x-2+5(x-3)=18
a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
c; \(\dfrac{7}{9}\) : (2 + \(\dfrac{3}{4}\)\(x\)) + \(\dfrac{5}{9}\) = \(\dfrac{23}{27}\)
\(\dfrac{7}{9}\): (2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{23}{27}\) - \(\dfrac{5}{9}\)
\(\dfrac{7}{9}\):(2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{8}{27}\)
2 + \(\dfrac{3}{4}\)\(x\) = \(\dfrac{7}{9}\) : \(\dfrac{8}{27}\)
2 + \(\dfrac{3}{4}\)\(x\) = \(\dfrac{21}{8}\)
\(\dfrac{3}{4}x\) = \(\dfrac{21}{8}\) - 2
\(\dfrac{3}{4}\)\(x\) = \(\dfrac{5}{8}\)
\(x\) = \(\dfrac{5}{8}\) : \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{5}{6}\)
Vậy \(x=\dfrac{5}{6}\)
d; - \(\dfrac{2}{3}\)\(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{10}\)
- \(\dfrac{2}{3}\)\(x\) = \(\dfrac{3}{10}\) - \(\dfrac{1}{5}\)
- \(\dfrac{2}{3}\)\(x\) = \(\dfrac{1}{10}\)
\(x\) = \(\dfrac{1}{10}\) ; (- \(\dfrac{2}{3}\))
\(x\) = - \(\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\)
f; |\(x\)| - \(\dfrac{3}{4}\) = \(\dfrac{5}{3}\)
|\(x\)| = \(\dfrac{5}{3}\) + \(\dfrac{3}{4}\)
|\(x\)| = \(\dfrac{29}{12}\)
\(\left[{}\begin{matrix}x=-\dfrac{29}{12}\\x=\dfrac{29}{12}\end{matrix}\right.\)
Vậy \(x\in\left(-\dfrac{29}{12};\dfrac{29}{12}\right)\)
f; |2\(x\) - \(\dfrac{1}{3}\)| + \(\dfrac{5}{6}\) = 1
|2\(x\) - \(\dfrac{1}{3}\)| = 1 - \(\dfrac{5}{6}\)
|2\(x\) - \(\dfrac{1}{3}\)| = \(\dfrac{1}{6}\)
\(\left[{}\begin{matrix}2x-\dfrac{1}{3}=-\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{1}{6}\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-\dfrac{1}{6}+\dfrac{1}{3}\\2x=\dfrac{1}{6}+\dfrac{1}{3}\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=\dfrac{1}{6}\\2x=\dfrac{1}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{1}{6}:2\\x=\dfrac{1}{2}:2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{1}{12}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\) \(\in\) {\(\dfrac{1}{12}\) ; \(\dfrac{1}{4}\)}