Tìm x ∈ z biết:
a) -(12-5+9-x)=3-|5-8+2|
b) 13-2.|x-1|= -3
c) (2x-1)3=27
d) -5.(x-2)+4.(x-3)=1
e) (x-1).(x+5)<0
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22 tháng 6 2016
\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
= \(\frac{1}{4}+\frac{1}{2}\)
= \(\frac{3}{4}\)
b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)
=\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)
= \(-\frac{35}{27}+\frac{47}{21}\)
= \(\frac{178}{189}\)
c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)
= \(\frac{117}{13}-\frac{311}{65}\)
= \(\frac{274}{65}\)
d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)
= \(\frac{1}{3}+\frac{5}{2}\)
= \(\frac{17}{6}\)
KV
12 tháng 8 2023
a) 3/8 . x = 9/8 - 1
3/8 . x = 1/8
x = 1/8 : 3/8
x = 1/3
b) 4/5 . x = 7/5 - 1/5
4/5 . x = 6/5
x = 6/5 : 4/5
x = 3/2
c) 12/7 : x + 2/3 = 7/5
12/7 : x = 7/5 - 2/3
12/7 : x = 11/15
x = 12/7 : 11/15
x = 180/77
d) 3.(x + 7) - 15 = 27
3.(x + 7) = 27 + 15
3.(x + 7) = 42
x + 7 = 42 : 3
x + 7 = 14
x = 14 - 7
x = 7
T
12 tháng 8 2023
a) \(\dfrac{3}{8}x=\dfrac{9}{8}-1\)
\(\Rightarrow\dfrac{3}{8}x=\dfrac{1}{8}\)
\(\Rightarrow x=\dfrac{1}{8}:\dfrac{3}{8}=\dfrac{1}{3}\)
b) \(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{1}{5}\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{6}{5}\)
\(\Rightarrow x=\dfrac{6}{5}:\dfrac{4}{5}=\dfrac{3}{2}\)
c) \(\dfrac{12}{7}:x+\dfrac{2}{3}=\dfrac{7}{5}\)
\(\Rightarrow\dfrac{12}{7}:x=\dfrac{7}{5}-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{12}{7}:x=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{12}{7}:\dfrac{11}{15}=\dfrac{180}{77}\)
d) \(3\left(x+7\right)-15=27\)
\(\Leftrightarrow3\left(x+7\right)=42\)
\(\Leftrightarrow x+7=14\Leftrightarrow x=7\)
3 tháng 6 2021
Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
a) \(-\left(12-5+9-x\right)=3-\left|5-8+2\right|\)
\(\Leftrightarrow-12+5-9-x=3-1\)
\(\Leftrightarrow-16-x=2\)
\(\Leftrightarrow x=-16-2\)
\(\Leftrightarrow x=-18\)
b) \(13-2\left|x-1\right|=-3\)
\(\Leftrightarrow2\left|x-1\right|=13+3\)
\(\Leftrightarrow2\left|x-1\right|=16\)
\(\Leftrightarrow\left|x-1\right|=8\)
\(\Rightarrow\left[{}\begin{matrix}x-1=8\Rightarrow x=9\\x-1=-8\Rightarrow x=-7\end{matrix}\right.\)
c) \(\left(2x-1\right)^3=27\)
\(\Leftrightarrow\left(2x-1\right)^3=3^3\)
\(\Rightarrow2x-1=3\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=4:2=2\)
d) \(-5.\left(x-2\right)+4\left(x-3\right)=1\)
\(\Leftrightarrow-5x+10+4x-12=1\)
\(\Leftrightarrow-x-2=1\)
\(\Leftrightarrow-x=1+2=3\)
\(\Leftrightarrow x=-3\)
e) \(\left(x-1\right)\left(x+5\right)< 0\)
Do \(\left(x-1\right)\left(x+5\right)< 0\) nên \(x-1\) và \(x+5\) phải trái dấu.
Mà \(x-1< x+5\)
\(\Rightarrow\left\{{}\begin{matrix}x-1< 0\\x+5>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x< 1\\x>-5\end{matrix}\right.\)
\(\Rightarrow-5< x< 1\)
Vậy \(x\in\left\{-4;-3;-2;-1;0\right\}\)