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1 tháng 1 2021
a, \(\frac{3}{4}-x=\frac{1}{2}\Leftrightarrow x=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\)Vậy \(x=\frac{1}{4}\)
b, \(\left|x+\frac{2}{3}\right|=\frac{5}{6}\)
TH1 : \(x+\frac{2}{3}=\frac{5}{6}\Leftrightarrow x=\frac{5}{6}-\frac{2}{3}=\frac{1}{6}\)
TH2 : \(x+\frac{2}{3}=-\frac{5}{6}\Leftrightarrow x=-\frac{5}{6}-\frac{2}{3}=\frac{-9}{6}=\frac{-3}{2}\)
Vậy \(x=\left\{\frac{1}{6};-\frac{3}{2}\right\}\)
NC
1 tháng 1 2021
a,\(\frac{3}{4}-x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{3}{4}-\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{4}\)
b,\(\left|x+\frac{2}{3}\right|=\frac{5}{6}\)
\(\Leftrightarrow x+\frac{2}{3}=\pm\frac{5}{6}\)
TH1:\(x+\frac{2}{3}=\frac{5}{6}\)
\(\Leftrightarrow x=\frac{5}{6}-\frac{2}{3}\)
\(\Leftrightarrow x=\frac{1}{6}\)
TH2:\(x+\frac{2}{3}=-\frac{5}{6}\)
\(\Leftrightarrow x=-\frac{5}{6}-\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{3}{2}\)
19 tháng 7 2018
Vì \(\left|2x+1\right|\ge0;\left|x+y-\frac{1}{2}\right|\ge0\)
Mà \(\left|2x+1\right|+\left|x+y-\frac{1}{2}\right|\le0\Rightarrow\orbr{\begin{cases}2x+1=0\\x+y-\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\y=\frac{1}{4}\end{cases}}\)(1)
Thế (1) vào A
\(\Rightarrow A=4.\left(-\frac{1}{2}\right)^3.\left(\frac{1}{4}\right)^2-\frac{1}{4}.\left(-\frac{1}{2}\right)+2.\frac{1}{4}-5\)
\(\Rightarrow A=-\frac{1}{2}+\frac{1}{8}+\frac{1}{2}-5\)
\(\Leftrightarrow A=\frac{1}{8}-5=\frac{1}{8}-\frac{40}{8}=-\frac{39}{8}\)
6 tháng 2 2018
a, => |5/3.x| = 1/6
=> 5/3.x = -1/6 hoặc 5/3.x = 1/6
=> x = -1/10 hoặc x = 1/10
Tk mk nha
N
13 tháng 6 2017
a)\(\left(3-2x\right)\left(x+1\right)\ge0\)
TH1: \(\hept{\begin{cases}3-2x\ge0\\1+x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\x\ge-1\end{cases}}}\)\(\Leftrightarrow-1\le x\le\frac{3}{2}\)
TH2: \(\hept{\begin{cases}3-2x\le0\\1+x\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\x\le-1\end{cases}}}\)( vô lí)
b) \(\left(2x-4\right)\left(x+3\right)\le0\)
\(\Leftrightarrow2\left(x-2\right)\left(x+3\right)\le0\)
TH1:\(\hept{\begin{cases}x-2\le0\\x+3\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le2\\x\ge-3\end{cases}\Leftrightarrow}-3\le x\le2}\)
TH2:\(\hept{\begin{cases}x-2\ge0\\x+3\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge2\\x\le-3\end{cases}}}\)(vô lí)
BM
0
11 tháng 9 2020
\(\frac{2x-3}{x+1\frac{3}{4}}< 0\)
<=> \(\frac{2x-3}{x+\frac{7}{4}}< 0\)
ĐKXĐ : \(x\ne-\frac{7}{4}\)
Xét hai trường hợp :
1. \(\hept{\begin{cases}2x-3>0\\x+\frac{7}{4}< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x>3\\x< -\frac{7}{4}\end{cases}}\Leftrightarrow\hept{\begin{cases}x>\frac{3}{2}\\x< -\frac{7}{4}\end{cases}}\)( loại )
2. \(\hept{\begin{cases}2x-3< 0\\x+\frac{7}{4}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x< 3\\x>-\frac{7}{4}\end{cases}}\Leftrightarrow\hept{\begin{cases}x< \frac{3}{2}\\x>-\frac{7}{4}\end{cases}}\Leftrightarrow-\frac{7}{4}< x< \frac{3}{2}\)
Vậy ...
DD
Đoàn Đức Hà
Giáo viên
29 tháng 7 2021
\(\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2003}\right)\left(-1\frac{1}{2004}\right)\)
\(=-\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2004}{2003}.\frac{2005}{2004}\)
\(=-\frac{3.4.5.....2004.2005}{2.3.4.....2003.2004}=\frac{-2005}{2}\)
3 tháng 9 2020
a)\(\left(\frac{-3}{4}\right)^{10}\cdot x=\left(-\frac{3}{4}\right)^{12}\)
\(x=\left(\frac{-3}{4}\right)^{12}:\left(-\frac{3}{4}\right)^{10}\)
\(x=\left(-\frac{3}{4}\right)^{12-10}\)
\(x=\left(-\frac{3}{4}\right)^2=\frac{9}{16}\)
b)\(x:\left(\frac{2}{3}\right)^8=\left(\frac{9}{4}\right)^4\)
\(x=\left(\frac{9}{4}\right)^4\cdot\left(\frac{2}{3}\right)^8\)
\(x=\left(\frac{9}{4}\right)^4\cdot\left[\left(\frac{2}{3}\right)^2\right]^4\)
\(x=\left(\frac{9}{4}\right)^4\cdot\left(\frac{4}{9}\right)^4\)
\(x=\left(\frac{9}{4}\cdot\frac{4}{9}\right)^4\)
\(x=1^4=1\)
c)\(\left(x-1\right)^3=-64\)
\(\Rightarrow\left(x-1\right)^3=\left(-4\right)^3\)
\(\Rightarrow x-1=-4\)
\(\Rightarrow x=-3\)
d)\(\left(x+1\right)^4=81\)
\(\Rightarrow\left(x+1\right)^4=\left(\pm3\right)^4\)
\(\Rightarrow\orbr{\begin{cases}x+1=3\\x+1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)
TT
3 tháng 9 2020
a)\(\left(\frac{-3}{4}\right)^{10}\cdot x=\left(-\frac{3}{4}\right)^{12}\)
\(\Leftrightarrow x=\left(\frac{-3}{4}\right)^{12}:\left(-\frac{3}{4}\right)^{10}\)
\(\Leftrightarrow x=\left(-\frac{3}{4}\right)^{12-10}\)
\(\Leftrightarrow x=\left(-\frac{3}{4}\right)^2=\frac{9}{16}\)
b)\(x:\left(\frac{2}{3}\right)^8=\left(\frac{9}{4}\right)^4\)
\(\Leftrightarrow x=\left(\frac{9}{4}\right)^4\cdot\left(\frac{2}{3}\right)^8\)
\(\Leftrightarrow x=\left(\frac{9}{4}\right)^4\cdot\left[\left(\frac{2}{3}\right)^2\right]^4\)
\(\Leftrightarrow x=\left(\frac{9}{4}\right)^4\cdot\left(\frac{4}{9}\right)^4\)
\(\Leftrightarrow x=1^4\Leftrightarrow x=1\)
c) \(\left(x-1\right)^3=-64\)
\(\Leftrightarrow\left(x-1\right)^3=\left(-4\right)^3\)
\(\Leftrightarrow x-1=-4\Leftrightarrow x=-3\)
d)\(\left(x+1\right)^4=81\)
\(\Leftrightarrow\left(x+1\right)^4=\left(3\right)^4\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=3\\x+1=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)
\(20.\dfrac{3}{4}+\dfrac{1}{4}:x=-3\\ \Rightarrow15+\dfrac{1}{4}:x=-3\\ \Rightarrow\dfrac{1}{4}:x=-18\\ \Rightarrow x=-\dfrac{1}{72}\)