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\(\left(27x+6\right):3-11=9\)
\(\Leftrightarrow27x+6=9\cdot11+3=102\)
hay \(x=\dfrac{32}{9}\)
Bài 1 d)
\(1-2+3-4+5-6+...+2013\)
\(=1+\left(-2+3\right)+\left(-4+5\right)+...+\left(-2012+2013\right)\)
\(=1+1+1+...+1\left(1006s\right)\)
\(=1006.1=1006\)
a)\(29.\left(-13\right)-27.\left(-29\right)-14.\left(-29\right)\)
\(=13\left(-29\right)-27.\left(-29\right)-14.\left(-29\right)\)
\(=-29.\left(13-27-14\right)\)
\(=\left(-28\right).\left(-29\right)\)
\(=812\)
Tìm x biết:
a,x-5/7=1/9
b,2x/5=6/2x+1
c,11/8+13/6=85/x
d,2x-2/11=1.1/5
e,x/15=3/5+-2/3
f,x/182=-6/14.35/91
a, \(x\) - \(\dfrac{5}{7}\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) + \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{52}{63}\)
b, \(\dfrac{2x}{5}\) = \(\dfrac{6}{2x+1}\)
2\(x\).(2\(x\) + 1) = 30
4\(x^2\)+ 2\(x\) - 30 = 0
4\(x^2\) + 12\(x\) - 10\(x\) - 30 = 0
(4\(x^2\) + 12\(x\)) - (10\(x\) + 30) =0
4\(x\).(\(x\) + 3) - 10.(\(x\) +3) = 0
2 (\(x\) + 3).(2\(x\) - 5) = 0
\(\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-3; \(\dfrac{5}{2}\)}
a) \(\left(2x+3\right)^3=\left(2x+3\right)^8\)
TH1 \(2x+3=1\)
\(2x=1-3=-2\)
\(x=-1\)
TH2 \(2x+3=0\)
\(2x=-3\Rightarrow x=-\frac{3}{2}\)
b) ? sai đề
c) \(\left|5-3\right|=\left|11+2x\right|\Rightarrow\left|2\right|=\left|11+2x\right|\)
\(\hept{\begin{cases}11+2x=-2\\11+2x=2\end{cases}\Rightarrow}\hept{\begin{cases}2x=13\\2x=9\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{13}{2}\\x=\frac{9}{2}\end{cases}}\)
d) \(\left(x-5\right)^4=\left(x-5\right)^6\Rightarrow\hept{\begin{cases}x-5=0\\x-5=1\end{cases}}\Rightarrow\hept{\begin{cases}x=5\\x=6\end{cases}}\)
( 27x + 6 ) : - 11 = 9
\(\left(27x+6\right)=9.\left(-11\right)\)
\(\left(27x+6\right)=-99\)
\(27x=\left(-99\right)-6\)
\(27x=-105\)
\(x=\left(-105\right):27\)
\(x=-\frac{35}{9}\)
\(\left(27x+6\right):\left(-11\right)=9\)
\(27x+6=9.\left(-11\right)=-99\)
\(27x=\left(-99\right)-6=-105\)
\(x=\left(-105\right):27=\frac{-35}{9}\)