Tìm x
\(\left(x-2\right)^3=-27\)
\(\left|3,5-x\right|+\dfrac{2}{7}=\dfrac{16}{7}\)
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a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
1:
a: =7/5(40+1/4-25-1/4)-1/2021
=21-1/2021=42440/2021
b: =5/9*9-1*16/25=5-16/25=109/25
|3,5-x|\(+\dfrac{2}{7}=\dfrac{16}{7}\)
\(\Leftrightarrow\left|3,5-x\right|=\dfrac{16}{7}-\dfrac{2}{7}\)
\(\Leftrightarrow\left\{{}\begin{matrix}3,5-x=2\\3,5-x=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}1,5\\x=5,5\end{matrix}\right.\)
\(|3,5-x|=\dfrac{16}{7}-\dfrac{2}{7}\)
\(|3,5-x|=2\)
\(\Rightarrow|\left[{}\begin{matrix}3,5-x=2\\3,5-x=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3,5-2\\x=3,5-\left(-2\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,5\\x=5,5\end{matrix}\right.\)
\(\Rightarrow x\in\left\{1,5;5,5\right\}\)
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
=>3x-9-10x+2=-4
=>-7x-7=-4
=>-7x=3
=>x=-3/7
b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)
=>10-2x+7x-14=4x-4+x
=>5x-4=5x-4
=>0x=0(luôn đúng)
Vậy: S=R\{0;2}
a, \(\left(x-2\right)^3=-27\)
\(\Rightarrow\left(x-2\right)^3=-3^3\)
\(\Rightarrow x-2=-3\)
\(\Rightarrow x=-3+2=-1\)
b, \(\left|3,5-x\right|+\dfrac{2}{7}=\dfrac{16}{7}\)
\(\Rightarrow\left|3,5-x\right|=\dfrac{16}{7}-\dfrac{2}{7}=\dfrac{14}{7}=2\)
\(\Rightarrow\left[{}\begin{matrix}3,5-x=2\\3,5-x=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,5\\x=5,5\end{matrix}\right.\)
\(\left(x-2\right)^3=-27\)
\(\left(x-2\right)^3=-3^3\)
=> \(x-2=-3\)
\(x=-3+\left(-2\right)\)
\(x=-5\)
Vậy x=5
\(\left|3,5-x\right|+\dfrac{2}{7}=\dfrac{16}{7}\)
\(\left|3,5-x\right|=\dfrac{16}{7}-\dfrac{2}{7}\)
\(\left|3,5-x\right|=\dfrac{14}{7}\)
\(\left|3,5-x\right|=2\)
=> \(3,5-x=2hoặc3,5-x=-2\)
\(x=3,5-2\) hoặc \(x=3,5-\left(-2\right)\)
\(x=1,5\) hoặc \(x=3,5+2\)
\(x=1,5\) hoặc \(x=5,5\)
Vậy x=1,5 hoặc x=5,5