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\(A=1+2^2+2^3+...+2^{2018}\)
\(2A=2+2^2+...+2^{2019}\)
\(2A-A=\left(2+2^2+...+2^{2019}\right)-\left(1+2^2+2^3+...+2^{2018}\right)\)
\(A=2^{2019}-1\)
\(\Rightarrow A+1=2^{2019}-1+1=2^{2019}\)
\(\Rightarrow A+1\)là một lũy thừa
đpcm
\(\left[{}\begin{matrix}\dfrac{1}{2}+2x=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)
50=1
12.35+35.182-35.66
=35.(12+182-66)
=35.(194-66)
=35.128
=4480
x+2x=6
3x=6
x=6:3
x=2
vậy x=2
x+x=30
2x=30
x=30:2
x=15
vậy x=15
2x-25+5x=45
2x+5x-25=45
2x+5x=45+25
7x=70
x=70:7
x=10
vậy x=10
x-7+2x=8
x+2x-7=8
x+2x=8+7
3x=15
x=15:3
x=5
vậy x=5
(2x+1)(y+2)=4
⇒(2x+1) và (y+2) ∈ Ư (4) = { 1,-1,2,-2,4,-4 }
⇒2x+1=1 ⇒2x=1-1=0 ⇒x=0:2=0
y+2=4 y=4-2=2 y=2
⇒2x+1=-1 ⇒2x=-1-1=-2 ⇒x=-2:2=-1
y+2=-4 y=-4-2=-6 y=-6
⇒2x+1=2 ⇒2x=2-1=1 ⇒x=1:2=0,5
y+2=-2 y=-2-2=-4 y=-4
\(\left(2x-1\right)\left(y-2\right)=4\)
\(\Rightarrow2x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Mà \(2x+1\) lẻ
\(\Rightarrow2x+1=\pm1\)
Xét \(2x+1=1\Rightarrow x=0\)
\(\Rightarrow y-2=4\Rightarrow y=6\)
Xét \(2x+1=-1\Rightarrow x=-1\)
\(\Rightarrow y-2=-4\Rightarrow y=-2\)
\(\dfrac{1}{2}\) \(\times\) ( \(x\) - \(\dfrac{2}{3}\)) - \(\dfrac{1}{3}\) \(\times\) ( 2\(x\) - 3) = \(x\)
\(\dfrac{1}{2}\) \(\times\) \(\dfrac{3x-2}{3}\) - \(\dfrac{2x-3}{3}\) = \(x\)
\(\dfrac{3x-2}{6}\) - \(\dfrac{4x-6}{6}\) = \(\dfrac{6x}{6}\)
3\(x-2-4x\) + 6 = 6\(x\)
-\(x\) + 4 - 6\(x\) = 0
7\(x\) = 4
\(x\) = \(\dfrac{4}{7}\)
=>(2x-7)2=9
=>2x-7 =3 hoặc 2x-7=-3
=>2x=10 hoặc 2x=4
=>x=5 hoặc x=2
Vậy ....................
\(2\cdot\left(2x-7\right)^2=18\)
\(\Rightarrow\left(2x-7\right)^2=9\)
\(\Rightarrow\hept{\begin{cases}2x-7=3\\2x-7=-3\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x=10\\2x=4\end{cases}\Rightarrow}\hept{\begin{cases}x=5\\x=2\end{cases}}\)