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a) \(|x|+x=\frac{1}{3}\)
\(|x|=\frac{1}{3}-x\)
Ta có: \(|x|\ge0\)
\(\Rightarrow\frac{1}{3}-x\ge0\)
\(\Rightarrow x\ge\frac{1}{3}>0\)
\(\Rightarrow|x|=x\)
\(\Rightarrow x+x=\frac{1}{3}\)
\(\Rightarrow2x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{6}\)
Vậy \(x=\frac{1}{6}\)
-4x (x-5) - 2x (8-2x) = -3
= -4x^2 + 20x - 16x + 4x^2 = -3
= 20x - 16x = -3
= 4x = -3
=> x = -3/4
Bài 1 :
\(8^7-2^{18}\)
\(=\left(2^3\right)^7-2^{18}\)
\(=2^{21}-2^{18}\)
\(=2^{18}\left(2^3-1\right)\)
\(=2^{18}\cdot7\)
\(=2^{17}\cdot2\cdot7\)
\(=2^{17}\cdot14⋮14\left(đpcm\right)\)
\(\left|3x-2018\right|+\left|x-2017\right|=\left|2x-1\right|\)
\(\Rightarrow\orbr{\begin{cases}3x-2018+x-2017=2x-1\\-\left(3x-2018\right)+\left[-\left(x-2017\right)\right]=2x-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x-4035=2x-1\\\left(-3x-x\right)+\left(2018+2017\right)=2x-1\end{cases}}\)
Làm tiếp
TH2:
\(\left|3x-2018\right|+\left|x-2017\right|=\left|2x-1\right|\)
\(\Rightarrow\orbr{\begin{cases}3x-2018+x-2017=-2x+1\\-\left(3x-2018\right)+\left[-\left(x-2017\right)\right]=-2x+1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x-4035=-2x+1\\\left(-3x-x\right)+\left(2018+2017\right)=-2x+1\end{cases}}\)
Tự tiếp tiếp nha bạn
Bài sau cũng tg tự vậy mà làm
TH1 : \(91-3x< 7+x\Rightarrow3x+x>91-7\Rightarrow4x>84\Rightarrow x>21\left(1\right)\)
TH2 : \(7+x\ge64\Rightarrow x\ge57\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow x\ge57\)
91 - 3\(x\) < 7 + \(x\) ≥ 64
⇒ \(\left\{{}\begin{matrix}91-3x< 7+x\\7+x\ge64\end{matrix}\right.\)
\(\left\{{}\begin{matrix}7+x+3x>91\\x\ge64-7\end{matrix}\right.\)
\(\left\{{}\begin{matrix}4x>91-7\\x\ge64-7\end{matrix}\right.\)
\(\left\{{}\begin{matrix}4x>84\\x\ge57\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x>84:4\\x\ge57\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x>21\\x\ge57\end{matrix}\right.\)
\(x\ge\) 57
/x/=2013
=>X\(\hept{\begin{cases}2013\\-2013\end{cases}}\)
(x-2).3=60
=>x-2=60:3=20
=>x=20+2=22
7+x=8-(-7)
=>7+x=15
=>x=15-7=8
/x/=2013=>x=2013
(x-2).3=60=>x-2=20=>x=20-2=18
7+x=8-(-7)=>7+x=15=>x=15-7=8
k mik nhé, hok tốt