Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
(3x-1)^100=1024^10
(3x-1)^100 = (2^10)^10
(3x-1)^100 = 2^100
3x-1=2
3x=3
x=1
a: 2x-3>5x+10
=>-3x>13
hay x<-13/3
b: \(2x^2-3x>x+7x\)
\(\Leftrightarrow2x^2-10x>0\)
=>2x(x-5)>0
=>x>5 hoặc x<0
c: (x-1)(x+3)<0
=>x+3>0 và x-1<0
=>-3<x<1
a) \(2x-3>5x+10\) \(\Leftrightarrow\) \(2x-5x>10 +3\Leftrightarrow-3x>13\Leftrightarrow x< \dfrac{13}{-3}\) vậy \(x< \dfrac{13}{-3}\)
b) \(2x^2-3x>x+7x\) \(\Leftrightarrow\) \(2x^2-3x-x-7x>0\)
\(\Leftrightarrow\) \(2x^2-11x>0\) \(\Leftrightarrow\) \(x\left(2x-11\right)>0\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x>0\\2x-11>0\end{matrix}\right.\)\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x>0\\x>\dfrac{11}{2}\end{matrix}\right.\)
\(\Rightarrow\) \(x>\dfrac{11}{2}\) vậy \(x>\dfrac{11}{2}\)
c) \(\left(x-1\right)\left(x+3\right)< 0\) \(\Leftrightarrow\) \(x^2+3x-x-3< 0\)
\(\Leftrightarrow\) \(x^2+2x-3>0\) \(\Leftrightarrow\) \(x^2-x+3x-3>0\)
\(\Leftrightarrow\) \(x\left(x-1\right)+3\left(x-1\right)\) \(\Leftrightarrow\) \(\left(x+3\right)\left(x-1\right)\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x-1>0\\x+3>0\end{matrix}\right.\)\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x>1\\x>-3\end{matrix}\right.\) \(\Rightarrow\) \(x>1\) vậy \(x>1\)
\(M=\left(-x-8\right)-\left(-3x+10\right)-\left(x-10\right)\\ =-x-8+3x-10-x+10\\ =\left(-x+3x-x\right)+\left(-8-10+10\right)\\ =x-8\)
\(N=-\left(x-100\right)+\left(-3x+10\right)-\left(-x-100\right)\\ =-x+100+-3x+10+x+100\\ =\left(-x+-3x+x\right)+\left(100+10+100\right)\\ =-3x+210\\ =3\left(-x+70\right)\)
\(Q=100-\left(-4x+1\right)-\left(99+x\right)-\left(x-1\right)\\ =100+4x-1-99-x-x+1\\ =\left(4x-x-x\right)+\left(100-1-99+1\right)\\ =2x+1\)
Bài 2:
b: =>x-1>-4 và x-1<4
=>-3<x<5
c: =>x-2011>2012 hoặc x-2011<-2012
=>x>4023 hoặc x<-1
d: \(\left(3x-1\right)^{2016}+\left(5y-3\right)^{2018}>=0\forall x,y\)
mà \(\left(3x-1\right)^{2016}+\left(5y-3\right)^{2018}< 0\)
nên \(\left(x,y\right)\in\varnothing\)