Taco : 3/10 = 0,3

0,2999<x<0,3

x= 0,29991; 0,29992; 0,29993

0,29991:0,2992......:0,29999

a: |2x-3|=|1-x|

=>\(\left[{}\begin{matrix}2x-3=1-x\\2x-3=x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+x=3+1\\2x-x=-1+3\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}3x=4\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

b: \(x^2-4x< =5\)

=>\(x^2-4x-5< =0\)

=>\(x^2-5x+x-5< =0\)

=>\(x\left(x-5\right)+\left(x-5\right)< =0\)

=>\(\left(x-5\right)\left(x+1\right)< =0\)

TH1: \(\left\{{}\begin{matrix}x-5>=0\\x+1< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=5\\x< =-1\end{matrix}\right.\)

=>\(x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}x-5< =0\\x+1>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =5\\x>=-1\end{matrix}\right.\)

=>-1<=x<=5

c: 2x(2x-1)<=2x-1

=>\(\left(2x-1\right)\cdot2x-\left(2x-1\right)< =0\)

=>\(\left(2x-1\right)^2< =0\)

mà \(\left(2x-1\right)^2>=0\forall x\)

nên \(\left(2x-1\right)^2=0\)

=>2x-1=0

=>2x=1

=>\(x=\dfrac{1}{2}\)

6,011;6,012;6,013;6,014

6,011 ; 6,012 ; 6,013 ; 6,014

a)0

b)-5

c)15

a, 0
b, -5
c, 15