a) 3(x+2) - ( x + 2 ) = 0
3(x + 2 ) = 0 + x + 2
3(x+2) = x+2
=> 3x + 6 = x + 2
3x - x = -6 + 2
2x = -4
x = -4 : 2
x = -2
b) (x\(^2\) + 1 ) . (x+2019) = 0
=> \(x^2+1=0\) hoặc x + 2019 = 0
\(x^2\) = -1 x = 0 -2019
vô lí x = -2019
vậy x = -2019
1a) \(3\left(x+2\right)-\left(x+2\right)=0\)
=> \(\left(3-1\right)\left(x+2\right)=0\)
=> \(2\left(x+2\right)=0\)
=> \(x+2=0\)
=> \(x=-2\)
b) \(\left(x^2+1\right)\left(x+2019\right)=0\)
=> \(\left[{}\begin{matrix}x^2+1=0\\x+2019=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x^2=-1\left(l\right)\\x=-2019\end{matrix}\right.\)
Vậy x =-2019
1,a) 3(x+2)-(x+2)=0
3(x+2) =0+x+2
3(x+2) =x+2
3x+6 =x+2
3x-x =(-6)+2
2x =-4
x =(-4):2
x =-2
b) (x2+1).(x+2019)=0
=>\(\left\{{}\begin{matrix}x^2+1=0\\x+2019=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2=-1\\x=-2019\end{matrix}\right.\)
Xét x2=-1(vô lí)
=>x=-2019
2. + TH1 : \(x< -\frac{3}{2}\) ta có :
\(A=-\left(x+\frac{3}{2}\right)+\left(x-1\right)-\left(x-\frac{3}{4}\right)\)
\(=-x-\frac{7}{4}\)
+ TH2 : \(-\frac{3}{2}\le x< \frac{3}{4}\) ta có :
\(A=x+\frac{3}{2}+x-1-\left(x-\frac{3}{4}\right)\)
\(=x+\frac{5}{4}\)
+ TH3 : \(\frac{3}{4}\le x< 1\) ta có :
\(A=x+\frac{3}{2}+x-1+x-\frac{3}{4}\)
\(=3x-\frac{1}{4}\)
+ TH4 : \(x\ge1\) ta có :
\(A=x+\frac{3}{2}-\left(x-1\right)+x-\frac{3}{4}\)
\(A=x+\frac{7}{4}\)
