\(2x^3+5x^2+3x=0\\ < =>x\left(2x^2+5x+3\right)=0\\ < =>x\left[2x\left(x+1\right)+3\left(x+1\right)\right]=0\\< =>x\left(2x+3\right)\left(x+1\right)=0\\ < =>\left[{}\begin{matrix}x=0\\2x+3=0\\x+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=0\\x=\frac{-3}{2}\\-1\end{matrix}\right.\)
\(\left(x+5\right)\left(x-3\right)+x^2-25=0\\ < =>\left(x+5\right)\left(x+3\right)+\left(x-5\right)\left(x+5\right)=0\\ < =>\left(x+5\right)\left(x-3+x-5\right)=0\\ < =>\left(x+5\right)\left(2x-8\right)=0\\ < =>\left[{}\begin{matrix}x+5=0\\2x-8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=-5\\x=4\end{matrix}\right.\)
\(x\left(x-2\right)-3x+6=0\\ < =>x\left(x-2\right)-3\left(x-2\right)=0\\ < =>\left(x-2\right)\left(x-3\right)=0\\< =>\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
@Mốc
2x3 + 5x2 + 3x = 0
⇔ x.(2x2 + 5x + 3) = 0
⇔ x.(x + 1).(2x + 3) = 0
TH1: x = 0
TH2: x + 1 = 0
⇔ x = - 1
TH3: 2x + 3 = 0
⇔ x = \(\dfrac{-3}{2}\)
Vậy S = {0;- 1;\(\dfrac{-3}{2}\)}
(x + 5).(x - 3) + x2 - 25 = 0
⇔ (x + 5).(x - 3) + (x - 5).(x + 5) = 0
⇔ (x + 5).(x - 3 + x - 5) = 0
⇔ (x + 5).(2x - 8) = 0
TH1: x + 5 = 0
⇔ x = - 5
TH2: 2x - 8 = 0
⇔ x = 4
Vậy S = {- 5; 4}
x.(x - 2) - 3x + 6 = 0
⇔ x.(x - 2) - 3.(x - 2) = 0
⇔ (x - 2).(x - 3) = 0
TH1: x - 2 = 0
⇔ x = 2.
TH2: x - 3 = 0
⇔ x = 3
Vậy S = {2;3}
#chucbanhoctot:)
