Toán Giải phương trình (x^2+3x+2)(x^2+3x+3)-2=0 16/07/2021 By Rose Giải phương trình (x^2+3x+2)(x^2+3x+3)-2=0
(x²+3x+2)(x²+3x+3)-2=0 Đặt: y=x²+3x+2 ⇒y.(y+1)-2=0 ⇒y²+y-2=0 ⇒y²+2y-y-2=0 ⇒y(y+2)-(y+2)=0 ⇒(y+2)(y-1)=0 ⇒\(\left[ \begin{array}{l}y=2\\y=1\end{array} \right.\) =>\(\left[ \begin{array}{l}x^2+3x+2=2\\x^2+3x+2=1\end{array} \right.\) =>\(\left[ \begin{array}{l}x(x+3)=0\\x^2+3x+1=0\end{array} \right.\) (loại TH2)=>\(\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.\) Vậy S={0;-3} Trả lời
( x²+3x+2).( x²+3x+3)-2= 0 Đặt x²+3x+2= t ⇒ t.( t+1)-2= 0 ⇔ t²+t-2= 0 ⇔ t²-t+2t-2= 0 ⇔ t.( t-1)+2.( t-1)= 0 ⇔ ( t-1).( t+2)= 0 ⇔ t-1= 0⇔ t= 1 hoặc t+2= 0⇔ t= -2 Nếu t= 1 thì: x²+3x+2= 1 ⇔ x²+3x+1= 0 Bấm máy tính được: x= $\frac{-3+\sqrt[]{5}}{2}$ hoặc x= $\frac{-3-\sqrt[]{5}}{2}$ Nếu t= -2 thì x²+3x+2= -2 ⇔ x²+3x+4= 0 ⇔ x²+2.$\frac{3}{2}$.x+$\frac{9}{4}$+$\frac{7}{4}$= 0 ⇔ ( x+$\frac{3}{2}$)²+$\frac{7}{4}$= 0 ( vô lí) Trả lời
(x²+3x+2)(x²+3x+3)-2=0
Đặt: y=x²+3x+2
⇒y.(y+1)-2=0
⇒y²+y-2=0
⇒y²+2y-y-2=0
⇒y(y+2)-(y+2)=0
⇒(y+2)(y-1)=0
⇒\(\left[ \begin{array}{l}y=2\\y=1\end{array} \right.\) =>\(\left[ \begin{array}{l}x^2+3x+2=2\\x^2+3x+2=1\end{array} \right.\) =>\(\left[ \begin{array}{l}x(x+3)=0\\x^2+3x+1=0\end{array} \right.\) (loại TH2)=>\(\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.\)
Vậy S={0;-3}
( x²+3x+2).( x²+3x+3)-2= 0
Đặt x²+3x+2= t
⇒ t.( t+1)-2= 0
⇔ t²+t-2= 0
⇔ t²-t+2t-2= 0
⇔ t.( t-1)+2.( t-1)= 0
⇔ ( t-1).( t+2)= 0
⇔ t-1= 0⇔ t= 1
hoặc t+2= 0⇔ t= -2
Nếu t= 1 thì: x²+3x+2= 1
⇔ x²+3x+1= 0
Bấm máy tính được: x= $\frac{-3+\sqrt[]{5}}{2}$
hoặc x= $\frac{-3-\sqrt[]{5}}{2}$
Nếu t= -2 thì x²+3x+2= -2
⇔ x²+3x+4= 0
⇔ x²+2.$\frac{3}{2}$.x+$\frac{9}{4}$+$\frac{7}{4}$= 0
⇔ ( x+$\frac{3}{2}$)²+$\frac{7}{4}$= 0 ( vô lí)