Câu hỏi của Erza Scarlet

Question

Tìm x, biết:x2+5x+6=02(x+3)-x2-3x=0x2+4x+3=02x2-3x-5=0

qwerty · Accepted Answer

1/ x² - 5x + 6 = 0 ⇔ x² - 2x - 3x + 6 = 0 ⇔ x(x - 2) - 3(x - 2) = 0 ⇔ (x - 2)(x - 3) = 0 ⇒S = {2 ; 3}.Câu trả lời: 1/ x² - 5x + 6 = 0 ⇔ x² - 2x - 3x + 6 = 0 ⇔ x(x - 2) - 3(x - 2) = 0 ⇔ (x - 2)(x - 3) = 0 ⇒S = {2 ; 3}.

Trần Việt Linh · Answer

1) $x^2+5x+6=0$$\Leftrightarrow x^2+2x+3x+6=0$$\Leftrightarrow x\left(x+2\right)+3\left(x+2\right)=0$$\Leftrightarrow\left(x+2\right)\left(x+3\right)=0$$\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\x+3=0\end{array}\right.$$\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\x=-3\end{array}\right.$2) $2\left(x+3\right)-x^2-3x=0$$\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0$$\Leftrightarrow\left(x+3\right)\left(2-x\right)=0$$\Leftrightarrow\left[\begin{array}{nghiempt}x+3=0\2-x=0\end{array}\right.$$\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\x=2\end{array}\right.$3) $x^2+4x+3=0$$\Leftrightarrow x^2+x+3x+3=0$$\Leftrightarrow x\left(x+1\right)+3\left(x+1\right)=0$$\Leftrightarrow\left(x+1\right)\left(x+3\right)=0$$\Leftrightarrow\left[\begin{array}{nghiempt}x+1=0\x+3=0\end{array}\right.$$\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\x=-3\end{array}\right.$4) $2x^2-3x-5=0$$\Leftrightarrow2x^2+2x-5x-5=0$$\Leftrightarrow2x\left(x+1\right)-5\left(x+1\right)=0$$\Leftrightarrow\left(x+1\right)\left(2x-5\right)=0$$\Leftrightarrow\left[\begin{array}{nghiempt}x+1=0\2x-5=0\end{array}\right.$$\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\x=\frac{5}{2}\end{array}\right.$Câu trả lời: 1) $x^2+5x+6=0$$\Leftrightarrow x^2+2x+3x+6=0$$\Leftrightarrow x\left(x+2\right)+3\left(x+2\right)=0$$\Leftrightarrow\left(x+2\right)\left(x+3\right)=0$$\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\x+3=0\end{array}\right.$$\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\x=-3\end{array}\right.$2) $2\left(x+3\right)-x^2-3x=0$$\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0$$\Leftrightarrow\left(x+3\right)\left(2-x\right)=0$$\Leftrightarrow\left[\begin{array}{nghiempt}x+3=0\2-x=0\end{array}\right.$$\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\x=2\end{array}\right.$3) $x^2+4x+3=0$$\Leftrightarrow x^2+x+3x+3=0$$\Leftrightarrow x\left(x+1\right)+3\left(x+1\right)=0$$\Leftrightarrow\left(x+1\right)\left(x+3\right)=0$$\Leftrightarrow\left[\begin{array}{nghiempt}x+1=0\x+3=0\end{array}\right.$$\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\x=-3\end{array}\right.$4) $2x^2-3x-5=0$$\Leftrightarrow2x^2+2x-5x-5=0$$\Leftrightarrow2x\left(x+1\right)-5\left(x+1\right)=0$$\Leftrightarrow\left(x+1\right)\left(2x-5\right)=0$$\Leftrightarrow\left[\begin{array}{nghiempt}x+1=0\2x-5=0\end{array}\right.$$\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\x=\frac{5}{2}\end{array}\right.$

Duong Thi Nhuong · Answer

1. $x^2+5x+6=0$

$\Delta=5^2-4.1.6=1>0\left(PTC2NPB\right)$

$X_1=\frac{-5+\sqrt{1}}{2}=-2$

$X_2=\frac{-5-\sqrt{1}}{2}=-2$

2. $2\left(x+3\right)-x^2-3x=0$

$\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0$

$\Leftrightarrow\left(x+3\right)\left(2-x\right)=0$

$\Leftrightarrow\left[\begin{matrix}x+3=0\2-x=0\end{matrix}\right.\Leftrightarrow}\left[\begin{matrix}x=0-3=-3\x=1-0=2\end{matrix}\right.$3. $x^2+4x+3=0$

$\Delta=4^2-4.1.3=4>0\left(PTC2NPB\right)$

$X_1=\frac{-4+\sqrt{4}}{2}=-1$

$X_1=\frac{-4-\sqrt{4}}{2}=-3$

4. $2x^2-3x-5=0$

$\Delta=\left(-3\right)^2-4.2.\left(-5\right)=49>0\left(PTC2NPB\right)$

$X_1=\frac{3+\sqrt{49}}{4}=2,5$

$X_1=\frac{3-\sqrt{49}}{4}=-2$Câu trả lời: 1. $x^2+5x+6=0$