Question

[Trần Khánh Huyền]

ko lm tắt nha mn

Answer 1

[Phương Nora kute]

2) Tìm x:

a) \(x^2+5x+6=0\)

⇒\(x^2+2x+3x+6=0\)

⇒\(\left(x^2+2x\right)+\left(3x+6\right)=0\)

⇒\(\left(x.x+2.x\right)+\left(3.x+3.2\right)=0\)

⇒\(x.\left(x+2\right)+3.\left(x+2\right)=0\)

⇒\(\left(x+2\right).\left(x+3\right)=0\)

⇒\(x+2=0\) \(hoặc\) \(x+3=0\)

\(+\))\(x+2=0\)                    \(+\))\(x+3=0\)

⇔\(x=-2\)                         ⇔\(x=-3\)

\(Vậy\) \(x\in\left\{-3;-2\right\}\)

c)\(x^2+6x+8=0\)

⇒ \(x^2+4x+2x+8=0\)

⇒ \(\left(x^2+4x\right)+\left(2x+8\right)=0\)

⇒ \(\left(x.x+4.x\right)+\left(2.x+2.4\right)=0\)

⇒ \(x.\left(x+4\right)+2.\left(x+4\right)=0\)

⇒ \(\left(x+4\right).\left(x+2\right)=0\)

⇒\(x+4=0\) \(hoặc\) \(x+2=0\)

\(+\)) \(x+4=0\)                   \(+\)) \(x+2=0\)

⇔\(x=-4\)                        ⇔\(x=-2\)

\(Vậy\) \(x\in\left\{-4;-2\right\}\)

Answer 2

[ILoveMath]

a) x²+5x+6=0

⇒x²+2x+3x+6=0

⇒(x²+2x)+(3x+6)=0

⇒x(x+2)+3(x+2)=0

⇒(x+2)(x+3)=0

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)

b) 9x³=x

⇒9x³-x=0

⇒9x(x²-1)=0

⇒\(\left[{}\begin{matrix}9x=0\\x^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

c) x²+6x+8=0

⇒ x²+2x+4x+8=0

⇒ (x²+2x)+(4x+8)=0

⇒ x(x+2)+4(x+2)=0

⇒ (x+2)(x+4)=0

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)

d) x(x-3)-(x-2)²=x+1

⇒x²-3x-x²+4x-4=x+1

⇒x²-3x-x²+4x-4-x-1=0

⇒-5=0(vô lí)

e) (x+2)(x+3)-(x+1)²=0

⇒x²+5x+6-x²-2x-1=0

⇒3x+5=0

⇒3x=-5

⇒x=\(-\dfrac{5}{3}\)

f)x(x+1)-(x+1)²=0

⇒(x-x-1)(x+1)=0

⇒-1(x+1)=0

⇒x+1=0

⇒x=-1

g) (x-2)²-4(x+3)²=0

⇒x²-4x+4-4(x²+6x+9)=0

⇒x²-4x+4-4x²-24x-36=0

⇒-3x²-28x-32=0

đến đây mik chx bt lm