Question

[Phùng Phương Linh]

Tìm x biết:

a, x(x+3)-2x-6=0

b, 4x²-1+x(2x-1)=0

Làm hộ mk nhé.Cảm ơn các cậu nhìu 😊

Answer 1

[Nguyễn Lê Phước Thịnh]

a) Ta có: \(x\left(x+3\right)-2x-6=0\)

\(\Leftrightarrow x\left(x+3\right)-\left(2x+6\right)=0\)

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

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

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

Vậy: \(x\in\left\{-3;2\right\}\)

b) Ta có: \(4x^2-1+x\left(2x-1\right)=0\)

\(\Rightarrow\left(2x\right)^2-1^2+x\left(2x-1\right)=0\)

\(\Rightarrow\left[\left(2x\right)^2-1^2\right]+x\left(2x-1\right)=0\)

\(\Rightarrow\left(2x-1\right)\left(2x+1\right)+x\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1+x\right)=0\)

\(\Rightarrow\left(2x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{1}{2};\frac{-1}{3}\right\}\)

Answer 2

[trần thị thảo mai]

a, x(x+3)-2x-6=0

⇔x(x+3)-2(x+3)=0

⇔(x+3)(x-2)=0

⇔x+3=0 hoặc x-2=0

⇔x= -3 hoặc x=2

b, 4x²-1+x(2x-1)=0

⇔(2x-1)(2x+1)+x(2x-1)=0

⇔(2x-1)(2x+1+x)=0

⇔(2x-1)(3x+1)=0

⇔2x-1=0 hoặc 3x+1=0

⇔x=1/2 hoặc x= -1/3