Question

2. Giải phương trình

(x + 2)(2x - 3) = x²- 4

X² + 3x +2= 0

2x² +5x +3 = 0

X³ + x² - 12x = 0

 

 

Answer 1

1.\(\left(x+2\right)\left(2x-3\right)=x^2-4\)

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

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

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

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

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

\(\Leftrightarrow x^2+x+2x+2=0\)

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

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

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

3.\(2x^2+5x+3=0\)

\(\Leftrightarrow2x^2+2x+3x+3=0\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)

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

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

4.\(x^3+x^2-12x=0\)

\(\Leftrightarrow x\left(x^2+x-12\right)=0\)

\(\Leftrightarrow x\left(x+4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=3\end{matrix}\right.\)

Answer 2

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

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

=>(x+2)(x-1)=0

=>x=-2 hoặc x=1

b: =>(x+1)(x+2)=0

=>x=-1 hoặc x=-2

c: =>(2x+3)(x+1)=0

=>x=-1 hoặc x=-3/2

d: =>x(x+4)(x-3)=0

hay \(x\in\left\{0;-4;3\right\}\)