a: (5x-4)(4x+6)=0
=>\(\left[{}\begin{matrix}5x-4=0\\4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
c: \(x^2\left(2x+1\right)+4x+2=0\)
=>\(x^2\left(2x+1\right)+2\left(2x+1\right)=0\)
=>\(\left(2x+1\right)\left(x^2+2\right)=0\)
mà \(x^2+2>=2>0\forall x\)
nên 2x+1=0
=>2x=-1
=>\(x=-\dfrac{1}{2}\)
e: \(\left(2x+5\right)^2=16\)
=>\(\left(2x+5\right)^2-16=0\)
=>(2x+5+4)(2x+5-4)=0
=>(2x+9)(2x+1)=0
=>\(\left[{}\begin{matrix}2x+9=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
`a) (5x - 4)(4x + 6) = 0`
`<=> 5x - 4 = 0` hoặc `4x + 6 = 0`
`<=> 5x = 4` hoặc `4x = -6`
`<=> x = 4/5` hoặc `x = -3/2`
Vậy ...
``
`c) x^2 . (2x + 1) + 4x + 2 = 0`
`<=> x^2 . (2x + 1) + 2(2x + 1) = 0`
`<=> (x^2 + 2) . (2x + 1)= 0`
Do `x^2 + 2 > 0` nên `2x + 1 = 0`
`<=> x = -1/2`
Vậy ...
``
`e) (2x + 5)^2 = 16 `
`<=> (2x + 5)^2 = 4^2`
`<=> 2x + 5 = 4` hoặc `2x + 5 = -4`
`<=> 2x = -1` hoặc `2x = -9`
`<=> x = -1/2` hoặc `x = -9/2`
Vậy ...
