`d)`
`(2x+3)^2-(2x+1)(2x-1)=22`
`<=> 4x^2+12x+9-(4x^2-1)=22`
`<=> 4x^2+12x+9-4x^2+1=22`
`<=> 12x+10=22`
`<=> 12x=12`
`<=> x=1`
Vậy `x=1`
`e)`
`(4x+3)(4x-3)-(4x-5)^2=46`
`<=> 16x^2-9-(16x^2-40x+25)=46`
`<=> 16x^2-9-16x^2+40x-25=46`
`<=> 40x-34=46`
`<=> 40x=80`
`<=> x=2`
Vậy `x=2`
`f)`
`25x^2-9=0`
`<=> (5x)^2-3^2=0`
`<=> (5x-3)(5x+3)=0`
`<=>` \(\left[ \begin{array}{l}5x-3=0\\5x+3=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{array} \right.\)
Vậy `x in {3/5;-3/5}`
