Question

BT8: Tìm x, biết

\(a,\left(x-2\right)\left(x-3\right)=0\)

\(b,x^2-5x=0\)

\(c,x^2-9=0\)

\(d,4x^2-25=0\)

Answer 1

`@` `\text {Ans}`

`\downarrow`

`a,`

`(x - 2)(x - 3) =0`

`<=>`\(\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)

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

`<=>`\(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy, `S = {2; 3}`

`b,`

`x^2 - 5x = 0`

`<=> x(x - 5) = 0`

`<=>`\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\x=0+5\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

Vậy, `S = {0; 5}`

`c,`

`x^2 - 9 = 0`

`<=> x^2 = 0 + 9`

`<=> x^2 = 9`

`<=> x^2 = (+-3)^2`

`<=> x = +-3`

Vậy, `S = {3; -3}`

`d,`

`4x^2 - 25 = 0`

`<=> 4x^2 = 25`

`<=> x^2 = 25/4`

`<=> x^2 = (+-5/2)^2`

`<=> x = +-5/2`

Vậy,` S = {5/2; -5/2}.`

Answer 2

a: =>x-2=0 hoặc x-3=0

=>x=2 hoặc x=3

b: =>x(x-5)=0

=>x=0 hoặc x=5

c: =>(x-3)(x+3)=0

=>x=3 hoặc x=-3

d: =>(2x-5)(2x+5)=0

=>x=5/2 hoặc x=-5/2