Đáp án:

`a)` `x ∈ { 20 ; - 20 }`

`b)` `x ∈ { 3 ; 6 }`

`c)` `x ∈ { - 2 ; 1/5 }`

`d)` `x = 2`

Giải thích các bước giải:

Bài `2 :`

`a)` `25 . ( x^2 - 1 ) - 75 = 9900`

`=> 25 . ( x^2 - 1 ) = 9900 + 75`

`=> 25 . ( x^2 - 1 ) = 9975`

`=> x^2 - 1 = 9975 : 25`

`=> x^2 - 1 = 399`

`=> x^2 = 399 + 1`

`=> x^2 = 400`

`=> x^2 ∈ { 20^2 ; ( - 20 )^2 }`

`=> x ∈ { 20 ; - 20 }`

Vậy `x ∈ { 20 ; - 20 }`

`b) ( x - 6 )( 2x - 6 ) = 0`

`=> 2 . ( x - 3 )( x - 6 ) = 0`

`=>` \(\left[ \begin{array}{l}x-3=0\\x-6=0\end{array} \right.\) 

`=>`  \(\left[ \begin{array}{l}x=3\\x=6\end{array} \right.\) 

Vậy `x ∈ { 3 ; 6 }`

`c) ( 5x - 1 )( 6x + 12 ) = 0`

`=> ( 6x + 12 )(5x - 1 ) = 0`

`=> 6 . ( x + 2 )( 5x - 1 ) = 0`

`=>` \(\left[ \begin{array}{l}x+2=0\\5x - 1 =0\end{array} \right.\) 

`=>`  \(\left[ \begin{array}{l}x= - 2\\5x = 1\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}x= - 2\\x = \frac{1}{5}\end{array} \right.\) 

Vậy `x ∈ { - 2 ; 1/5 }`

`d)` `( x^2 + 1 )( x - 2 ) = 0 `

Trường hợp `1 :`

`x^2 + 1 = 0`

`=> x^2 = - 1` `(` Vô lý vì `x^2 ≥ 0 ; - 1 < 0 )`

Trường hợp `2 :`

`x - 2 = 0`

`=> x = 2`

Vậy `x = 2`

`@`TSA

Lời giải `:`

`a)` `25×(x² -1)-75=9900`

`<=>25×(x² -1)=9900+75`

`<=>25×(x² -1)=9975`

`<=>x² -1=9975:25`

`<=>x^2-1=399`

`<=>x^2=399+1`

`<=>x^2=400`

`<=>x=+-20`

Vậy `x in {20;-20}`

`b)` `(x-6).(2x-6)=0`

$⇔\left[ \begin{array}{l}x-6=0\\2x-6=0\end{array} \right.$

$⇔\left[ \begin{array}{l}x=6\\2x=6\end{array} \right.$

$⇔\left[ \begin{array}{l}x=6\\x=3\end{array} \right.$

Vậy `x in {6;3}`

`c)` `( 5x - 1 )( 6x + 12 ) = 0`

`=> 6 . ( x + 2 )( 5x - 1 ) = 0`

`=>` $\left[ \begin{array}{l}x+2=0\\5x - 1 =0\end{array} \right.$

`=>` $\left[ \begin{array}{l}x= - 2\\5x = 1\end{array} \right.$

`=>` $\left[ \begin{array}{l}x= - 2\\x = \frac{1}{5}\end{array} \right.$

Vậy `x ∈ { - 2 ; 1/5 }`

`d)` `( x^2 + 1 )( x - 2 ) = 0`

`TH1:`

`x^2 + 1 = 0`

`=> x^2 = - 1` `(` loại vì `x^2 ≥ 0 ; - 1 < 0 )`

`TH2:`

`x - 2 = 0`

`<=> x=2`

Vậy `x=2`