Giải phương trình e) (x + 1)^2 = 4(x^2 – 2x + 1)^2 f) (x^2 – 9)^2 – 9(x – 3)^2 = 0

Đáp án:

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`e) (x+1)^2 = 4(x^2 - 2x + 1)^2`

`<=> (x+1)^2 - 4(x^2 - 2x + 1)^2 = 0`

`<=> (x+1)^2 -2^2(x^2-2x+1)^2 = 0 `

`<=> (x+1)^2- [2(x-1)]^2 = 0 `

`<=> (x+1)^2 - (2x-2)^2 = 0`

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

`<=> (-x+3)(3x-1) = 0`

`<=>` $\left[\begin{matrix} -x+3=0\\ 3x-1=0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = 3\\ x =\dfrac{1}{3} \end{matrix}\right.$

Vậy `S = { 1/3 ; 1 }`

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`f) (x^2 - 9)^2 - 9(x-3)^2 = 0`

`<=> (x^2 - 9)^2 - 3^2(x-3)^2 = 0`

`<=> (x^2 - 9)^2 - [3(x-3)]^2 = 0`

`<=> (x^2 - 9)^2 - (3x-9)^2 = 0`

`<=> (x^2 - 9 - 3x + 9)(x^2 -9+3x-9) = 0 `

`<=> (x^2-3x)(x^2+3x-18) = 0`

`<=>x(x-3)(x^2 + 3x - 18) = 0`

`<=>x(x-3)(x-3)(x+6) = 0`

`<=> x(x-3)(x+6) = 0`

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

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

Vậy `S = {-6;0;3}`

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