Giải thích các bước giải:

a/. 4x² - 4x + 1 = 0   

⇔ (2x)² - 2.2x. 1 + 1² = 0   

⇔ (2x - 1)² = 0

⇒ 2x - 1 = 0

⇒  2x = 1

⇒ x = `1/2`

Vậy S = { `1/2`}

b/. 9x² + 6x + 1 = 0

⇔ x² + `6/9`x + `1/9` = 0

⇔ x² + `2/3`x + `1/(3²)` = 0

⇔ x² + 2. x . `1/3` + (`1/3`)² = 0

⇔  (x + `1/3`)² = 0

⇒ x + `1/3` = 0

⇒ x = -`1/3`

Vậy S = { -`1/3`}

c/. 25x² - 10x = 48   

⇔ 25x² - 10x - 48  = 0

⇔ 25x² - 40x + 30x - 48 = 0

⇔ 5x(5x - 8) + 6(5x - 8) = 0

⇔ (5x - 8) (5x + 6) = 0

⇒ 5x - 8 = 0 hay 5x + 6 = 0

⇒ x = `8/5` hay 5x = - 6

⇒ x = `8/5` hay x = - `6/5`

Vậy S = { `8/5`; - `6/5`}   

d/. `1/4`x² + x = 120   

⇔ `1/4`x² + x - 120 = 0  

⇔ [(`1/2`x)² + 2.`1/2`x + 1²[ - 121 = 0

⇔ (`1/2`x + 1)² - 11² = 0

⇔ (`1/2`x + 1 - 11)(`1/2`x + 1 + 11) = 0

⇔ (`1/2`x - 10)(`1/2`x + 12) = 0

⇒ (`1/2`x - 10) = 0 hay (`1/2`x + 12) = 0

⇒ `1/2`x - 10 = 0 hay `1/2`x + 12 = 0

⇒ `1/2`x = 10 hay `1/2`x = - 12

⇒  x = 10 : `1/2` hay x = -12 : `1/2`

⇒ x = 20 hay x = - 24

Vậy S = { 20; -24}

e/. `4/9`x² - `4/3`x = 143

⇔ `4/9`x² - `4/3`x - 143 = 0

⇔ (`2/3`x)² - 2.`2/3`x. 1 + 1² - 1 - 143  = 0

⇔ (`2/3`x)² - 2.`2/3`x. 1 + 1² - 144  = 0

⇔ (`2/3`x - 1)² - 12² = 0

⇔ (`2/3`x - 1 - 12)(`2/3`x - 1 +12) = 0

⇔ (`2/3`x - 13)(`2/3`x +11) = 0

⇒ `2/3`x - 13 = 0 hay `2/3`x +11= 0

⇒ `2/3`x = 13  hay `2/3`x = - 11 

⇒ x = 13 : `2/3` hay x = - 11 : `2/3`

⇒ x = `(39)/2` hay x = - `(33)/2`

Vậy S = { `(39)/2`; - `(33)/2`}

Đáp án+Giải thích các bước giải:

Bài `7:`

`a) 4x^2-4x+1=0`

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

`<=> 2x-1=0`

`<=>2x=1`

`<=> x=1/2`

  Vậy `x=1/2`

`b) 9x^2+6x+1=0`

`<=> (3x+1)^2=0`

`<=> 3x+1=0`

`<=> 3x=-1`

`<=> x = -1/3`

  Vậy `x=-1/3`

`c) 25x^2-10x=48`

`<=> 25x^2-10x+1=49`

`<=> (5x-1)^2=49`

`<=> (5x-1)^2=7^2`

`<=>`\(\left[ \begin{array}{l}5x-1=7\\5x-1=-7\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}5x=8\\5x=-6\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{8}{5}\\x=-\dfrac{6}{5}\end{array} \right.\) 

  Vậy `x∈{8/5; -6/5}`

`d) 1/4x^2+x=120`

`<=> (1/2x)^2+2. 1/2x . 1 +1 =120+1`

`<=> (1/2x+1)^2=121`

`<=> (1/2x+1)=11^2`

`<=>`\(\left[ \begin{array}{l}\dfrac{1}{2}x+1=11\\\dfrac{1}{2}x+1=-11\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}\dfrac{1}{2}x=10\\\dfrac{1}{2}x=-12\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=20\\x=-24\end{array} \right.\) 

  Vậy `x∈{20; -24}`

`e) 4/9x^2-4/3x = 143`

`<=> (2/3x)^2 - 2. 2/3x . 1+1= 143+1`

`<=> (2/3x - 1)^2 = 144`

`<=> (2/3x - 1)^2 =12^2`

`<=>`\(\left[ \begin{array}{l}\dfrac{2}{3}x-1=12\\\dfrac{2}{3}x-1=-12\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}\dfrac{2}{3}x=13\\\dfrac{2}{3}x=-11\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{39}{2}\\x=\dfrac{-33}{2}\end{array} \right.\) 

    Vậy `x∈{(39)/2; (-33)/2}`