`3.Tìm x ,biết:`

`a)x.(x-120)+x-120=0`

`⇔x.(x-120)+(x-120)=0`

`⇔(x-120)+(x+1)=0`

\(⇔\left[ \begin{array}{l}x-120=0\\x+1=0\end{array} \right.\)

\(⇔\left[ \begin{array}{l}x=120\\x=-1\end{array} \right.\)

Vậy `x∈{120 ;-1}`

`b)4x.(3x-1)-3x+1=0`

`⇔4x.(3x-1)-(3x-1)=0`

`⇔(3x-1).(4x-1)=0`

\(⇔\left[ \begin{array}{l}3x-1=0\\4x-1=0\end{array} \right.\)

\(⇔\left[ \begin{array}{l}x=\frac{1}{3}\\x=\frac{1}{4}\end{array} \right.\)

Vậy `x∈ {1/3 ;1/4}`

`c)(5x-3)^2-(2x-1)^2=0`

`⇔(5x-3-2x+1).(5x-3+2x-1)=0`

`⇔(3x-2).(7x-4)=0`

\(⇔\left[ \begin{array}{l}3x-2=0\\7x-4=0\end{array} \right.\)

\(⇔\left[ \begin{array}{l}x=\frac{2}{3}\\x=\frac{4}{7}\end{array} \right.\)

Vậy `x∈{2/3 ;4/7}`

`d)x^2.(x+2)-18-9x=0`

`⇔x^2.(x+2)-(18+9x)=0`

`⇔x^2.(x+2)-9.(2+x)=0`

`⇔x^2.(x+2)-9.(x+2)=0`

`⇔(x+2).(x^2-9)=0`

`⇔(x+2).(x-3).(x+3)=0`

\(⇔\left[ \begin{array}{l}x+2=0\\x+3=0\\x-3=0\end{array} \right.\)

\(⇔\left[ \begin{array}{l}x=-2\\x=-3\\x=3\end{array} \right.\)

Vậy `x∈{ -2; 3;-3}`

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Đáp án:

Giải thích các bước giải:

 `3,`

`a, x(x - 120) + x - 120 = 0`

`⇔ x(x - 120) + (x - 120) = 0`

`⇔ (x - 120)(x + 1) = 0`

`⇒` $\left[\begin{matrix} x - 120 = 0\\ x + 1 = 0\end{matrix}\right.$ `⇒` $\left[\begin{matrix} x = 120\\ x = -1\end{matrix}\right.$

Vậy `x ∈ {-1; 120}`

`b, x(3x - 1) - 3x + 1 = 0`

`⇔ x(3x - 1) - (3x - 1) = 0`

`⇔ (3x - 1)(x - 1) = 0`

`⇒` $\left[\begin{matrix} 3x - 1= 0\\ x - 1 = 0\end{matrix}\right.$ `⇒` $\left[\begin{matrix} 3x = 1\\ x = 1 \end{matrix}\right.$ `⇒` $\left[\begin{matrix} x = \dfrac{1}{3}\\ x = 1\end{matrix}\right.$

Vậy `x ∈ {1/3 ; 1}`

`c, (5x - 3)^2 - (2x - 1)^2 = 0`

`⇔ [(5x - 3) + (2x - 1)][(5x - 3) - (2x - 1)] = 0`

`⇔ (5x - 3 + 2x - 1)(5x - 3 - 2x + 1) = 0`

`⇔ (7x - 4)(3x - 2) = 0`

`⇒` $\left[\begin{matrix} 7x-4= 0\\ 3x-2 = 0\end{matrix}\right.$ `⇒` $\left[\begin{matrix} 7x = 4\\ 3x = 2 \end{matrix}\right.$ `⇒` $\left[\begin{matrix} x = \dfrac{4}{7}\\ x = \dfrac{2}{3}\end{matrix}\right.$

Vậy `x = 4/7` hoặc `x = 2/3`

`d, x^2 (x + 2) - 18 -9x = 0`

`⇔ x^2 (x + 2) - 9(2 + x) = 0`

`⇔ (x + 2)(x^2 - 9) = 0`

`⇔ (x + 2)(x - 3)(x + 3) = 0`

`⇒` $\left[\begin{matrix} x+2= 0\\ x - 3 = 0\\ x + 3 = 0\end{matrix}\right.$ `⇒` $\left[\begin{matrix} x = -2\\ x = 3\\x = -3 \end{matrix}\right.$ 

Vậy `x ∈ {-3;-2;3}`