Toán Lớp 8: Tìm x biết a,X²-3x=0 b,X²-2x+1=0 c,4x²-4x=-1 d,36²=(5-x)² e,X(x-1)-x+1=0 g,X²+5x+6=0

TRẢ LỜI

1. Giải đáp:

   a, x^2 – 3x = 0

   ↔ x.(x – 3) = 0

   ↔ $\left[\begin{matrix} x = 0\\ x- 3 = 0\end{matrix}\right.$ ↔ $\left[\begin{matrix} x = 0\\ x = 3\end{matrix}\right.$ 

   Vậy x ∈ {0; 3}

   ————–

   b, x^2 – 2x + 1 = 0

   ↔ x^2 – 2.x.1 + 1^2 = 0

   ↔ (x – 1)^2 = 0

   ↔ x – 1 = 0

   ↔ x = 1

   Vậy x = 1

   —————

   c, 4x^2 – 4x = -1

   ↔ 4x^2 – 4x + 1 = 0

   ↔ (2x)^2 – 2.2x.1 + 1^2 =0

   ↔ (2x – 1)^2 = 0

   ↔ 2x – 1 = 0

   ↔ 2x = 1

   ↔ x = 1/2

   Vậy x = 1/2

   —————

   d, 36^2 = (5 – x)^2

   ↔ 36^2 – (5 – x)^2 = 0

   ↔ [36 – (5 – x)].[36 + (5 – x)] = 0

   ↔ (36 – 5 + x).(36 + 5 – x) = 0

   ↔ (31 + x).(41 – x) = 0

   ↔ $\left[\begin{matrix} 31 + x = 0\\ 41 – x = 0\end{matrix}\right.$ ↔ $\left[\begin{matrix} x = – 31\\ x = 41\end{matrix}\right.$

   Vậy x ∈ { -31; 41}

   —————

   e, x.(x – 1) – x + 1 = 0

   ↔ x.(x – 1) – (x – 1) = 0

   ↔ (x – 1).(x – 1) = 0

   ↔ (x – 1)^2 = 0

   ↔ x – 1 = 0

   ↔ x = 1

   Vậy x = 1

   —————–

   g, x^2 + 5x + 6 = 0

   ↔ x^2 +2x + 3x + 6 = 0

   ↔ x.(x + 2) + 3.(x + 2) = 0

   ↔ (x + 2).(x + 3) = 0

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

   Vậy x ∈ {-2; -3}

   #dariana

   Trả lời
2. a) x^2 – 3x = 0

   ⇔ x ( x – 3 ) = 0

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

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

   Vậy x = 0 ; x = 3

   b) x^2 – 2x + 1 = 0

   ⇔ ( x – 1 )^2 = 0

   ⇔ x – 1 = 0

   ⇔ x = 1

   Vậy x = 1

   c) 4x^2 – 4x = – 1

   ⇔ 4x^2 – 4x + 1 = 0

   ⇔ ( 2x – 1 )^2 = 0

   ⇔ 2x – 1 = 0

   ⇔ 2x = 1

   ⇔ x = 1/2

   Vậy x = 1/2

   d) 36^2 = ( 5 – x )^2

   ⇔ ( 5 – x )^2 = 36^2

   ⇔ 5 – x = ± 36

   ⇔ \(\left[ \begin{array}{l}5-x=36\\5-x=-36\end{array} \right.\) 

   ⇔ \(\left[ \begin{array}{l}x=-31\\x=41\end{array} \right.\) 

   Vậy x = – 31 ; x = 41

   e) x ( x – 1 ) – x + 1 = 0

   ⇔ x ( x – 1 ) – ( x – 1 ) = 0

   ⇔ ( x – 1 ) ( x – 1 ) = 0

   ⇔ ( x – 1 )^2 = 0

   ⇔ x – 1 = 0

   ⇔ x = 1

   Vậy x = 1

   g) x^2 + 5x + 6 = 0

   ⇔ x^2 + 3x + 2x + 6 = 0

   ⇔ x ( x + 3 ) + 2 ( x + 3 ) = 0

   ⇔ ( x + 2 ) ( x + 3 ) = 0

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

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

   Vậy x = – 2 ; x = – 3

   Trả lời