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222-9+11+12:2*14+14 = ? ( )

## 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

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

1. 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

2. 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}
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