Toán Lớp 8: Tìm x:
a, x(3-2x)+2x^2=12
b, 4x^2-25-(x+7)(2x-5)=0
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Toán Lớp 8: Tìm x:
a, x(3-2x)+2x^2=12
b, 4x^2-25-(x+7)(2x-5)=0
Comments ( 2 )
Giải đáp:
a, x.(3 – 2x) + 2x^2 = 12
↔ 3x – 2x^2 + 2x^2 = 12
↔ 3x + [(-2x^2) + 2x^2] = 12
↔ 3x = 12
↔ x = 4
Vậy x = 4
———-
b, 4x^2 – 25 – (x + 7).(2x – 5) = 0
↔ [(2x)^2 – 5^2] – (x + 7).(2x – 5) = 0
↔ (2x – 5).(2x + 5) – (x + 7).(2x – 5) = 0
↔ (2x – 5).[(2x + 5) – (x + 7)] = 0
↔ (2x – 5).(2x + 5 – x -7) = 0
↔ (2x – 5).(x – 2)= 0
↔ $\left[\begin{matrix} 2x – 5 = 0\\ x – 2 = 0\end{matrix}\right.$
↔ $\left[\begin{matrix} x = \dfrac{5}{2}\\ x = 2\end{matrix}\right.$
Vậy x ∈ {5/2; 2}
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a)
x.(3-2x)+2x^2=12
->x.3-x.2x+2x^2=12
->3x-2x^2+2x^2=12
->3x=12
->x=12:3
->x=4
Vậy x-4
$\\$
b)
4x^2-25-(x+7).(2x-5)=0
->(2x)^2-5^2-(x+7).(2x-5)=0
->(2x-5).(2x+5)-(x+7).(2x-5)=0
->(2x-5).[(2x+5)-(x+7)]=0
->(2x-5).[2x+5-x-7]=0
->(2x-5).(x-2)=0
->\(\left[ \begin{array}{l}2x-5=0\\x-2=0\end{array} \right.\)
->\(\left[ \begin{array}{l}2x=5\\x=2\end{array} \right.\)
->\(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=2\end{array} \right.\)
Vậy x=\frac{5}{2}; x=2