Toán Lớp 8: d) (4x + 3)(x – 2) = 2(x ^ 2 – 4)
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Toán Lớp 8: d) (4x + 3)(x – 2) = 2(x ^ 2 – 4)
Comments ( 2 )
d) (4x + 3)(x – 2) = 2(x ^ 2 – 4)
⇔(4x + 3)(x – 2)=2(x – 2)(x+2)
⇔(4x + 3)(x – 2) – 2(x – 2)(x+2)=0
⇔(x – 2)(4x+3-2x-4)=0
⇔(x-2)(2x-1)=0
⇔x-2=0 ⇒x=2 hoặc 2x-1=0⇒2x=1⇒x=$\frac{1}{2}$
Giải đáp:
(4x + 3)(x – 2) = 2(x^2 – 4)
=> 4x^2−8x+3(x−2)=2(x^2−4)
=> 4x^2−5x−6=2(x^2−4)
=> 4x^2−5x−6−2x^2+8=0
=> 2x^2−5x+2=0
=> x=$\frac{-(-5)± \sqrt{(-5)^2-4.2.2}}{2.2}$
=> x= \(\left[ \begin{array}{l}\frac{5+3}{4}\\\frac{5-3}{4}\end{array} \right.\)
=> x={2;1/2}
Vậy x={2;1/2}