Toán Lớp 8: Tìm x:
a.(x+2)^3-x^2(x+6)=4
b.9x^2-4=(2x-1)(3x+2)
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Toán Lớp 8: Tìm x:
a.(x+2)^3-x^2(x+6)=4
b.9x^2-4=(2x-1)(3x+2)
Comments ( 2 )
Giải đáp:
#Clickbim
a) (x+3)^3 – x^2 (x+6) = 4
<=> x³ + 6x² + 12x + 8 – x³ – 6x² = 4
<=> (x³ – x³) + (6x^2 -6x^2) + 12x + 8 = 4
<=> 12x = 4 -8
<=> 12x = -4
<=> x = (-4) : 12
<=> x = -1/3
Vậy x = -1/3
b) 9x^2 – 4 = (2x-1)(3x+2)
<=> (3x)^2 -2^2 – (2x-1)(3x+2) = 0
<=> (3x-2)(3x+2) – (2x-1)(3x+2) = 0
<=> (3x+2)(3x-2-2x+1) =0
<=> (3x+2)(x-1) =0
<=> $\left[\begin{matrix} 3x+2=1\\ x-1=0\end{matrix}\right.$
<=> $\left[\begin{matrix} x=-2/3\\ x=1\end{matrix}\right.$
Vậy x in {-2/3;1}
Lời giải và giải thích chi tiết:
a)(x+3)3−x2(x+6)=4a)(x+3)3-x2(x+6)=4
⇔x³+6x²+12x+8−x³−6x²=4⇔x³+6x²+12x+8-x³-6x²=4
⇔(x³−x³)+(6×2−6×2)+12x+8=4⇔(x³-x³)+(6×2-6×2)+12x+8=4
⇔12x=4−8⇔12x=4-8
⇔12x=−4⇔12x=-4
⇔x=(−4):12⇔x=(-4):12
⇔x=−13⇔x=-13
Vậy x=−13x=-13
b)9×2−4=(2x−1)(3x+2)b)9×2-4=(2x-1)(3x+2)
⇔(3x)2−22−(2x−1)(3x+2)=0⇔(3x)2-22-(2x-1)(3x+2)=0
⇔(3x−2)(3x+2)−(2x−1)(3x+2)=0⇔(3x-2)(3x+2)-(2x-1)(3x+2)=0
⇔(3x+2)(3x−2−2x+1)=0⇔(3x+2)(3x-2-2x+1)=0
⇔(3x+2)(x−1)=0⇔(3x+2)(x-1)=0
⇔⇔ [3x+2=1x−1=0[3x+2=1x−1=0
⇔⇔ [x=−2/3x=1[x=−2/3x=1
Vậy x∈{−23;1}x∈{-23;1}
Chúc bạn học tốt .