Toán Lớp 8: Tìm x: a, (x+2)^2-x^2-3x= 21 b, x^2+x-12=0 c, x^3-3x^2+3x-126=0

Question

Toán Lớp 8:  Tìm x: a, (x+2)^2-x^2-3x= 21 b, x^2+x-12=0 c, x^3-3x^2+3x-126=0

haianhtu97 · Answer

a) (x+2)^2-x^2-3x=21       x^2+4x+4-x^2-3x=21       x+4=21       x=17       b)    x^2+x-12=0     x^2+4x-3x-12=0     x(x+4)-3(x+4)=0     (x+4)(x-3)=0       $ left[ begin{matrix} x+4=0   x-3=0 end{matrix} right.$      $ left[ begin{matrix} x=-4   x=3 end{matrix} right.$       c)x^3-3x^2+3x-126=0       x^3-3x^2+3x-1-125=0       (x-1)^3-5^3=0       (x-1-5)[(x-1)^2+5x-5+25]=0       (x-6)(x^2-2x+1+5x-5+25)=0       (x-6)(x^2+3x+21)=0       $ left[ begin{matrix} x-6=0   x^2+3x+21=0(v&ocirc; nghiệm) end{matrix} right.$       x=6

lanhuongthu · Answer

Lời giải :
a, (x+2)^2-x^2-3x= 21
<=>x^2+4x+4-x^2-3x=21
<=>x=21-4
<=>x=17
Vậy x=17
b) x^2+x-12=0
<=>x^2+4x-3x-12=0
<=>(x^2+4X)-(3x+12)=0
<=>x(x+4)-3(x+4)=0
<=>(x-3)(x+4)=0
<=>$\left[ \begin{array}{l}x-3=0\\x+4=0\end{array} \right.$
<=>$\left[ \begin{array}{l}x=3\\x=-4\end{array} \right.$
Vậy x in {3;-4}
c)x^3-3x^2+3x-126=0
<=>(x^2-3x^2+3x-1)-125=0
<=>(x-1)^3-5^3=0
<=>(x-1-5)[(x-1)^2+5(x-1)+5^2]=0
<=>(x-6)(x^2-2x+1+5x-5+25)=0
<=>(x-6)(x^2+3x+21)=0
<=>x-6=0 (Vì 2x+3x+21=0(vô nghiệm))
<=>x=6
Vậy x=6