13) (x – 2)^2 – x (x – 5) – 3 = 0
14) (3x + 1) (3x – 1) – (x – 2)(x^2 + 2x + 4) = x (6 – x^2)
15) 5x(x – 3)^2 + 5(x – 4) (x + 4) – 5x^3 = -5x(6x – 1)
16) x^2 (3x + 1) – (x – 3)^2 = 3x^3
13) (x – 2)^2 – x (x – 5) – 3 = 0
14) (3x + 1) (3x – 1) – (x – 2)(x^2 + 2x + 4) = x (6 – x^2)
15) 5x(x – 3)^2 + 5(x – 4) (x + 4) – 5x^3 = -5x(6x – 1)
16) x^2 (3x + 1) – (x – 3)^2 = 3x^3
Đáp án:
Giải thích các bước giải:
13) $(x-2)^{2}-x(x-5)-3=0_{}$
⇔ $x^{2}-4x+4-x^{2}+5x-3=0_{}$
⇔ $x+1=0_{}$
⇔ $x=-1_{}$
14) $(3x+1)(3x-1)-(x-2)(x^{2}+2x+4)=x(6-x^{2})_{}$
⇔ $9x^{2}-1-(x^{3}-8)=6x-x^{3}_{}$
⇔ $9x^{2}-1-x^{3}+8=6x-x^{3}_{}$
⇔ $9x^{2}+7=6x_{}$
⇔ $9x^{2}-6x+7=0_{}$
⇔ $x∈R_{}$
15) $5x(x-3)^{2}+5(x-4)(x+4)-5x^{3}=-5x(6x-1)_{}$
⇔ $x(x-3)^{2}+(x-4)(x+4)-x^{3}=-x(6x-1)_{}$
⇔ $x(x^{2}-6x+9)+x^{2}-16-x^{3}=-6x^{2}+x_{}$
⇔ $x^{3}-6x^{2}+9x+x^{2}-16-x^{3}=-6x^{2}+x_{}$
⇔ $-6x^{2}+9x+x^{2}-16=-6x^{2}+x_{}$
⇔ $9x+x^{2}-16=x_{}$
⇔ $9x+x^{2}-16-x=0_{}$
⇔ $x^{2}+8x-16=0_{}$
⇔ $x_{1}=-4-4\sqrt{2}$; $x_{2}=-4+4\sqrt{2}$
16) $x^{2}.(3x+1)-(x-3)^{2}=3x^{3}_{}$
⇔ $3x^{3}+x^{2}-(x-3)^{2}=3x^{3}_{}$
⇔ $x^{2}-(x^{2}-6x+9)=0_{}$
⇔ $x^{2}-x^{2}+6x-9=0_{}$
⇔ $6x-9=0_{}$
⇔ $6x=9_{}$
⇔ $x=\frac{3}{2}_{}$