Câu trong ảnh làm như nào các bạn ơi...

a) $(x+3)(x^2-3x+9)-x(x-1)(x+1)=14$ $(x+3)(x^2-3x+9)-x(x^2-1)=14$ $x(x^2-3x+9)+3(x^2-3x+9)-x^3+x=14$ $x^3-3x^2+9x+3x^2-9x+27-x^3+x=14$ $x+27=14$ $x=14-27$ $x=-13$ Vậy $x=-13$ b) $(x+1)(x^2-x+1)-x(x-3)(x+3)=-27$ $(x+1)(x^2-x+1)-x(x^2-9)=-27$ $x(x^2-x+1)+(x^2-x+1)-x^3+9x=-27$ $x^3-x^2+x+x^2-x+1-x^3+9x=-27$ $9x+1=-27$ $9x=-27-1$ $9x=-28$ $x=-\frac{28}{9}$ Vậy $x=-\frac{28}{9}$ c) $3(x-1)^2-3x(x-5)=1$ $3(x^2-2x+1)-3x^2+15x=1$ $3x^2-6x+3-3x^2+15x=1$ $9x+3=1$ $9x=1-3$ $9x=-2$ $x=-\frac{2}{9}$ Vậy $x=-\frac{2}{9}$ d) $(x-3)(x^2+3x+9)+x(x+2)(2-x)=1$ $(x-3)(x^2+3x+9)+x(2+x)(2-x)=1$ $(x-3)(x^2+3x+9)+x(4-x^2)=1$ $x(x^2+3x+9)-3(x^2+3x+9)+4x-x^3=1$ $x^3+3x^2+9x-3x^2-9x-27+4x-x^3=1$ $4x-27=1$ $4x=1+27$ $4x=28$ $x=7$ Vậy $x=7$