Question

Answer 1

Bài 1:

a: \(\left(x-1\right)^3+\left(2-x\right)\left(x^2+2x+4\right)+3x\left(x+2\right)=17\)

=>\(x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)

=>9x+7=17

=>9x=10

=>x=10/9

b: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x_{}^2-2\right)=15\)

=>\(x^3+8-x^3+2x=15\)

=>2x=15-8=7

=>\(x=\frac72\)

c: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2-3x+9\right)+9\left(x+1\right)^2=15\)

=>\(x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=15\)

=>\(-9x^2+27x+9x^2+18x+9=15\)

=>45x=15-9=6

=>\(x=\frac{6}{45}=\frac{2}{15}\)

Bài 2:

a: \(A=2\left(x^3-y^3\right)-3\left(x+y\right)^2\)

\(=2\left\lbrack\left(x-y\right)^3+3xy\left(x-y\right)\right\rbrack-3\left\lbrack\left(x-y\right)^2+4xy\right\rbrack\)

=2[8+6xy]-3(4+4xy]

=16+12xy-12-12xy

=4

b:

Sửa đề: x-y=1

\(B=x^2\left(x+1\right)-y^2\left(y-1\right)+xy-3xy\left(x-y+1\right)\)

\(=x^3+x^2-y^3+y^2+xy-3x^2y+3xy^2-3xy\)

\(=x^3-3x^2y+3xy^2-y^3+x^2-2xy+y^2\)

\(=\left(x-y\right)^3+\left(x-y\right)^2\)

=1+1

=2