\(=3x^2\left(x^2-1\right)+\left(x^8-3x^4+3x^2-1\right)-\left(x^8-1\right)\)

\(=3x^4-3x^2+x^8-3x^4+3x^2+1-x^8+1\)

\(=2\)

=2 nha ban

(con cach lam ban nhan dang thuc len rui rut gon lai)

a) \(x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)

\(=\left(x+1\right)\cdot\left[x\cdot\left(x-1\right)-\left(x^2-x+1\right)\right]\)

\(=\left(x+1\right)\left(x^2-x-x^2+x-1\right)\)

\(=\left(x+1\right)\cdot\left(-1\right)\)

\(=-1\left(x+1\right)\)

b) \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x+4\right)\left(x-4\right)\)

\(=x^3-3x^2+3x-1-\left(x^3+8\right)+\left(3x+12\right)\left(x-1\right)\)

\(=x^3-3x^2+3x-1-\left(x^3+8\right)+3x^2-3x+12x-12\)

\(=x^3-1-x^3-8+12x-12\)

\(=-21+12x\)

c) \(3x^2\left(x+1\right)\left(x-1\right)+\left(x^2-1\right)^3-\left(x^2-1\right)\left(x^4+x^2+1\right)\)

\(=3x^2\left(x^2-1\right)+x^6-3x^4+3x^2-1-\left(x^6-1\right)\)

\(=3x^4-3x^2+x^6-3x^4+3x^2-1-x^6+1\)

\(=0\)

câu b bạn làm sai rồi í!

a: \(=-\left[\left(\dfrac{1}{3}ab^2+2a^3b\right)^3\right]\)

\(=\dfrac{-1}{27}a^3b^6-3\cdot\dfrac{1}{9}a^2b^4\cdot2a^3b-3\cdot\dfrac{1}{3}ab^2\cdot4a^6b^2-8a^9b^3\)

\(=\dfrac{-1}{27}a^3b^6-\dfrac{2}{3}a^5b^5-4a^7b^4-8a^9b^3\)

b: \(=x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-1\right)\)

\(=6x^2+2-6x^2+6\)

=8

\(a.\left(2xy-3\right)^2=4x^2y^2-12xy+9\)

\(b.\left(\dfrac{1}{2}x+\dfrac{1}{3}\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}x+\dfrac{1}{9}\)

a) (2xy)²-2(2xy-3)+3²

\(a,\left(-4xy-5\right)\left(5-4xy\right)=\left(4xy+5\right)\left(4xy-5\right).\)

\(=\left(4xy\right)^2-5^2=16x^2y^2-25\)

\(b,\left(a^2b+ab^2\right)\left(ab^2-a^2b\right)=\left(ab^2+a^2b\right)\left(ab^2-a^2b\right)\)

\(=\left(ab^2\right)^2-\left(a^2b\right)^2=a^2b^4-a^4b^2\)

\(c,\left(3x-4\right)^2+2\left(3x-4\right)\left(4-x\right)+\left(4-x\right)^2\)

\(=\left[\left(3x-4\right)+\left(4-x\right)\right]^2\)

\(=\left(3x-4+4-x\right)^2=\left(2x\right)^2=4x^2\)

\(d,\left(a^2+ab+b^2\right)\left(a^2-ab+b^2\right)-\left(a^4+b^4\right)\)

\(=\left[\left(a^2+b^2\right)+ab\right]\left[\left(a^2+b^2\right)-ab\right]-\left(a^4+b^4\right)\)

\(=\left(a^2+b^2\right)^2-\left(ab\right)^2-a^4-b^4\)

\(=a^4+2a^2b^2+b^4-a^2b^2-a^4-b^4=a^2b^2\)

a) \(12\left(2x-5\right)^2-3\left(1+4x\right)\left(4x-1\right)\)

\(=12\left[\left(2x\right)^2-2.2x.5+5^2\right]-3\left(4x+1\right)\left(4x-1\right)\)

\(=12\left(4x^2-20x+25\right)-3\left[\left(4x\right)^2-1\right]\)

\(=48x^2-240x+300-3\left(16x^2-1\right)\)

\(=48x^2-240x+300-48x^2+3\)

\(=-240x+303\)

a,\(\left(2x-1\right)\left(4x^2+2x+1\right)=\left(2x-1\right)\left[\left(2x\right)^2+2x.1+1^2\right]\)

\(=\left(2x\right)^3-1=8x^3-1\)

b,\(\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2\)

\(=x^2+2.x.2y+\left(2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)

`a)(2x-1)(4x^2+2x+1)`

`=(2x-1)[(2x)^2+2x.1+1^2]`

`=(2x)^3-1^3`

`=8x^3-1`

Áp dụng HĐT:`A^3-B^3=(A-B)(A^2+AB+B^2)`

`b)(x+2y+z)(x+2y-z)`

`=[(x+2y)+z][(x+2y)-z]`

`=(x+2y)^2-z^2`

`=x^2+2.x.2y+(2y)^2-z^2`

`=x^2+4xy+4y^2-z^2`

Áp dụng HĐT:`A^2-B^2=(A+B)(A-B)`

                      `(A+B)^2=A^2+2AB+B^2`