1/(x+2)² -(3x-1)²=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x²+12x

2/(x⁴+x²)(-2x³-2x)=x²(x²+1)-2x(x²+1)=(x²+1)(x²-2x)

\(8x^3+12x^2y+6xy^2+y^3-z^3\)

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

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

a, 8a3 - 36a2 +54ab2 - 27b3

=(8a3-36a2b +54ab2 - 27b3)

=(2a-3b)2

=(2a-3b)(2a-3b)(2a-3b)

b, 8x3 + 12x2y + 6xy2 + y3 - z 3

=(8x3 + 12x2y + 6xy2 + y3) - z3

=(2x + y)3 - y3

=(2x + y +z) . [ (2x + Y)2 + 2(2x + y)+ z2

= (2x + y + z)(4x2 + 4xy + y2 + 4x + 2y + z2

Bạn ơi sai đề ko vậy????

ko bạn ơi thầy in đề vậy đó

Đề bài sai rồi bạn nhé ~

Tham khảo : https://olm.vn/hoi-dap/detail/92085650467.html

Bài khó quá

4x⁴+4x-3= (2x)²+2(2x)+1-4

=(2x+1)²-2²=(2x+1-2)(2x+1+2)

=(2x-1)(2x+3)

bn làm đ rùi, tui đ cho bn

\(\dfrac{xy}{2}-x+\dfrac{x^2}{4}=x\left(\dfrac{y}{2}-1+\dfrac{x}{4}\right)\)

a, 4\(x^3\).y + \(\dfrac{1}{2}\)yz

  =y.(4\(x^3\) + \(\dfrac{1}{2}\)z)

b, (a² + b² - 5)² - 2.(ab + 2)²

 = [a² + b² - 5  - \(\sqrt{2}\)(ab + 2) ].[ a² + b² - 5 + \(\sqrt{2}\)(ab +2)]

a) \(4x^3y+\dfrac{1}{2}yz=y\left(4x^3+\dfrac{1}{2}z\right)\)

b) \(\left(a^2+b^2-5\right)^2-2.\left(ab+2\right)^2\)

\(=\left[\left(a^2+b^2-5\right)+2\left(ab+2\right)\right]\left[\left(a^2+b^2-5\right)-2\left(ab+2\right)\right]\)

\(=\left[a^2+b^2-5+2ab+4\right]\left[a^2+b^2-5-2ab-4\right]\)

\(=\left[a^2+b^2+2ab-1\right]\left[a^2+b^2-2ab-9\right]\)

\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)

\(=\left[\left(a+b+1\right)\left(a+b-1\right)\right]\left[\left(a-b+3\right)\left(a-b-3\right)\right]\)

Mình đã làm sai, cô Hoài Thương đúng rồi.

Lỗi?

Mình nghĩ là đề thiếu đó bạn :)

đề đáng lẽ phải là: \(x^7+x^2+1\)

\(x^7+x^2+1=\left(x^7-x\right)+\left(x^2+x+1\right)\)

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

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

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

\(=\left[x\left(x-1\right)\left(x+3\right)+1\right]\left(x^2+x+1\right)\)

\(=\left[\left(x^2-x\right)\left(x^3+1\right)+1\right]\left(x^2+x+1\right)\)

\(=\left(x^5-x^4-x^2-x+1\right)\left(x^2+x+1\right)\)

Vay chac minh dua nam de XD

Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)

\(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)

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

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

\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)

\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)