Giải hộ bài 2 cảm ơn

Giải thích các bước giải:

Bài 2:

5.1.Ta có:

$((x+m)^2+(x-m)^2)^3$

$=\left(x^2+2xm+m^2+x^2-2xm+m^2\right)^3$

$=\left(2x^2+2m^2\right)^3$

5.2.Ta có:

$(x-y)^3+3xy(x+y)$

$=x^3-3x^2y+3xy^2-y^3+3x^2y+3xy^2$

$=x^3+6xy^2-y^3$

5.3.Ta có:

$(\dfrac{3x}y+\dfrac{2y}x)^3+(\dfrac{3x}y-\dfrac{2y}x)^3$

$=\dfrac{\left(3x^2+2y^2\right)^3}{x^3y^3}+\dfrac{\left(3x^2-2y^2\right)^3}{x^3y^3}$

$=\dfrac{\left(3x^2+2y^2\right)^3+\left(3x^2-2y^2\right)^3}{x^3y^3}$

$=\dfrac{27x^6+54x^4y^2+36x^2y^4+8y^6+27x^6-54x^4y^2+36x^2y^4-8y^6}{x^3y^3}$

$=\dfrac{54x^6+72x^2y^4}{x^3y^3}$

5.4.Ta có:

$(\dfrac{x}2-y)^3-y(y-\dfrac{x}2)^2+\dfrac1{24}(2-3x^3)-\dfrac13$

$=\dfrac{\left(-2y+x\right)^3}{8}-\dfrac{y\left(2y-x\right)^2}{4}+\dfrac{2-3x^3}{24}-\dfrac{1}{3}$

$=\dfrac{\left(-2y+x\right)^3\cdot \:3-y\left(2y-x\right)^2\cdot \:6+2-3x^3-8}{24}$

$=\dfrac{-24y^3+36y^2x-18yx^2+3x^3-6y\left(2y-x\right)^2-3x^3-6}{24}$

$=\dfrac{-48y^3+60y^2x-24yx^2-6}{24}$