\(K=\left(x^2y-3\right)^2-\left(2x-y\right)^3+xy^2\left(6-x^3\right)+8x^3-6x^2y-y^3\)

\(=x^4y^2-6x^2y+9-4x^2+4xy-y^2+6xy^2-x^4y^2+8x^3-6x^2y-y^3\)

\(=-12x^2y+9-4x^2+4xy-y^2+6xy^2+8x^3-y^3\)

a: \(=\dfrac{12xy^3z^4}{24x^2y^3z^3}=\dfrac{1}{2}\cdot\dfrac{1}{x}\cdot z=\dfrac{z}{2x}\)

b: \(=\dfrac{3\left(x-2\right)}{6x\left(x-2\right)}=\dfrac{1}{2x}\)

\(B=\left(5x-4y\right)^2-\left(6x+4y\right)\left(5x-4y\right)+\left(3x+2y\right)^2\)

\(B=\left(5x-4y\right)\left(5x-4y-6x-4y\right)+\left(3x+2y\right)^2\)

\(B=\left(5x-4y\right)\left(-x-8y\right)+\left(3x+2y\right)^2\)

\(B=-5x^2-40xy+4xy+32y^2+9x^2+12xy+4y^2\)

\(B=4x^2-24xy+36y^2\)

\(B=x^2-6xy+6y^2\)

Bài chưa đc ktra lại đâu . Có gì sai sót thì bỏ qua

1/

x² - 3x - 4 

= \(x^2-3x+\frac{9}{4}-\frac{9}{4}-4\)

\(=\left(x^2-3x+\frac{9}{4}\right)-\frac{25}{4}\)

\(=\left(x-\frac{3}{2}\right)^2-\left(\frac{5}{2}\right)^2\)

\(=\left(x-\frac{3}{2}-\frac{5}{2}\right)\left(x-\frac{3}{2}+\frac{5}{2}\right)\)

\(=\left(x-4\right)\left(x+1\right)\)

Bài 1 :

\(x^2-3x-4\)

\(=x^2+x-4x-4\)

\(=x\left(x+1\right)-4\left(x+1\right)\)

\(=\left(x+1\right)\left(x-4\right)\)

\(a,\dfrac{21x^2y^3}{24x^3y^2}=\dfrac{7y}{8x}\)

\(b,\dfrac{15xy^3\left(x^2-y^2\right)}{20x^2y\left(x+y\right)^2}=\dfrac{15xy^3\left(x-y\right)\left(x+y\right)}{20x^2y\left(x+y\right)^2}=\dfrac{3y^2\left(x-y\right)}{4x\left(x+y\right)}=\dfrac{3xy^2-3y^3}{4x^2+4xy}\)

a) Ta có: \(\dfrac{21x^2y^3}{24x^3y^2}\)

\(=\dfrac{21x^2y^3:3x^2y^2}{24x^3y^2:3x^2y^2}\)

\(=\dfrac{7y}{8x}\)

a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)

b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)

c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)

d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)

e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)