sau bạn đăng tách ra cho mn cùng giúp nhé 

a, \(\left(-2x^5+3x^2-4x^3\right):2x^2=-x^3+\frac{3}{2}-2x\)

b, \(\left(x^3-2x^2y+3xy^2\right):\left(-\frac{1}{2}x\right)=-\frac{x^2}{2}+xy-\frac{3y^2}{2}\)

c, \(\left(3x^2y^2+6x^3y^3-12xy^2\right):3xy=xy+2x^2y^2-4y\)

d, \(\left(4x^3-3x^2y+5xy^2\right):\frac{1}{2}x=2x^2-\frac{3xy}{2}+\frac{5y^2}{2}\)

e, \(\left(18x^3y^5-9x^2y^2+6xy^2\right):3xy^2=6x^2y^3-3x+2\)

f, \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)=x^2+y^2\)

viết sai đề hết rồi

rưa mi chỉnh t cấy

b: \(\frac{2x-1}{x^2+x}+\frac{5-x}{x^2-1}\)

\(=\frac{2x-1}{x\left(x+1\right)}+\frac{5-x}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{\left(2x-1\right)\left(x-1\right)+x\left(5-x\right)}{x\left(x+1\right)\left(x-1\right)}=\frac{2x^2-3x+1+5x-x^2}{x\left(x+1\right)\left(x-1\right)}\)

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

a: \(\frac{3xy-4y^2}{2x^2y}-\frac{xy-4y^2}{2x^2y}\)

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

\(=\frac{2xy}{2x^2y}=\frac{1}{x}\)

Ói , hoa mắt chóng mặt nhức đầu ,

sao giống có chữa quá z

a: \(N=\dfrac{3x^5-4x^4+6x^3}{-2x^2}=-\dfrac{3}{2}x^3+2x^2-3x\)

b: \(N=\dfrac{\left(6x^4y^5-3x^3y^4+\dfrac{1}{2}x^4y^3z\right)}{-\dfrac{1}{3}x^2y^3}=-18x^2y^2+9xy-\dfrac{3}{2}x^2z\)

c: \(\Leftrightarrow N\cdot\left(y-x\right)=\left(x-y\right)^3\)

\(\Leftrightarrow N=\dfrac{\left(x-y\right)^3}{y-x}=-\left(y-x\right)^2\)

d: \(\Leftrightarrow N\cdot\left(y^2-x^2\right)=\left(y^2-x^2\right)^2\)

hay \(N=y^2-x^2\)

a: \(x^2\cdot x-2x^3=x^3-2x^3=-x^3\)

b: \(6x^2y\cdot3xy-2y^2\left(x+y\right)=18x^3y^2-2xy^2-2y^3\)

c: \(\left(4x^2+5x-1\right)\left(2x^3-3x\right)\)

\(=8x^5-12x^3+10x^4-15x^2-2x^3+3x\)

\(=8x^5+10x^4-14x^3-15x^2+3x\)