a: \(=\dfrac{1}{2}\cdot xy^5\cdot a^2b^4\cdot-x^3z^7=-\dfrac{1}{2}a^2b^4\cdot x^4y^5z^7\)

b: \(=x^3y\left(-1+\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{5}{4}x^3y\)

c: \(=4x^2+\dfrac{1}{2}x-7-3x^2-\dfrac{1}{2}x+\dfrac{1}{2}=x^2-\dfrac{13}{2}\)

d: \(=-243x^5y^{10}\cdot\left(-x^3y^6\right)=243x^8y^{16}\)

\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)

\(A=x^3.\left(-\dfrac{5}{4}x^2y\right).\left(\dfrac{2}{5}x^3y^4\right).\\ A=-\dfrac{1}{2}x^8y^5.\)

- Bậc: 8.

- Hệ số: \(-\dfrac{1}{2}.\)

- Biến: \(x;y.\)

\(B=\left(-\dfrac{3}{4}x^5y^4\right).\left(xy^2\right).\left(-\dfrac{8}{9}x^2y^3\right).\\ B=\dfrac{2}{3}x^8y^9.\)

- Bậc: 9.

- Hệ số: \(\dfrac{2}{3}.\)

- Biến: \(x;y.\)

ai giúp em vs

a: \(5x^2y^4:10x^2y=\dfrac{1}{2}y^3\)

c: \(\left(-xy\right)^{10}:\left(-xy\right)^5=-x^5y^5\)

a)\(\dfrac{32x^5\left(3y-7\right)^5}{-4x\left(7-3y\right)^4}=\dfrac{-4x.\left(-8x^4\right)\left(3y-7\right)^4\left(3y-7\right)}{-4x\left(3y-7\right)^4}\)

\(=\dfrac{\left(-8x^4\right)\left(3y-7\right)}{1}=\left(-8x^4\right)\left(3y-7\right)\)

\(=-32x^4y+56x^4\)

b) \(\dfrac{12x^3\left(3x-5\right)^2}{4x\left(3x-5\right)^2}-\dfrac{2x\left(x+7\right)}{\left(x+7\right)^3}=\dfrac{12x^3}{4x}-\dfrac{2x}{\left(x+7\right)^2}\)

\(=3x^2-\dfrac{2x}{\left(x+7\right)^2}\)

\(\)

\(\left(\dfrac{x^2}{x+y}+y\right).\left(\dfrac{1}{x^2-xy}-\dfrac{3y^3}{x^4-xy^3}-\dfrac{y}{x^3+x^2y+xy^2}\right)\)

\(=\left(\dfrac{x^2+xy+y^2}{x+y}\right).\left(\dfrac{1}{x\left(x-y\right)}-\dfrac{3y^2}{x\left(x^3-y^3\right)}-\dfrac{y}{x\left(x^2+xy+y^2\right)}\right)\)\(=\left(\dfrac{x^2+xy+y^2}{x+y}\right).\left(\dfrac{x^2+xy+y^2}{x\left(x^3-y^3\right)}-\dfrac{3y^2}{x\left(x^3-y^3\right)}-\dfrac{xy-y^2}{x\left(x^3-y^3\right)}\right)\)

\(=\dfrac{x\left(x^3-y^3\right)}{x^3-xy^2}.\dfrac{x^2+xy+y^2-3y^2-xy+y^2}{x\left(x^3-y^3\right)}\\ =\dfrac{x^2-y^2}{x\left(x^2-y^2\right)}=\dfrac{1}{x}\)

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