4: \(-5x^5y\left(B-xy\right)=-10x^5y^5+5x^6y^5\)
=>\(B-xy=\dfrac{-10x^5y^5+5x^6y^5}{-5x^5y}=2y^4-xy^4\)
=>\(B=2y^4-xy^4+xy\)
5: \(\left(2x^2y-5xy^3\right)\cdot3xy=B-\dfrac{5}{3}\)
=>\(B-\dfrac{5}{3}=6x^3y^2-15x^2y^4\)
=>\(B=6x^3y^2-15x^2y^4+\dfrac{5}{3}\)
6: \(\left(4x^4y^4-\dfrac{5}{4}x^5y^5\right):3x^2y^2=B-\dfrac{5}{12}x^3y^3\)
=>\(\dfrac{4}{3}x^2y^2-\dfrac{5}{12}x^3y^3=B-\dfrac{5}{12}x^3y^3\)
=>\(B=\dfrac{4}{3}x^2y^2\)
4) \(-5x^5y\left(-xy+B\right)=-10x^5y^5+5x^6y^5\)
\(\Rightarrow-xy+B=\dfrac{-10x^5y^5+5x^6y^5}{-5x^5y}=\dfrac{-5x^5y\left(2y^4-xy^4\right)}{-5x^5y}=2y^4-xy^4\\ \Rightarrow B=2y^4-xy^4+xy\)
5) \(\left(2x^2y-5xy^3\right):3xy=B-\dfrac{5}{3}\)
\(\Rightarrow B=2x^2y:3xy-5xy^3:3xy+\dfrac{5}{3}=\dfrac{2}{3}x-\dfrac{5}{3}y^2+\dfrac{5}{3}\)
6) Đoạn này ở vế trái là nhân hay chia \(3x^2y^2\) vậy bạn?