a: B=-4x^3y^2/2xy=-2x^2y
b: \(\Leftrightarrow H-2xy=\dfrac{4x^3y^2-3x^2y^3}{-2x^2y}=-2xy+\dfrac{3}{2}y^2\)
=>H=-2xy+3/2y^2+2xy=3/2y^2
\(a,4x^3y^2:B=-2xy\\ B=\dfrac{4x^3y^2}{-2xy}=4:\left(-2\right).\left(x^3:x\right).\left(y^2:y\right)=-2x^2y\\ b,\left(4x^3y^2-3x^2y^3\right):B=-2xy+H\\ \Leftrightarrow\left(4x^3y^2-3x^2y^3\right):\left(-2x^2y\right)=-2xy+H\\ \Leftrightarrow4x^3y^2:\left(-2x^2y\right)-3x^2y^3:\left(-2x^2y\right)=-2xy+H\\ \Leftrightarrow-2xy+\dfrac{3}{2}y^2=-2xy+H\\ \Rightarrow H=-2xy+2xy+\dfrac{3}{2}y^2=\dfrac{3}{2}y^2\\ Vậy:H=\dfrac{3}{2}y^2\)
a) Ta có:
\(4x^3y^2:B=2xy\)
\(\Rightarrow B=\dfrac{4x^3y^2}{-2xy}\)
\(\Rightarrow B=-2x^2y\)
b) \(\left(4x^3y^2-3x^2y^3\right):B=-2xy+H\)
\(\Rightarrow\dfrac{4x^3y^2-3x^2y^3}{-2x^2y}=-2xy+H\)
\(\Rightarrow\dfrac{-2x^2y\left(-2xy+\dfrac{3}{2}y^2\right)}{-2x^2y}=-2xy+H\)
\(\Rightarrow-2xy-\dfrac{3}{2}y^2=-2xy+H\)
\(\Rightarrow H=-2xy-\dfrac{3}{2}y^2+2xy\)
\(\Rightarrow H=-\dfrac{3}{2}y^2\)
a. Ta có: \(4x^3y^2:B=-2xy\)
\(4x^3y^2=-2xy:B\)
Đặt A = - 2xy, ta có:
\(4x^3y^2=AB\)
\(\Leftrightarrow B=\dfrac{AB}{4x^3y^2}\)
Ta thấy \(B=\dfrac{A}{4x^3y^2}\) đã thõa mãn pt ban đầu.
Vật đơn thức \(B=\dfrac{A}{4x^3y^2}\)
b. Ta có: \(B=-2xy\)
\(H=4x^3y^2-3x^3y^2:B\)
\(H=4x^3y^2-3x^3y^2:\left(-2xy\right)\)
\(H=4x^3y^2+6x^3y^2\)
\(H=10x^3y^2\)
Vậy đơn thức H là 10x3y2
