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\(-\dfrac{2}{5}x^2y^5\left(5xy^2-\dfrac{1}{2}x^2y-\dfrac{7}{5}x^3\right)\)
\(-2x^3y^7+\dfrac{1}{5}x^4y^6+\dfrac{14}{25}x^5y^5\)
a)\((\dfrac{5}{7}x^2y)^3:(\dfrac{1}{7}xy)^3\)
=\((\dfrac{5}{7}x^2y:\dfrac{1}{7}:x:y)^3\)
=(\(\dfrac{5}{7}.7.x^2:x.y:y)^3\)
=(5x)\(^3\)
=5\(^3\).x\(^3\)
=125.x\(^3\)
1: \(=\left(\dfrac{1}{2}:\dfrac{-2}{3}\right)\cdot\left(a^m:a^2\right)\cdot\left(b^n:b\right)\cdot\left(c^2:c\right)\)
\(=\dfrac{-3}{4}a^{m-2}b^{n-1}c\)
2: \(=\dfrac{2}{7}\left(x-2y\right)^{2m+1}\cdot\dfrac{-14}{5\left(2y-x\right)^{2m-1}}\)
\(=\dfrac{-4}{5}\cdot\dfrac{-\left(2y-x\right)^{2m+1}}{\left(2y-x\right)^{2m-1}}=\dfrac{4}{5}\cdot\left(2y-x\right)^2\)
Bài 45: (SBT/12):
a. (5x4 - 3x3 + x2) : 3x2
= (5x4 : 3x2) + (-3x3 : 3x2) + (x2 : 3x2)
=\(\dfrac{5}{2}\)x2 - x + \(\dfrac{1}{3}\)
b. (5xy2 + 9xy - x2y2) : (-xy)
= [5xy2 : (-xy)] + [9xy : (-xy)] + [(-x2y2) : (-xy)]
= -5y - 9 + xy
c. (x3y3 : \(\dfrac{1}{3}\)x2y3 - x3y2) : \(\dfrac{1}{3}\)x2y2
= (x3y3 : \(\dfrac{1}{3}\)x2y2) + (-\(\dfrac{1}{2}\)x2y3 : \(\dfrac{1}{3}\)x2y2) + (-x3y2 : \(\dfrac{1}{3}\)x2y2)
= 3xy - \(\dfrac{3}{2}\)y - 3x
a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
\(=-2x^3y^7+\dfrac{1}{5}x^4y^6+\dfrac{14}{25}x^5y^5\)