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Bài giải:
a) 3x - 6y = 3 . x - 3 . 2y = 3(x - 2y)
b) 2525x2 + 5x3 + x2y = x2 (2525 + 5x + y)
c) 14x2y – 21xy2 + 28x2y2 = 7xy . 2x - 7xy . 3y + 7xy . 4xy = 7xy(2x - 3y + 4xy)
d) 2525x(y - 1) - 2525y(y - 1) = 2525(y - 1)(x - y)
e) 10x(x - y) - 8y(y - x) =10x(x - y) - 8y[-(x - y)]
= 10x(x - y) + 8y(x - y)
= 2(x - y)(5x + 4y)
a,\(3x-6y=3\left(x-2y\right)\)
b,\(x^2(\dfrac{2}{5}+5x+y)\)
c,\(7xy\left(2x-3y+4xy\right)\)
d,\(\dfrac{2}{5}x\left(y-1\right)-\dfrac{2}{5}y\left(y-1\right)\)
=\(\dfrac{2}{5}\left(y-1\right)\left(x-y\right)\)
e,\(10x\left(x-y\right)-8y\left(y-x\right)=10x\left(x-y\right)+8y\left(x-y\right)\)
\(2\left(x-y\right)\left(5x+4y\right)\)
a: \(=\left(\dfrac{x}{y\left(x-y\right)}-\dfrac{2x-y}{x\left(x-y\right)}\right):\dfrac{x+y}{xy}\)
\(=\dfrac{x^2-2xy+y^2}{xy\left(x-y\right)}\cdot\dfrac{xy}{x+y}\)
\(=\dfrac{\left(x-y\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x-y}{x+y}\)
b: \(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x-y\right)\left(x+y\right)}\cdot\dfrac{x-y}{2y}\)
\(=\dfrac{4xy+4y^2}{2\left(x+y\right)}\cdot\dfrac{1}{2y}=\dfrac{4y\left(x+y\right)}{4y\left(x+y\right)}=1\)
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)
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: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
a: \(B=\left(x^2+y\right)\left(y+\dfrac{1}{4}\right)+x^2y^2+\dfrac{3}{4}\left(y+\dfrac{1}{3}\right)\)
\(=x^2y+\dfrac{1}{4}x^2+y^2+\dfrac{1}{4}y+x^2y^2+\dfrac{3}{4}y+\dfrac{1}{4}\)
\(=x^2y+x^2y^2+y^2+y+\dfrac{1}{4}x^2+\dfrac{1}{4}\)
\(=y\left(x^2+1\right)+y^2\left(x^2+1\right)+\dfrac{1}{4}\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(y+\dfrac{1}{2}\right)^2\)
\(C=x^2y^2+1+\left(x^2-y\right)\left(1-y\right)\)
\(=x^2y^2+1+x^2-x^2y-y+y^2\)
\(=x^2y^2-y+x^2+y^2-x^2y+1\)
\(=y^2\left(x^2+1\right)-y\left(x^2+1\right)+x^2+1\)
\(=\left(x^2+1\right)\left(y^2-y+1\right)\)
=>\(A=\dfrac{y^2+y+\dfrac{1}{4}}{y^2-y+1}\)
b: \(=\dfrac{y^2-y+1+2y-\dfrac{3}{4}}{y^2-y+1}=1+\dfrac{2y-\dfrac{3}{4}}{y^2-y+1}>=1\)
Dấu = xảy ra khi y=3/8
đáp án: a là đúng