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a,\(\dfrac{x^2-9}{2x+6}:\dfrac{3-x}{2}=\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}.\dfrac{2}{3-x}=\dfrac{x-3}{3-x}=\dfrac{-\left(3-x\right)}{3-x}=-1\)
b, \(\dfrac{2x}{x-y}-\dfrac{2y}{x-y}=\dfrac{2x-2y}{x-y}=\dfrac{2\left(x-y\right)}{x-y}=2\)
\(a,=\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}\cdot\dfrac{2}{-\left(x-3\right)}=\dfrac{x-3}{2}\cdot\dfrac{2}{-\left(x-3\right)}=-1\\ b,=\dfrac{2x-2y}{x-y}=\dfrac{2\left(x-y\right)}{\left(x-y\right)}=2\)
Bài 3:
a: \(\frac{x}{x-3}+\frac{9-6x}{x^2-3x}\)
\(=\frac{x}{x-3}+\frac{-6x+9}{x\left(x-3\right)}\)
\(=\frac{x^2-6x+9}{x\left(x-3\right)}=\frac{\left(x-3\right)^2}{x\left(x-3\right)}=\frac{x-3}{x}\)
b: \(\frac{6x-3}{x}:\frac{4x^2-1}{3x^2}\)
\(=\frac{3\left(2x-1\right)}{x}\cdot\frac{3x^2}{\left(2x-1\right)\left(2x+1\right)}=\frac{3\cdot3x}{2x+1}=\frac{9x}{2x+1}\)
Bài 2:
a: \(\frac{x^3-x}{3x+3}\)
\(=\frac{x\left(x^2-1\right)}{3\left(x+1\right)}=\frac{x\left(x-1\right)\left(x+1\right)}{3\left(x+1\right)}=\frac{x\left(x-1\right)}{3}\)
b: \(\frac{x^2+3xy}{x^2-9y^2}=\frac{x\left(x+3y\right)}{\left(x-3y\right)\left(x+3y\right)}=\frac{x}{x-3y}\)
Bài 1:
a: \(\frac{x^2-9}{2x+6}:\frac{3-x}{2}\)
\(=\frac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}\cdot\frac{2}{-\left(x-3\right)}=\frac{-2}{2}=-1\)
b: \(\frac{2x}{x-y}-\frac{2y}{x-y}=\frac{2x-2y}{x-y}=\frac{2\left(x-y\right)}{x-y}=2\)
c: \(\frac{x+15}{x^2-9}+\frac{2}{x+3}\)
\(=\frac{x+15}{\left(x-3\right)\left(x+3\right)}+\frac{2}{x+3}\)
\(=\frac{x+15+2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{x+15+2x-6}{\left(x-3\right)\left(x+3\right)}=\frac{3x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3}{x-3}\)
d: \(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{y^2+x^2}{y^2-x^2}\)
\(=\frac{x+y}{2\left(x-y\right)}-\frac{x-y}{2\left(x+y\right)}+\frac{x^2+y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\frac{\left(x+y\right)^2-\left(x-y\right)^2+2\left(x^2+y^2\right)}{2\left(x-y\right)\left(x+y\right)}=\frac{x^2+2xy+y^2-x^2+2xy-y^2+2x^2+2y^2}{2\left(x-y\right)\left(x+y\right)}\)
\(=\frac{2x^2+4xy+2y^2}{2\left(x-y\right)\left(x+y\right)}=\frac{2\left(x^2+2xy+y^2\right)}{2\left(x-y\right)\left(x+y\right)}=\frac{\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)}=\frac{x+y}{x-y}\)
b) \(=\left(y^2-9\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(=y^4-81-y^4+4\)\(=-77\)
a) \(\dfrac{4}{9}x + \dfrac{2}{3}x = (\dfrac{4}{9} + \dfrac{2}{3})x = (\dfrac{4}{9} + \dfrac{6}{9})x = \dfrac{{10}}{9}x\);
b) \( - 12{y^2} + 0,7{y^2} = ( - 12 + 0,7){y^2} = - 11,3{y^2}\);
c) \( - 21{t^3} - 25{t^3} = ( - 21 - 25){t^3} = - 46{t^3}\).
a: \(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\)
\(=\dfrac{5xy+y^3-x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}=\dfrac{x^3+y^3}{x^2y^2}\)
b: \(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x+3}{x^2-3x}\)
a) =(x-y)5+(x-y)3=(x-y)3[(x-y)2+1]
b) =33(y-2x)3:-9(y-2x)=-3(y-2x)2
c) =(x-y)2 [3(x-y)3-2(x-y)2+3]:5(x-y)2=[3(x-y)3-2(x-y)2+3]/5
a: \(\left(4x^5-8x^3\right):\left(-2x^3\right)\)
\(=-4x^5:2x^3+8x^3:2x^3\)
\(=-2x^2+4\)
b: \(\left(9x^3-12x^2+3x\right):\left(-3x\right)\)
\(=-\frac{9x^3}{3x}+\frac{12x^2}{3x}-\frac{3x}{3x}\)
\(=-3x^2+4x-1\)
c: \(\left(xy^2+4x^2y^3-3x^2y^4\right):\left(-\frac12x^2y^3\right)\)
\(=-xy^2:\frac12x^2y^3-4x^2y^3:\frac12x^2y^3+3x^2y^4:\frac12x^2y^3\)
\(=-\frac{2}{xy}-8+6y\)
d: \(\left\lbrack2\left(x-y\right)^3-7\left(y-x\right)^2-\left(y-x\right)\right\rbrack:\left(x-y\right)\)
\(=\left\lbrack2\left(x-y\right)^3-7\left(x-y\right)^2+\left(x-y\right)\right\rbrack:\left(x-y\right)\)
\(=\frac{2\left(x-y\right)^3}{x-y}-\frac{7\left(x-y\right)^2}{x-y}+\frac{\left(x-y\right)}{x-y}\)
\(=2\left(x-y\right)^2-7\left(x-y\right)+1\)
Chọn đáp án B