2010000901 Level: BAssuming \(xy\neq 1\), simplify the expression: \[ \frac{ \frac{x+y} {1-xy} -x} {1 +\frac{x(x+y)} {1-xy} } \]\( y\)\(\frac{y(1+x^{2})} {1-x^{2}} \)\(\frac{y} {1+x^{2}} \)\( y(1+x^2)\)
2010000814 Level: BAssuming \(x\neq 0\), \(y\neq 0\), \(y\neq \pm 1\), simplify the expression: \[\left [\left ( \frac{y-1} {y}\right )^{2} : \left (\frac{x} {y+1} \right )^{2}\right ] : \frac{2(y^2-1)} {xy}\]\(\frac{y^2-1} {2xy}\)\( 2\)\(\frac{y^2-1} {2}\)\(\frac{y-1} {2}\)
2010000813 Level: CAssuming \(x\neq -3\), \(x\neq 4\), simplify the expression: \[\frac{x^{3} + 27} {x^2-x-12}\]\(\frac{x^2-3x+9} {x-4}\)\(\frac{x^2+3x+9} {x-4}\)\(\frac{x^2+6x+9} {x-4}\)\(\frac{x^2-6x+9} {x-4}\)
2010000812 Level: AAssuming \( y \neq 1\), \(x\neq \pm y\), simplify the expression: \[\frac{y^2-2xy+x^2}{(1-y)(y-x)}\cdot\frac{3y^2-6y+3}{x^2-y^2}\]\(\frac{3(y-1)}{x+y}\)\(\frac{3(1-y)}{x+y}\)\(\frac{3}{x+y}\)\( 3\)
2010000811 Level: AEvaluate the following expression at \(x = 9\). \[\frac{\frac{1}{x}+\frac{1}{x^2}}{\frac{1}{\sqrt{x}}}\]\( \frac{10}{27}\)\( -\frac{10}{9}\)\(-30\)\(30\)
2010000810 Level: AEvaluate the following expression at \(x = 4\). \[\frac{\frac{1}{\sqrt{x}}}{\frac{1}{x^2}-\frac{1}{x}}\]\(- \frac{8}{3}\)\(\frac{31}{3}\)\( \frac{8}{3}\)\( 6\)
2010000809 Level: AAssuming \( x \notin \{-4;0;3;4\}\), simplify the following expression. \[\frac{x^2-3x}{x^2-16}:\frac{x-3}{x^2+4x}\]\( \frac{x^2}{x-4} \)\( \frac{x-4}{x^2} \)\( \frac{x-4}{x} \)\( \frac{x}{x-4} \)
2010000808 Level: AAssuming \( x \notin \{0;1;3\}\), simplify the following expression. \[\frac{x^2-9}{x^2-x}:\frac{x^2-3x}{x-1}\]\( \frac{x+3}{x^2} \)\( \frac{x-3}{x^2}\)\( \frac{x+3}{2x}\)\( \frac{x+3}{x} \)
2010000807 Level: AAssuming \(x\neq 0\), \(y\neq 0\), \(x\neq -y\), simplify the following expression. \[ { \frac{1} {y^{2}} - \frac{1} {x^{2}} \over -\frac{1} {x} - \frac{1} {y}} \]\(\frac{y-x} {xy} \)\(\frac{x-y} {xy} \)\(x-y\)\(y-x\)
2010000806 Level: CExpress \( d_1 \) from the formula \( v=\frac{v_1v_2(d_1+d_2 )}{d_1v_2+d_2v_1} \).\( d_1=-\frac{d_2v_1(v-v_2)}{v_2(v-v_1)} \)\( d_1=\frac{d_2v_1(v-v_2)}{v_2(v-v_1)} \)\( d_1=-\frac{d_2v_1(v_2-v)}{v_2(v-v_1)} \)\( d_1=\frac{d_2v_1(v_2-v)}{v_2(v_1-v)} \)