1003158601 Level: CThe number \( \sqrt[5]9 \) is bigger than:\( 3^{0.3} \)\( \sqrt3 \)\( \sqrt[9]{81} \)\( 3 \)
1003158602 Level: CIf \( a=\frac1{\sqrt3-1} \) and \( b=\frac1{\sqrt5-1}-\frac2{\sqrt5}+1 \), then:\( a > b \)\( a < b \)\( a = b \)\( a = -b \)
1003158603 Level: CAfter rationalizing the denominator \( \frac{\sqrt{\sqrt{13}+\sqrt{12}}}{\sqrt{\sqrt{13}-\sqrt{12}}} \) we get:\( \sqrt{13}+2\sqrt3 \)\( \sqrt{13}-\sqrt{12} \)\( 1 \)\( 25 \)
1003158604 Level: CAfter rationalizing the denominator \( \frac4{1-\sqrt2+\sqrt3} \) we get:\( \sqrt6-\sqrt2+2 \)\( -4 \)\( \sqrt6-\sqrt2 \)\( 1+\sqrt2-\sqrt3 \)
1003158605 Level: CUsing the formula for factoring \( a^3-b^3 \) to rationalize the denominator of \( \frac2{\sqrt[3]{25}+\sqrt[3]{15}+\sqrt[3]{9}} \) you get:\( \sqrt[3]5-\sqrt[3]3 \)\( 2 \)\( 2\sqrt[3]{49} \)\( 2\left( \sqrt[3]{25}-\sqrt[3]{15-\sqrt[3]9}\right) \)
1003158606 Level: CUsing the formula for factoring \( a^3-b^3 \) to find the multiplicative inverse of \( x=\sqrt[3]3-\sqrt[3]4 \) you get:\( -\sqrt[3]9-\sqrt[3]{12}-\sqrt[3]{16} \)\( \sqrt[3]9-\sqrt[3]{12}+\sqrt[3]{16} \)\( \sqrt[3]3-\sqrt[3]4 \)\( \sqrt[3]4-\sqrt[3]3 \)
1003158607 Level: CAs a result of this mathematical action \( \sqrt[3]{\left[\left(\sqrt2+1\right)^3-\left(\sqrt2-1\right)^3\right]^2-13^2} \) we get:\( 3 \)\( \sqrt[3]{-170} \)\( \sqrt[3]{13} \)\( \sqrt[3]{-168} \)
1003158608 Level: CAs a result of this mathematical action \( \frac1{\sqrt2+1}+\frac1{\sqrt3+\sqrt2}+\frac1{\sqrt4+\sqrt3}+\dots+\frac1{\sqrt{100}+\sqrt{99}} \) we get:\( 9 \)\( 12 \)\( 100 \)\( 99 \)
1003158609 Level: CChoose the correct relation.\( \frac{4^{2\sqrt3}\cdot8}{16^{\sqrt3+1}} =\frac12 \)\( \frac{9^{2\sqrt5}\cdot27^{\frac13}}{81^{1-\sqrt5}} =\frac1{27} \)\( \frac{\pi^{1-\sqrt2}\cdot\pi^{1+\sqrt2}}{\pi^2} < \frac1{\pi^3} \)\( \frac{2^{2\pi}\cdot9^{\pi}}{36^{\pi-1}} < 6 \)
1003164001 Level: CAfter simplifying the expression \[ \sqrt[3]{5\sqrt2+7}-\sqrt[3]{5\sqrt2-7} \] we get:\( 2 \)\( 2\sqrt[3]{57} \)\( 1 \)\( 2\sqrt[3]{97} \)