Eerst $\sqrt{9}$ en $2^3$, dan vermenigvuldigen en tenslotte de optelling.

$8+3\cdot 8=8+24=32$

$88$

$2\cdot 3^3=2\cdot 27=54$

$\sqrt{36}/4=6/4=\mathrm{1,5}$

$\sqrt{9}+\sqrt{16}=3+4=7$

$36/4+2^3=36/4+8=9+8=17$

$6^5-6^3=7776-216=7560$

${(2+3)}^4=5^4=625$

$4\cdot 2^5-400/\sqrt{16}=128-100=28$

$\left(2^3+3^2\right)/17-\sqrt[3]{\mathrm{0,125}}=17/17-\mathrm{0,5}=\mathrm{0,5}$

`4^3 // 8 + (sqrt(25))^2 = 64 // 8 + 5^2 = 8 + 25 = 33`

$3^5/3^2+3^4=108$

$3^4\cdot 2^3=648$

${\left(\sqrt{196}-3^2\right)}^3=125$

${\left(2\cdot \sqrt[3]{27}\right)}^3=216$

$\sqrt{2\cdot 70+4}=12$

$\frac{12\cdot 3}{2^3-4}=9$

$\frac{2^{4+\sqrt{16}}}{2^5}=8$

$\sqrt[3]{\frac{1}{3}-{\left(\frac{1}{3}\right)}^3}=\frac{2}{3}$  

Correct, want $2^3\cdot 2^4=2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2=2^7$.

Niet correct, want $2^6/2^3=\frac{2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2}{2\cdot 2\cdot 2}=\frac{2\cdot 2\cdot 2}{1}=2^3$.

Niet correct, want ${\left(2^2\right)}^3={\left(2\cdot 2\right)}^3=2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2=2^6$.

Bijvoorbeeld: `3+4+8+9 = 24` , `(3+9)*8//4 = 24` , `(3 + sqrt(9))*8//sqrt(4) = 24` , ...

Bijvoorbeeld: `1*2*3*4 = 24` , `2^3 * (4-1) = 24` , `3//1 * 4 * 2 = 24` , ...

Bijvoorbeeld: `(7-4)*(9-1) = 24` , ...

$\sqrt{5^3-100}-\frac{125}{{10}^2+5^2}=\sqrt{25}-\frac{125}{125}=5-1=4$

$\frac{2^{1+\sqrt{9}}}{2^5-2^4}=\frac{2^4}{16}=1$