Lutasin ang 3/2sqrt{96}+sqrt[2]{486}-2sqrt[3]{6}+9sqrt[4]{243}

$$\frac{ 3 }{ 2 } \sqrt{ 96 } + \sqrt[ 2 ]{ 486 } -2 \sqrt[ 3 ]{ 6 } +9 \sqrt[ 4 ]{ 243 }$$

I-evaluate

$$15\sqrt{6}+27\sqrt[4]{3}-2\sqrt[3]{6}\approx 68.642103306$$

Mga Hakbang ng Solution

I-factor out ang 96=4^{2}\times 6. I-rewrite ang square root ng product na \sqrt{4^{2}\times 6} bilang product ng mga square root na \sqrt{4^{2}}\sqrt{6}. Kunin ang square root ng 4^{2}.

$$\frac{3}{2}\times 4\sqrt{6}+\sqrt[2]{486}-2\sqrt[3]{6}+9\sqrt[4]{243}$$

I-multiply ang \frac{3}{2} at 4 para makuha ang 6.

$$6\sqrt{6}+\sqrt[2]{486}-2\sqrt[3]{6}+9\sqrt[4]{243}$$

I-factor

$$15 \sqrt{6} + 27 \sqrt[4]{3} - 2 \sqrt[3]{6} = 68.642103306$$