Question

kak tolong dong gmn caranya​

Answer 1

a

[tex]3 \sqrt{2} + 2 \sqrt{2} = [/tex]

[tex](3 + 2) \sqrt{2} = [/tex]

[tex]5 \sqrt{2} [/tex]

b

[tex]4 \sqrt{3} + 5 \sqrt{12} = [/tex]

[tex]4 \sqrt{3} + 5 \sqrt{4·3} = [/tex]

[tex]4 \sqrt{3} + 5 \sqrt{4} \sqrt{3} = [/tex]

[tex]4 \sqrt{3} + 5·2 \sqrt{3} = [/tex]

[tex]4 \sqrt{3} + 10 \sqrt{3} = [/tex]

[tex](4 + 10) \sqrt{3} = [/tex]

[tex]14 \sqrt{3} [/tex]

c

[tex]5 \sqrt{5} - 4 \sqrt{5} = [/tex]

[tex](5 - 4) \sqrt{5} = [/tex]

[tex] \sqrt{5} [/tex]

d

[tex]4 \sqrt{32} - 3 \sqrt{2} = [/tex]

[tex]4 \sqrt{16·2} - 3 \sqrt{2} = [/tex]

[tex]4 \sqrt{16} \sqrt{2} - 3 \sqrt{2} = [/tex]

[tex]4·4 \sqrt{2} - 3 \sqrt{2} = [/tex]

[tex]16 \sqrt{2} - 3 \sqrt{2} = [/tex]

[tex](16 - 3) \sqrt{2} = [/tex]

[tex]13 \sqrt{2} [/tex]

e

[tex] \sqrt{50} - \sqrt{18} + \sqrt{32} = [/tex]

[tex] \sqrt{25·2} - \sqrt{9·2} + \sqrt{16·2 } = [/tex]

[tex] \sqrt{25} \sqrt{2} - \sqrt{9} \sqrt{2} + \sqrt{16} \sqrt{2} = [/tex]

[tex]5 \sqrt{2} - 3 \sqrt{2} + 4 \sqrt{2} = [/tex]

[tex](5 - 3 + 4) \sqrt{2} = [/tex]

[tex](2 + 4) \sqrt{2} = [/tex]

[tex]6 \sqrt{2} [/tex]

f

[tex]3 \sqrt{24} - \sqrt{150} + \sqrt{54} = [/tex]

[tex]3 \sqrt{4·6} - \sqrt{25·6} + \sqrt{9·6} = [/tex]

[tex]3 \sqrt{4} \sqrt{6} - \sqrt{25} \sqrt{6} + \sqrt{9} \sqrt{6} = [/tex]

[tex]3·2 \sqrt{6} - 5 \sqrt{6} + 3 \sqrt{6} = [/tex]

[tex]6 \sqrt{6} - 5 \sqrt{6 } + 3 \sqrt{6} = [/tex]

[tex](6 - 5 + 3) \sqrt{6} = [/tex]

[tex](1 + 3) \sqrt{6} = [/tex]

[tex]4 \sqrt{6} [/tex]

g

[tex] \sqrt{ \frac{2}{3} } + 3 \sqrt{6} - \sqrt{ \frac{3}{2} } = [/tex]

[tex] \frac{ \sqrt{2} }{ \sqrt{3} } + 3 \sqrt{6} - \frac{ \sqrt{3} }{ \sqrt{2} } = [/tex]

[tex] \frac{ \sqrt{2} }{ \sqrt{3} } · \frac{ \sqrt{3} }{ \sqrt{3} } + 3 \sqrt{6} - \frac{ \sqrt{3} }{ \sqrt{2} } · \frac{ \sqrt{2} }{ \sqrt{2} } = [/tex]

[tex] \frac{ \sqrt{6} }{3} + 3 \sqrt{6} - \frac{ \sqrt{6} }{2} = [/tex]

[tex] \frac{2 \sqrt{6} }{6} + \frac{18 \sqrt{6} }{6} - \frac{3 \sqrt{6} }{6} = [/tex]

[tex] \frac{2 \sqrt{6 } + 18 \sqrt{6} -3 \sqrt{6} }{6} = [/tex]

[tex] \frac{(2 + 18 - 3) \sqrt{6} }{6} = [/tex]

[tex] \frac{(20 - 3) \sqrt{6} }{6} = [/tex]

[tex] \frac{17 \sqrt{6} }{6} [/tex]