bantu, soal perpangkatan & bentuk akar.... Question from @andidarmawatiwati

AryaSudewa

= 2⁻² + 3⁻³ + 1⁻⁴

= 1/2² + 1/3³ + 1/1⁴

= 1/4 + 1/27 + 1

= 27/108 + 4/108 + 108/108

= 139/108

= 1 31/108

= (9x⁻²y³z⁻⁴)²

= 81x⁻⁴y⁶z⁻⁸

$= \sqrt{175}+4\sqrt{7}-\sqrt{63}\\= \sqrt{25\times7}+4\sqrt{7}-3\sqrt{7}\\=5\sqrt{7}+4\sqrt{7}-3\sqrt{7}\\=6\sqrt{7}$

$=7\sqrt{3}+\sqrt{48}-\sqrt{768}\\=7\sqrt{3}+\sqrt{16\times3}-\sqrt{256\times3}\\=7\sqrt{3}+4\sqrt{3}-16\sqrt{3}\\=-5\sqrt{3}$

$=3\sqrt{2}+5\sqrt{8}-\sqrt{32}\\=3\sqrt{2}+5\sqrt{4\times2}-\sqrt{16\times2}\\=3\sqrt{2}+5\times2\sqrt{2}-4\sqrt{2}\\=3\sqrt{2}+10\sqrt{2}-4\sqrt{2}\\=9\sqrt{2}$

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Gema0007

Jawaban

2. 1,287

3. $\frac{ {81y}^{6} }{ {x}^{4} {z}^{8} }$

4.a. 6√7

4.b. -5√3

4.c. 9√2

Penyelesaian

$2.) \: \: \: {2}^{ - 2} + {3}^{ - 3} + {1}^{ - 4} = \frac{1}{ {2}^{2} } + \frac{1}{ {3}^{3} } + 1 = \frac{1}{4} + \frac{1}{27} + 1 = \frac{139}{108} = 1.287 \\ \\ 3.) \: \: \: {( {9x}^{ - 2} {y}^{3} {z}^{ - 4} )}^{2} = {(9 \times \frac{1}{ {x}^{2} } \times {y}^{3} \times \frac{1}{ {z}^{4} } ) }^{2} = {( \frac{ {9y}^{3} }{ {x}^{2} {z}^{4} } )}^{2} = \frac{ {81y}^{6} }{ {x}^{4} {z}^{8} } \\ \\ 4.a.) \: \: \: \sqrt{175} + 4 \sqrt{7} - \sqrt{63} = 5 \sqrt{7} + 4 \sqrt{7} - 3 \sqrt{7} = 6 \sqrt{7} \\ \\ 4.b.) \: \: \: 7 \sqrt{3} + \sqrt{48} - \sqrt{768} = 7 \sqrt{3} + 4 \sqrt{3} - 16 \sqrt{3} = - 5 \sqrt{3} \\ \\ 4.c) \: \: \: 3 \sqrt{2} + 5 \sqrt{8} - \sqrt{32} = 3 \sqrt{2} + 10 \sqrt{2} - {2}^{2} \sqrt{2} = 3 \sqrt{2} + 10 \sqrt{2} - 4 \sqrt{2} = 9 \sqrt{2}$

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