kak bantu no 1 sampe 19 dongg lagi butuhh bangett

Berikut ini adalah pertanyaan dari jermansanda pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Kak bantu no 1 sampe 19 dongg lagi butuhh bangett

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawaban:

$1) \: \: {81}^{ \frac{3}{4} } = { {(3}^{4}) }^{ \frac{3}{4} } = {3}^{3} = 27 \: (b)$

$2) \: \sqrt{300} = \sqrt{ {10}^{2} \times 3 } = 10 \sqrt{3} \: \: (a)$

$\frac{28}{ \sqrt{7} } = \frac{28}{ \sqrt{7} } \times \frac{ \sqrt{7} }{ \sqrt{7} } = \frac{28 \sqrt{7} }{7} = 4 \sqrt{7} \: (c)$

$4) \: \: \sqrt{10} \: (b) \: karena \: tidak \: dapat \: disederhanakan \: lagi$

$5) \: \: 2 \sqrt{200} \div \sqrt{8} \times \sqrt{50} = \sqrt{ {10}^{2} \times 2 } \div \sqrt{ {2}^{2} \times 2 } \times \sqrt{ {5}^{2} \times 2} = 10 \sqrt{2} \div 2 \sqrt{2} \times 5 \sqrt{2} = 25 \sqrt{2} \: \: (b)$

$6) \: \: \frac{6}{ \sqrt{5} + \sqrt{3} } = \frac{6}{ \sqrt{5} + \sqrt{3} } \times \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} - \sqrt{3} } = \frac{6 \sqrt{5} - 6 \sqrt{3} }{5 - 3} = \frac{6 \sqrt{5} - 6 \sqrt{3} }{2} = 3 \sqrt{5} - 3 \sqrt{3} \: \: (c)$

$7) \: \: (64)^{ - \frac{1}{3} } = {2}^{6 \times - \frac{1}{3} } = {2}^{ - 2} = \frac{1}{4} \: \: (b)$

$9) = \frac{1}{ {2}^{2} } + \frac{1}{ {3}^{3} } + \frac{1}{ {1}^{4} } = \frac{1}{4} + \frac{1}{27} + \frac{1}{1} = \frac{27}{108} + \frac{4}{108} + \frac{108}{108} = \frac{139}{108} = 1\frac{39}{108} \: \: (d)$

10) x^2 -11x + 10 = (x-10) (x-1)

x - 10 = 0 x - 1 = 0

x = 10 x = 1 HP = {1, 10} (A)

11) X^2 - 21X - 72 = (X - 24) (X + 3)

X - 24 = 0 X + 3 = 0

X = 24 X = -3 HP = {-3,24} (B)

$12) \: \: {(32)}^{ \frac{1}{5} } = { {(2}^{5}) }^{ \frac{1}{5} } = {2}^{5 \times \frac{1}{5} } = {2}^{1} = 2 \: \: (d)$

$13) \: \: = 2.308 \times {10}^{7} \: \: (b)$

$14) \: \: \sqrt{175} + 4 \sqrt{7} - \sqrt{63} = \sqrt{ {5}^{2} \times 7 } + 4 \sqrt{7} - \sqrt{ {3}^{2} \times 7 } = 5 \sqrt{7} + 4 \sqrt{7} - 3 \sqrt{7} = 6 \sqrt{7} \: \: (a)$

$15) \: \: = \frac{(2 + \sqrt{8}) \sqrt{6} }{ \sqrt{6} \times \sqrt{6} } = \frac{2 \sqrt{6} + 4 \sqrt{3} }{6} = \frac{1}{3} \sqrt{6} + \frac{2}{3} \sqrt{3} \: \: (c)$

$16) \: \: {3}^{9 - 3x} = {3}^{3}$

9 - 3X = 3

3X = 6

X = 2 (A)

$17) \: \: = \frac{ \sqrt{3} }{ \sqrt{2} - \sqrt{5} } \times \frac{ \sqrt{2} + \sqrt{5} }{ \sqrt{2} + \sqrt{5} } = \frac{ \sqrt{6} + \sqrt{15} }{ - 3} = - \frac{1}{3} \sqrt{6} - \frac{1}{3} \sqrt{15} = - \frac{1}{3}( \sqrt{6} + \sqrt{15)} \: \: (c)$