BANTU SOAL MAT KELAS 9

BANTU SOAL MAT KELAS 9

Jawab:

1. 13 x 12⁵

2. 3√2 – 2√3

3. 2√6

4. $frac{3}{2}(3 - sqrt{5} )$

5. 9√2 + 12

Penjelasan dengan langkah-langkah:

1. gunakan sifat eksponen yaitu aⁿ x bⁿ = (a x b)ⁿ maka

4⁵ x 3⁵ + 12⁶

= (4 x 3)⁵ + 12⁶

= 12⁵ + 12⁶

= 12⁵ + 12 x 12⁵

= (1 + 12) x 12⁵

= 13 x 12⁵

Jadi, bentuk sederhanya adalah 13 x 12⁵

2. gunakan sifat akar √(a x b) = √a x √b maka

3√72 + √108 – 2√48 – 3√50

= 3 x √(36 x 2) + √(36 x 3) – 2 x √(16 x 3) – 3 x √(25 x 2)

= 3 x √36 x √2 + √36 x √3 – 2 x √16 x √3 – 3 x √25 x √2

= 3 x 6 x √2 + 6 x √3 – 2 x 4 x √3 – 3 x 5 x √2

= 18√2 + 6√3 – 8√3 – 15√2

= 3√2 – 2√3

3. kalikan sekawan

$frac{12}{sqrt{6} }$

= $frac{12}{sqrt{6}}$ x $frac{sqrt{6}}{sqrt{6}}$

= $frac{12sqrt{6}}{6}$

= 2√6

4. $frac{6}{3 + sqrt{5}}$

= $frac{6}{3 + sqrt{5}}$ x $frac{3 - sqrt{5} }{3 - sqrt{5}}$

= $frac{6(3 - sqrt{5})}{3^{2} - (sqrt{5} )^{2}}$

= $frac{6(3 - sqrt{5})}{9 - 5}$

= $frac{6(3 - sqrt{5})}{4}$

= $frac{3}{2}(3 - sqrt{5} )$

5. $frac{6}{3sqrt{2} - 4 }$

= $frac{6}{3sqrt{2} - 4}$ x $frac{3sqrt{2} + 4}{3sqrt{2} + 4}$

= $frac{6(3sqrt{2} + 4)}{(3sqrt{2} )^{2} - 4^{2}}$

= $frac{6(3sqrt{2} + 4)}{18 - 16}$

= $frac{6(3sqrt{2} + 4)}{2}$

= 3(3√2 + 4)

= 9√2 + 12