tolong di jawab sebelum jam 2 please.. semua nya.. makasih ;) ampun kalau mtk yang asal jawab hati aja.. :v

Bentuk Akar dan Pangkat.

Nomor 23.
[tex]\displaystyle \frac{3\sqrt{6}+\sqrt{5}}{4\sqrt{8}}\\ =\frac{3\sqrt{6}+\sqrt{5}}{4\sqrt{8}}~\frac{4\sqrt{8}}{4\sqrt{8}}\\ =\frac{48\sqrt{3}+8\sqrt{10}}{128}\\ =\frac{6\sqrt{3}+\sqrt{10}}{16}[/tex]

Nomor 25
[tex]\displaystyle \sqrt{a+b+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}\\ \sqrt{14+2\sqrt{24}}\\ =\sqrt{12+2+2\sqrt{12(2)}}\\ =\sqrt{2}+\sqrt{12}[/tex]

Nomor 1.a
[tex]\displaystyle \left ( \frac{a}{b} \right )^n=\frac{a^n}{b^n}\\\\ \frac{1}{a^n}=a^{-n}\\ \left ( -\frac{1}{2p} \right )^5=-\frac{1}{32p^5}=-32p^{-5}[/tex]

Nomor 1.c
[tex]\displaystyle \left ( -\frac{1}{3k} \right )^2\left ( -\frac{2}{3k} \right )^3\\ =\frac{1}{9k^2}~\frac{8}{27k^3}\\ =\frac{8}{243k^5}[/tex]

Nomor 2.b
(a - b)² = a² - 2ab + b²
(3√3 - √5)² = (3√3)² - 2(3√3)√5 + (√5)² = 32 - 6√15

Nomor 2.e
(√6 + √8)(√6 - √3)
= 6 - √18 + √48 - √24
= 6 - 3√2 + 4√3 + 2√6

Nomor 2.f
(2√3 - √6)(√8 - √2)
= 4√6 - 2√6 - 4√3 + 2√3
= 2(√6 - √3)

Nomor 3.b
[tex]\displaystyle (a^b)^n=a^{bn}\\ 5\left ( 64^{\frac{2}{3}} \right )-2\left ( 64^{\frac{2}{3}} \right )+4\left ( 64^{\frac{2}{3}} \right )\\ =\left ( 64^{\frac{2}{3}} \right )\left ( 5-2+4 \right )\\ =7\left ( 4^{3\times \frac{2}{3}} \right )\\=112[/tex]

Nomor 3.e
3 ∛40 + 4 ∛320 - 2 ∛125
= 6 ∛5 + 16 ∛5 - 10
= 22 ∛5 - 10