\(P=\dfrac{4\cdot36^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{4\cdot\left(2^2\cdot3^2\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

a.

=(5^7+5^9)(6^8+6^10)(16-4^2)

=(5^7+5^9)(6^8+6^10)(16-16)

=(5^7+5^9)(6^8+6^10).0

=0

a, =5^16 . 6^18 . 0 

=0

chúc bn hc tốt

a = 0

b = 24

c = 1032

d = 7

\(4^5\cdot9^4-2.6^9=\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9=2^{10}.3^8-2.2^9.3^9=2^{10}.3^8-2^{10}.3^9=3^8-3^9=-13122\)
\(2^{10}.3^8+6^8.20=2^{10}.3^8+\left(2.3\right)^8.2^2.5=2^{10}.3^8+2^8.3^8.2^2.5=2^{10}.3^8+2^{10}.3^8.5=2^{10}.3^8.\left(1+5\right)=2^{10}.3^8.6=2^{10}.3^8.2.3=2^{11}.3^9\)

a)\(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}=\sqrt{\dfrac{1}{2}\left(16+8\sqrt{3}\right)}-\sqrt{\dfrac{1}{2}\left(16-8\sqrt{3}\right)}\)

\(=\sqrt{\dfrac{1}{2}\left(2+2\sqrt{3}\right)^2}-\sqrt{\dfrac{1}{2}\left(2-2\sqrt{3}\right)^2}\)\(=\sqrt{\dfrac{1}{2}}\left(2+2\sqrt{3}\right)-\sqrt{\dfrac{1}{2}}\left(2\sqrt{3}-2\right)=2\sqrt{2}\)

b)\(=\dfrac{\sqrt{16+2.4\sqrt{5}+5}}{4+\sqrt{5}}.\sqrt{\left(2-\sqrt{5}\right)^2}\)\(=\dfrac{\sqrt{\left(4+\sqrt{5}\right)^2}}{4+\sqrt{5}}\left|2-\sqrt{5}\right|=\sqrt{5}-2\)

a) Ta có: \(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}\)

\(=\sqrt{6}+\sqrt{2}-\sqrt{6}+\sqrt{2}\)

\(=2\sqrt{2}\)

b) Ta có: \(\dfrac{\sqrt{21+8\sqrt{5}}}{4+\sqrt{5}}\cdot\sqrt{9-4\sqrt{5}}\)

\(=\left(4+\sqrt{5}\right)\left(4-\sqrt{5}\right)\)

=16-5=11