Question

[Bùi Ánh Ngọc]

đề hsg nên giúp vài câu vs ạaa

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 3:

a: \(\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}\)

\(=\dfrac{11\cdot3^{29}-3^{30}}{3^{28}\cdot2^2}=\dfrac{3^{29}\left(11-3\right)}{3^{28}\cdot4}\)

\(=3\cdot\dfrac{8}{4}=3\cdot2=6\)

b: \(\dfrac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}\)

\(=\dfrac{2^{10}\cdot3^9\cdot3-2^{10}\cdot3^9\cdot1}{2^9\cdot3^{10}}\)

\(=\dfrac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}\)

\(=\dfrac{2}{3}\cdot2=\dfrac{4}{3}\)

c: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

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

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

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

Bài 4:

a: \(\dfrac{10^2+11^2+12^2}{13^2+14^2}\)

\(=\dfrac{100+121+144}{169+195}\)

\(=\dfrac{365}{365}=1\)

b: \(\dfrac{2^3\cdot9^4+9^3\cdot45}{9^2\cdot10-9^2}\)

\(=\dfrac{2^3\cdot9^4+9^3\cdot9\cdot5}{9^2\left(10-1\right)}\)

\(=\dfrac{9^4\left(2^3+5\right)}{9^3}=9\cdot\left(8+5\right)=9\cdot13=117\)

c: \(\left[\dfrac{3^{14}\cdot69+3^{14}\cdot12}{3^{16}}-7\right]:2^4\)

\(=\left(\dfrac{3^{14}\left(69+12\right)}{3^{14}\cdot9}-7\right):16\)

\(=\dfrac{\left(\dfrac{81}{9}-7\right)}{16}=\dfrac{2}{16}=\dfrac{1}{8}\)

d: \(24^4:3^4-32^{12}:16^{12}\)

\(=\left(\dfrac{24}{3}\right)^4-\left(\dfrac{32}{16}\right)^{12}\)

\(=8^4-2^{12}\)

\(=2^{12}-2^{12}=0\)

e: \(2010^{2010}\left(7^{10}:7^8-3\cdot2^4-2^{2010}:2^{2010}\right)\)
\(=2010^{2010}\left(7^2-3\cdot16-1\right)\)

\(=2010^{2010}\cdot\left(49-48-1\right)\)

=0

f: \(\dfrac{2^{100}+2^{101}+2^{102}}{2^{97}+2^{98}+2^{99}}\)

\(=\dfrac{2^{100}\left(1+2+2^2\right)}{2^{97}\left(1+2+2^2\right)}\)

\(=\dfrac{2^{100}}{2^{97}}=2^3=8\)

Answer 2

[Bùi Ánh Ngọc]

thankss