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a.
A = 72^3.54^2/108^4
A = (72.54:108)^2:108^2.72
A = 36^2:108^2.72
A = (36:108)^2.72
A = (1/3)^2.72
A = 72/9
A = 8
B = \(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
B = \(\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\)
B = \(\frac{3^{10}.16}{3^9.16}\)
B = 3
a, \(A=\dfrac{3^{10}.11+3^{10}.5}{3^9.2^4}=\dfrac{3^{10}.\left(11+5\right)}{3^9.2^4}\)
\(=\dfrac{3^{10}.2^4}{3^9.2^4}=3\)
b, \(B=\dfrac{2^{10}.13+2^{10}.65}{2^8.104}=\dfrac{2^{10}.78}{2^8.104}\)
\(=\dfrac{2^2.3}{4}=3\)
c, \(C=\dfrac{4^9.36+64^4}{16^4.100}=\dfrac{\left(2^2\right)^9.36+\left(2^6\right)^4}{\left(2^4\right)^4.100}\)
\(=\dfrac{2^{18}.36+2^{24}}{2^{16}.100}=\dfrac{2^{18}.\left(36+2^6\right)}{2^{16}.100}\)
\(=\dfrac{2^4.100}{100}=2^4=16\)
Câu d làm tương tự! Chúc bạn học tốt!!!
b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
c ) Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
=> A < 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100= 1 - 1/100 = 99/100 < 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)< 1
b, \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\)\(\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
c,Ta thấy
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
e, \(E=\dfrac{4^6.3^4.9^5}{6^{12}}=\dfrac{\left(2^2\right)^6.3^4.\left(3^2\right)^5}{2^{12}.3^{12}}\)
\(=\dfrac{3^4.3^{10}}{3^{12}}=3^2=9\)
f, \(F=\dfrac{2^{13}+2^5}{2^{10}+2^2}=\dfrac{2^5.\left(2^8+1\right)}{2^2.\left(2^8+1\right)}=2^3=8\)
g, \(G=\dfrac{21^2.141.125}{35^{5.6}}=\dfrac{3^2.7^2.47.3.5^3}{5^{30}.7^{30}}\)
\(=\dfrac{3^3.47}{5^{27}.7^{28}}\)(bạn xem lại đề nha)
Các câu còn lại làm tương tự! Chúc bạn học tốt!!!
Câu a:
(7\(^{2017}\) - 7\(^{2015}\)) ⋮ (7\(^{2014}\).7)
= 7\(^{2015}\).(7\(^2\) - 1) ⋮ 7\(^{2015}\)
= (7\(^{2015}\) : 7\(^{2015}\)).(49 - 1)
= 1.48
= 48
Câu b:
(2\(^{13}\) + 2\(^5\)) : (2\(^{10}\) + 2\(^2\))
= 2\(^5.\)(2\(^8\) + 1):[2\(^2\).(2\(^8\) + 1)]
= (2\(^5:2^2\)).[(\(2^8+1):\left(2^8+1\right)\)]
= 2\(^3\).1
= 8
Câu b:
B = (3\(^{10}\).11 + 3\(^{10}\).5) : (3\(^9\).2\(^4\))
B = 3\(^{10}\).(11 + 5): (3\(^9\).2\(^4\))
B = 3\(^{10}\).16:(3\(^9\).2\(^4\))
B = (3\(^{10}\) : 3\(^9\)).(16 : 2\(^4\))
B = 3.1
B = 3
Câu c:
C = (2\(^{10}\).13 + 2\(^{10}\).65) : (2\(^8\).104)
C = 2\(^{10}\).(13 + 65):(2\(^8\).104)
C = 2\(^{10}\).78 : (2\(^8\).104)
C = (2\(^{10}\) : 2\(^8\)).(78 : 104)
C = 2\(^2\).\(\frac34\)
C = 3
d) D = (72^3 . 54^2) : 108^4
D = (72.54 :108) :(108^2).72
D = 36^2: 108^2.72
D = (36 : 108)^2.72
D = \(\frac{1}{3^2}\).72
D = 8
e)
E =(21^2 . 14 . 125) : (35^5 . 6)
E = 3^2.7^2.2.7.5^3: 5^5:7^5:6
E = (3^2.2:6).(7^2.7:7^5).(5^3:5^5)
E = (9.2:6).(7^3:7^5).\(\frac{1}{5^2}\)
E = 18 : 6 . \(\frac{1}{7^2}\).\(\frac{1}{5^2}\)
E = 3.\(\frac{1}{35^2}\)
E = \(\frac{3}{1225}\)