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AH
Akai Haruma
Giáo viên
21 tháng 10 2023
1.
$(5^{1986}-5^{1985}):5^{1985}=5^{1985}(5-1):5^{1985}=5-1=4$
2.
\((7^{846}+7^{847}):7^{846}=7^{846}(1+7):7^{846}=1+7=8\)
3.
\((9^{2018}-3^{4036}):6^{2006}=[(3^2)^{2018}-3^{4036}]:6^{2006}\)
$=(3^{4036}-3^{4036}):6^{2006}=0:6^{2006}=0$
4.
$(7^{80}.8^{70}-56^{70}):56^{70}$
$=[7^{10}(7.8)^{70}-56^{70}]:56^{70}$
$=[7^{10}.56^{70}-56^{70}]:56^{70}$
$=56^{70}(7^{10}-1):56^{70}=7^{10}-1$
AH
Akai Haruma
Giáo viên
21 tháng 10 2023
5.
$4^{4016}:(4^{4017}-4^{4016})=4^{4016}:[4^{4016}(4-1)]$
$=4^{4016}:4^{4016}:3=1:3=\frac{1}{3}$
6.
$(12^{206}.2^{207}-24^{206}):24^{206}$
$=(12^{206}.2^{206}.2-24^{206}):24^{206}$
$=[(12.2)^{206}.2-24^{206}]:24^{206}$
$=(24^{206}.2-24^{206}):24^{206}$
$=24^{206}(2-1):24^{206}=2-1=1$
7.
$(5^2-24)^{8946}+4^{30}:2^{60}=1^{8946}+(2^2)^{30}:2^{60}$
$=1+2^{60}:2^{60}=1+1=2$
8.
$(37.8^{1007}-7.2^{3021}):8^{1007}=[37.8^{1007}-7.(2^3)^{1007}]:8^{1007}$
$=[37.8^{1007}-7.8^{1007}]:8^{1007}$
$=8^{1007}(37-7):8^{1007}=37-7=30$
29 tháng 5 2018
\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)
7 tháng 5 2020
\(\frac{-7}{8}\cdot\frac{8}{9}=\frac{112}{x}\)
=> \(\frac{-7}{9}=\frac{112}{x}\)
=> \(\frac{-7}{9}=\frac{112\div\left(-16\right)}{x\div\left(-16\right)}\)
=> \(9=x\div\left(-16\right)\)
=> \(x=-144\)
Hoặc : \(-7x=9\cdot112\)
=> \(-7x=1008\)
=> \(x=-144\)
NP
1
HN
0
18 tháng 6 2017
Ta có : $16A=\dfrac{16}{6.10}+\dfrac{16}{7.9}+\dfrac{16}{8.8}+\dfrac{16}{9.7}+\dfrac{16}{10.6}$
$=>16A=\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{7}+\dfrac{1}{10}+\dfrac{1}{6}$
$=>16A=2.(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})$
$=>A=\dfrac{1}{8}(dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})$
18 tháng 6 2017
\(A=\dfrac{1}{6.10}+\dfrac{1}{7.9}+\dfrac{1}{8.8}+\dfrac{1}{9.7}+\dfrac{1}{10.6}\)
\(16A=\dfrac{16}{6.10}+\dfrac{16}{7.9}+\dfrac{16}{8.8}+\dfrac{16}{9.7}+\dfrac{16}{10.6}\)
\(16A=\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{7}+\dfrac{1}{10}+\dfrac{1}{6}\)
\(16A=2\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\)
\(A=2:16\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\)
\(A=\dfrac{1}{8}\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\left(đpcm\right)\)
DA
0
15