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PP
0
LH
1
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 12 2018
\(1^2+2^2+3^2+...+9^2=285\)
\(\Leftrightarrow\left(1^2+2^2+3^2+...+8^2+9^2\right).3^2=285.3^2\)
\(\Leftrightarrow1^2.3^2+2^2.3^2+3^2.3^2+...+8^2.3^2+9^2.3^2=285.9\)
\(\Leftrightarrow\left(1.3\right)^2+\left(2.3\right)^2+\left(3.3\right)^2+...+\left(8.3\right)^2+\left(9.3\right)^2=2565\)
\(\Leftrightarrow3^2+6^2+9^2+...+24^2+27^2=2565\)
DD
0
20 tháng 9 2019
\(a,\left[2^{17}+16^2\right]\cdot\left[9^{15}-3^{15}\right]\cdot\left[2^4-4^2\right]\)
\(=\left[2^{17}+16^2\right]\cdot\left[9^{15}-3^{15}\right]\cdot\left[16-16\right]\)
\(=\left[2^{17}+16^2\right]\left[9^{15}-3^{15}\right]\cdot0=0\)
\(b,\left[8^{2017}-8^{2015}\right]\cdot\left[8^{2014}\cdot8\right]\)
\(=8^{2015}\left[8^2-1\right]\cdot8^{2015}\)
\(=8^{2015}\cdot63\cdot8^{2015}=8^{4030}\cdot63\)sửa lại câu b , có vấn đề rồi
\(c,\frac{2^8+8^3}{2^5\cdot2^3}=\frac{2^8+\left[2^3\right]^3}{2^5\cdot2^3}=\frac{2^8+2^9}{2^8}=\frac{2^8\left[1+2\right]}{2^8}=3\)
2.a, \(2^6=\left[2^3\right]^2=8^2\)
Mà 8 = 8 nên 82 = 82 hay 26 = 82
b, \(5^3=5\cdot5\cdot5=125\)
\(3^5=3\cdot3\cdot3\cdot3\cdot3=243\)
Mà 125 < 243 nên 53 < 35
c, 26 = [23 ]2 = 82
Mà 8 > 6 nên 82 > 62 hay 26 > 62
d, 7200 = [72 ]100 = 49100
6300 = \(\left[6^3\right]^{100}\)= 216100
Mà 49 < 216 nên 49100 < 216100 hay 7200 < 6300
LT
0
EC
3 tháng 9 2016
Câu 1:
a) 2225 và 3150
Ta có:2225=(29)25=51225
3150=(36)25=72925
Vì 51225<72925
Suy ra: 2225<3150
EC
3 tháng 9 2016
Câu 2:
a)\(25^3:5^2=\left(5^2\right)^3:5^2=5^6:5^2=5^4\)
b)\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
c)\(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3+\frac{1}{4}:2=3+\frac{1}{8}=\frac{25}{8}\)
Câu 3:
a)\(9.3^3.\frac{1}{81}.3^2=3^2.3^3.3^2.\left(\frac{1}{3^4}\right)=3^7:3^4=3^3\)
b)\(4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:\left(2^3.\frac{1}{2^4}\right)=2^7:\frac{1}{2}=2^8\)
c)\(3^2.2^5.\left(\frac{2}{3}\right)^2=288.\frac{4}{9}=2^7\)
d)\(\left(\frac{1}{3}\right)^3.\frac{1}{3}.9^2=\left(\frac{1}{3}\right)^4.\left(3^2\right)^2=3^4.\left(\frac{1}{3}\right)^4=3^4:3^4=1\)
27 tháng 9 2016
- \(\frac{4^6.3^4.9^5}{6^{12}}=\frac{\left(2^2\right)^6.3^4.\left(3^2\right)^5}{\left(2.3\right)^{12}}=\frac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}=\frac{2^{12}.3^{14}}{2^{12}.3^{12}}=3^2=9\)
- \(\frac{3^{10}.11+9^5.5}{3^9.2^4}=\frac{3^{10}.11+\left(3^2\right)^5.5}{3^9.16}=\frac{3^{10}.11+3^{10}.5}{3^9.16}=\frac{3^{10}.\left(11+5\right)}{3^9.16}=\frac{3^{10}.16}{3^9.16}=3\)
- 2100 - 299 - 298 - ... - 22 - 2
= 2100 - (299 + 298 + ... + 22 + 2)
Đặt A = 299 + 298 + ... + 22 + 2
2A = 2100 + 299 + ... + 23 + 22
2A - A = (2100 + 299 + ... + 23 + 22) - (299 + 298 + ... + 22 + 2)
A = 2100 - 2
Ta có:
2100 - 299 - 298 - ... - 22 - 2
= 2100 - (2100 - 2)
= 2100 - 2100 + 2
= 0 + 2
= 2
- 38 : 36 + (22)4 : 29
= 32 + 28 : 29
\(=9+\frac{1}{2}\)
\(=\frac{18}{2}+\frac{1}{2}=\frac{19}{2}\)
DH
25 tháng 6 2017
a, \(\dfrac{4^2.4^3}{2^{10}}=\dfrac{4^5}{2^{10}}=\dfrac{\left(2^2\right)^5}{2^{10}}=\dfrac{2^{10}}{2^{10}}=1\)
b, \(\dfrac{2^7.9^3}{6^5.8^2}=\dfrac{2^7.\left(3^2\right)^3}{2^5.3^5.\left(2^3\right)^2}=\dfrac{2^7.3^6}{2^5.3^5.2^6}=\dfrac{3}{2^4}=\dfrac{3}{16}\)
c, \(\dfrac{9^7.5^6.125^9}{15^{15}.5^{18}}=\dfrac{3^{21}.5^6.5^{27}}{5^{15}.3^{15}.5^{18}}=\dfrac{3^{21}.5^{33}}{3^{15}.5^{33}}=3^6=729\)
d, \(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^9.\left(1+3.5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\dfrac{2.16}{3^2.5}=\dfrac{32}{45}\)
Chúc bạn học tốt!!!
2² + 4² + 6² + ... + 16² + 18²
= 4.(1 + 2² + 3² + ... + 8² + 9²)
= 4.285
= 1140
= 285 nha mình ghi nhầm thành 385