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NN
1
25 tháng 7 2016
160-(2^3.5^2-6.25)= 160-(8.25-150)=160-50=110
4.5^2-32.2^4= 4.25 - 32.16= 100-512=-412
5871:[928-(247-82).5]=5871:[(928-165).5=5871: ( 763.5)=5871:103=57
777:7+1331:11^3 = 111+1331:1331 = 111+1=112
17.85+15.17-120=17(85+15)-120=17.100-120=1700-120=1580
15.141+59.15=15(141+59)=15.200=3000
20-[30-(5-1)^2]=20-[30-4^2]=20-(30-16)=20-11=9
37-11.3.(24-23)+22.10= 37-33.1+220=37-253=-216
S
25 tháng 7 2016
160-(2^3.5^2-6.25)= 160-(8.25-150)=160-50=110
4.5^2-32.2^4= 4.25 - 32.16= 100-512=-412
5871:[928-(247-82).5]=5871:[(928-165).5=5871: ( 763.5)=5871:103=57
777:7+1331:11^3 = 111+1331:1331 = 111+1=112
17.85+15.17-120=17(85+15)-120=17.100-120=1700-120=1580 15.141+59.15=15(141+59)=15.200=3000
20-[30-(5-1)^2]=20-[30-4^2]=20-(30-16)=20-11=9
37-11.3.(24-23)+22.10= 37-33.1+220=37-253=-216
H9
HT.Phong (9A5)
CTVHS
9 tháng 7 2023
a) \(11^n=1331\)
\(\Rightarrow11^n=11^3\)
\(\Rightarrow n=3\)
b) \(n^3=125\)
\(\Rightarrow n^3=5^3\)
\(\Rightarrow n=5\)
c) \(5^4=n\)
\(\Rightarrow625=n\)
\(\Rightarrow n=625\)
d) \(\left(n+1^2\right)=9\)
\(\Rightarrow n+1=9\)
\(\Rightarrow n=9-1\)
\(\Rightarrow n=8\)
NM
9 tháng 7 2023
a) 11^n = 1331
⇒ 11^n = 11^3
⇔ n = 3
b) n^ 3 = 125
⇒ n^3 = 5^3
⇔ n = 5
c) 5^4 = n
⇒ n = 625
d) ( n + 1^2 ) = 9
⇒ ( n + 1 ) = 9
⇒ n = 8
LT
1
22 tháng 6 2015
1)\(8.2^n=128\Rightarrow2^n=128:8\Rightarrow2^n=16\Rightarrow2^n=2^4\Rightarrow n=4\)
2)\(121.11^n=1331\Rightarrow11^n=1331:121\Rightarrow11^n=11\Rightarrow n=1\)
3)\(7^n:49=343\Rightarrow7^n:7^2=7^3\Rightarrow7^n=7^3.7^2\Rightarrow7^n=7^5\Rightarrow n=5\)
nhớ **** cho mình nhé
TP
2
30 tháng 1
\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+3+...+20\right)\)
\(=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{20}\cdot\dfrac{20\cdot21}{2}\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{21}{2}\)
\(=\dfrac{2+3+4+...+21}{2}=\dfrac{\left(21+2\right)+\left(3+20\right)+...+\left(10+13\right)+\left(11+12\right)}{2}\)
\(=\dfrac{23+23+...+23}{2}=\dfrac{23\cdot10}{2}=23\cdot5=115\)
HH
0
\(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+20}\)
\(=1+\dfrac{1}{2\cdot\dfrac{3}{2}}+\dfrac{1}{3\cdot\dfrac{4}{2}}+...+\dfrac{1}{20\cdot\dfrac{21}{2}}\)
\(=\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{20\cdot21}\)
\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{20\cdot21}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{20}-\dfrac{1}{21}\right)\)
\(=2\left(1-\dfrac{1}{21}\right)=2\cdot\dfrac{20}{21}=\dfrac{40}{21}\)