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1) \(2^{x+1}\cdot2^{2014}=2^{2015}\)\(\Leftrightarrow2^{2014x+2014}=2^{2015}\)\(\Leftrightarrow2014x+2014=2015\)\(\Leftrightarrow x=\frac{1}{2014}\)
2) \(7x-2x=\frac{6^{17}}{6^{15}}+\frac{44}{11}\)\(\Leftrightarrow5x=6^2+4=36+4=40\)\(\Leftrightarrow x=\frac{40}{5}=8\)
3) \(3^x=9\)\(\Leftrightarrow3^x=3^2\)\(\Leftrightarrow x=2\)
4) \(7x-x=\frac{5^{21}}{5^{19}}+3\cdot2^2-7^0\)\(\Leftrightarrow6x=5^2+3\cdot4-1=25+12-1=36\)\(\Leftrightarrow x=6\)
5) \(4^x=64\)\(\Leftrightarrow4^x=4^3\)\(\Leftrightarrow x=3\)
6) \(9^{x-1}=9\)\(\Leftrightarrow x-1=1\)\(\Leftrightarrow x=0\)
7) \(\frac{2^x}{2^5}=1\)\(\Leftrightarrow2^{x-5}=2^0\)\(\Leftrightarrow x-5=0\)\(\Leftrightarrow x=5\)
8) \(\left(5x-9\right)^3=216\)\(\Leftrightarrow\left(5x-9\right)^3=6^3\)\(\Leftrightarrow5x-9=6\)\(\Leftrightarrow5x=15\)\(\Leftrightarrow x=3\)
9) \(5\cdot3^{7x-11}=135\)\(\Leftrightarrow5.3^{7x-11}=5.3^3\)\(\Leftrightarrow3^{7x-11}=3^3\)\(\Leftrightarrow7x-11=3\)\(\Leftrightarrow7x=14\Leftrightarrow x=2\)
10) \(2.3^x=19\cdot3^8-81^2\)\(\Leftrightarrow2.3^x=19\cdot3^8-3^8=18.3^8=2.3^{11}\)\(\Leftrightarrow3^x=3^{11}\Leftrightarrow x=11\)
Đây là cách làm của mình. Bạn có thể chỉnh sửa tuỳ ý theo cách làm của bạn nhé ^^
Học tốt ^3^

\(\frac{\left(\frac{5^9}{5^{17}}+3\right)}{7}\)= \(\frac{\left(\frac{1}{5^8}+3\right)}{7}\)= \(\frac{\left(\frac{1171876}{390625}\right)}{7}\)= 0.4285717943
\(\frac{7^9}{7^7}-3^2+2^3.5^2\)= \(7^2-3^2+2^3.5^2\)= \(7^2-3^2+200\)= \(40+200=240\)
\(\frac{5^9}{5^7}+\frac{70}{14}-20\)= \(5^2+5-20\)=\(30-20=10\)
\(\frac{2^{28}}{2^{26}}+5^1.3^2-7^2\)= \(2^2+5.3^2-7^2\)= \(4+45-49\)= \(0\)


\(A=2^0+2^1+2^2\)\(+2^3+...+\)\(2^{50}\)
\(2A=2+2^2+2^3+...+2^{51}\)
\(2A-A=A=2^{51}-2^0\)
\(B=5+5^2+5^3+...+5^{99}+5^{100}\)
\(5B=5^2+5^3+5^4+...+5^{100}+5^{101}\)
\(5B-B=4B=5^{101}-5\)
\(B=\frac{5^{101}-5}{4}\)
\(C=3-3^2+3^3-3^4+...+\)\(3^{2007}-3^{2008}+3^{2009}-3^{2010}\)
\(3C=3^2-3^3+3^4-3^5+...-3^{2008}+3^{2009}-3^{2010}+3^{2011}\)
\(3C+C=4C=3^{2011}+3\)
\(C=\frac{3^{2011}+3}{4}\)
\(S_{100}=5+5\times9+5\times9^2+5\times9^3+...+5\times9^{99}\)
\(S_{100}=5\times\left(1+9+9^2+9^3+...+9^{99}\right)\)
\(9S_{100}=5\times\left(9+9^2+9^3+...+9^{99}+9^{100}\right)\)
\(9S_{100}-S_{100}=8S_{100}=5\times\left(9^{100}-1\right)\)
\(S_{100}=\frac{5\times\left(9^{100}-1\right)}{8}\)
A=20+21+22+23+...++23+...+250250
2�=2+22+23+...+2512A=2+22+23+...+251
2�−�=�=251−202A−A=A=251−20
�=5+52+53+...+599+5100B=5+52+53+...+599+5100
5�=52+53+54+...+5100+51015B=52+53+54+...+5100+5101
<...

93.76+93.52-93
=93(76+52-1)
=729(76+25-1)
=729.100
=72900
vậy kết quả là 72900
\(9^3.76+9^3.5^2-9^3\)
\(=9^3.\left(76+5^2-1\right)\)
\(=9^3.\left(76+25-1\right)\)
\(=9^3.\left(101-1\right)\)
\(=9^3.100\)
\(=729.100\)
\(=72900\)

1.
B= 9+99+999+..+999...9(50 chữ số 9)
B= 10-1+100-1+1000-1+...+100...0(50 chữ số 0)-1
B=[10+100+1000+...+100...0(50 chữ số 0)]-(1+1+1+...+1)(50 số hạng 1)
B= 111...10(50 chữ số 1) - 50
B = 111...1060 (48 chữ số 1)
1. Tính
A = 9 + 99 + 999 + 9999
A = 108 + 999 + 9999
A = 1170 + 9999
A = 11106

a, 1024 : (17.25 + 15.25) = 210 : [25.(17+15)] = 210 : (25. 25) = 210 : 210 =1
b, (5.35 + 17.34) : 62 = 34.(5.3 + 17) : (2.3)2 = 34.25 : (22.32) = 32.23=72
c, (23.94 + 93.45) : (92.10 - 92) = 94.(23 + 5) : [92.(10-1)] = 94.13 : 93 = 9.13 =117
Đúng thì nha
\(9^5+3^{12}=\text{59049}+\text{531441}=\text{590490}\)