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a. \(5^{127}=5.5^{126}=5.125^{72}>119^{72}\)
\(\Rightarrow5^{217}>119^{72}\)
b. \(2^{1000}=\left(2^5\right)^{200}=32^{200}\)
\(5^{400}=\left(5^2\right)^{200}=25^{200}\)
\(\Rightarrow2^{1000}>5^{400}\)
c. \(9^{12}=\left(3^2\right)^{12}=3^{24}\)
\(27^7=\left(3^3\right)^7=3^{21}\)
\(\Rightarrow9^{12}>27^7\)
d. \(125^{80}=\left(5^3\right)^{80}=5^{240}\)
\(25^{118}=\left(5^2\right)^{118}=5^{236}\)
\(\Rightarrow125^{80}>25^{118}\)
e. \(5^{40}=\left(5^4\right)^{10}=625^{10}\)
\(\Rightarrow5^{40}>620^{10}\)
f. \(27^{11}=\left(3^3\right)^{11}=3^{33}\)
\(81^8=\left(3^4\right)^8=3^{32}\)
\(\Rightarrow27^{11}>81^8\)
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a) \(2^x=16=2^4\Rightarrow x=4\)
b) \(x^3=27=3^3\Rightarrow x=3\)
c) \(x^{50}=x\Rightarrow x\left(x^{49}-1\right)=0\Rightarrow x=0\) hay \(x=1\)
d) \(\left(x-2\right)^2=16=4^2\Rightarrow x-2=4\) hay \(x-2=-4\)
\(\Rightarrow x=6\) hay \(x=-2\)
a) \(2^{300}=2^{3.100}=8^{100}\)
\(3^{200}=3^{2.100}=9^{100}\)
vì \(8^{100}< 9^{100}\)
\(\Rightarrow2^{300}< 3^{200}\)
b) \(3^{500}=3^{5.100}=243^{100}\)
\(7^{300}=7^{3.100}=343^{100}\)
vì \(243^{100}< 343^{100}\)
\(\Rightarrow3^{500}< 7^{300}\)
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\(A=1+2+2^2+...+2^{2022}\)
\(\Rightarrow2A=2+2^2+...+2^{2023}\)
\(\Rightarrow2A-A=2^{2023}-1\)
\(\Rightarrow A=2^{2023}-1\)
\(\Rightarrow A< 2^{2023}=2^2.2^{2021}=4.2^{2021}< 5^{2021}\)
\(\Rightarrow A< B\)
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a)
Ta có : A = 275 = (33)5 = 315
B = 2433 = (35)3 = 315
Vì 315 = 315 => A = B
b )
Ta có : A = 2300 = (23)100 = 8100
B = 3200 = (32)100 = 9100
Vì 8100 < 9100 => A<B
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\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}\)
Có \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
......
\(\frac{1}{2011^2}< \frac{1}{2010.2011}\)
=> \(A< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2010.2011}\)
=> \(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2010}-\frac{1}{2011}\)
=> \(A< 1-\frac{1}{2011}< 1\)
=> A < 1
=> A < B
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+5+5^2+5^3+..+5^{100}\)
\(5A=5+5^2+5^3+..+5^{101}\)
\(A=\frac{5^{101}-1}{4}\)\(SUYRA\) \(A< B\)
\(A=5^0+5+5^2+...+5^{100}.\)
\(\Rightarrow5A=5+5^2+5^3+...+5^{101}\)
\(\Rightarrow5A-A=4A=\left(5+5^2+5^3+...+5^{101}\right)-\left(5^0+5+5^2+...+5^{100}\right)\)
\(=5^{101}-1\)
\(\Rightarrow A=\frac{5^{101}-1}{4}\)
Còn lại tự lm nha bn
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\(\text{#040911}\)
\(a,\)
\(202^{303}\text{ và }303^{202}\)
Ta có:
\(202^{303}=\left(202^3\right)^{101}=\left(101^3\cdot2^3\right)^{101}=\left(101^3\cdot8\right)^{101}\)
\(303^{202}=\left(303^2\right)^{101}=\left(101^2\cdot3^2\right)^{101}=\left(101^2\cdot9\right)^{101}\)
Ta có:
\(8\cdot101^3=8\cdot101\cdot101^2=808\cdot101^2\)
Vì \(808>9\)
\(\Rightarrow808\cdot101^2>9\cdot101^2\)
\(\Rightarrow202^{303}>303^{202}\)
\(b,\)
Ta có:
\(11^{1979}< 11^{1980}=\left(11^3\right)^{660}=1331^{660}\\ 37^{1320}=\left(37^2\right)^{660}=1369^{660}\\ \text{Vì }1331< 1369\\ \Rightarrow1331^{660}< 1369^{660}\\ \Rightarrow11^{1979}< 37^{1320}\)
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Vì B là phân số bé hơn 1 nên cộng cùng một số vào tử và mẫu của phân số đó thì giá trị của B sẽ tăng thêm, ta có:
\(B=\frac{2009^{2009}+1}{2009^{2010}+1}< \frac{2009^{2009}+1+2008}{2009^{2010}+1+2008}=\frac{2009^{2009}+2009}{2009^{2010}+2009}=\frac{2009\left(2009^{2008}+1\right)}{2009\left(2009^{2009}+1\right)}=\frac{2009^{2008}+1}{2009^{2009}+1}=A\)
Vậy B < A