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Ta có :
\(16^{15}=\left(4^2\right)^{15}=4^{30}\); \(4^{32}\)
Vì \(4^{30}< 4^{32}\)
=> \(16^{15}< 4^{32}\)
k mik nha
Ta có:
\(VT:\frac{4^{15}}{7^{30}}=\frac{\left(2^2\right)^{15}}{7^{30}}=\frac{2^{30}}{7^{30}}\)
\(VP:\frac{8^{10}\cdot3^{30}}{7^{30}.4^{15}}=\frac{\left(2^3\right)^{10}.3^{30}}{7^{30}.\left(2^2\right)^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{3^{30}}{7^{30}}\)
Ta thấy :\(\frac{2^{30}}{7^{30}}vs\frac{3^{30}}{7^{30}}\)có:
\(\orbr{\begin{cases}2^{30}< 3^{30}\\7^{30}=7^{30}\end{cases}\Rightarrow\frac{2^{30}}{7^{30}}< \frac{3^{30}}{7^{30}}\Leftrightarrow\frac{4^{15}}{7^{30}}< \frac{8^{10}.3^{30}}{7^{30}.4^{15}}}\)
Chúc bn hok tốt
Ta có:
3160 < 3260
3160 < (25)60
3160 < 2300
1774 > 1674
1774 > (24)74
1774 > 2296
Ta thấy:
2296 < 1714 < 3160 < 2300
Vậy 3160 > 1774
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
\(2^{24}=(2^3)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
vì \(8^8< 9^8\Rightarrow2^{24}< 3^{16}\)
\(2^{24}=\left(2^3\right)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
\(8< 9\)
\(\Rightarrow8^9< 9^9\)
\(\Rightarrow2^{24}< 3^{16}\)
ADTCDTSBN
có: \(\frac{a+2001}{b+2001}=\frac{a}{b}=\frac{2001}{2001}=1\)
\(\Rightarrow\frac{a}{b}=\frac{a+2001}{b+2001}\)
ta xét tích
a( b +2001) = ab + 2001a
b(a + 2001) = ab + 2001b
vì b > 0 => b+ 2001>0
+) a>b => ab + 2001a > ab + 2001b
=> \(\frac{a}{b}>\frac{a+2001}{b+2001}\)
+) a < b => ab + 2001a < ab + 2001b
=> \(\frac{a}{b}< \frac{a+2001}{b+2001}\)
+) a = b
=> \(\frac{a}{b}=\frac{a+2001}{b+2001}\)
a) \(\frac{45^{10}.5^{20}}{75^{15}}\)
=
\(\frac{\left(5.9\right)^{10}.5^{20}}{\left(5.15\right)^{15}}\)
= \(\frac{5^{10}.9^{10}.5^{20}}{5^{15}.15^{15}}\)
= \(\frac{5^{10}.3^{20}.5^{20}}{5^{15}.15^{15}}\)
= \(\frac{5^{10}.15^{20}}{5^{15}.15^{15}}\)
= \(\frac{15^5}{5^5}\)
= \(\frac{3^5.5^5}{5^5}\)
= \(3^5\)
b) \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}\)
= \(\frac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}\)
= \(\frac{2^5}{0,4}\)
= \(2^5\) : 0,4
(=) 32 : \(\frac{2}{5}\)
= 90
c) \(\frac{2^{15}.9^4}{6^6.8^3}\)
= \(\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}\)
= \(\frac{2^{15}.3^8}{2^6.3^6.2^9}\)
= \(3^2\)
\(\text{Ta có : A}=222^{555}=(222^5)^{111}\)
\(\text{B}=555^{222}=(555^2)^{111}\)
\(\text{Vì }222^{555}-555^{222}>0\Rightarrow A>B\)
Chúc bạn học tốt :>
\(\text{Có j thắc mắc thì cứ hỏi mk}\)
ta có:
A=222555=(2225)111
B=555222=(5552)111
=>A>B vì 2225>5552
vậy A>B
\(4^{32}< 6^{15}\) mik di
\(4^{32}< 6^{15}\) mik di nha