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a, Ta có: \(\left(\dfrac{1}{2}\right)^{300}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left(\dfrac{1}{3}\right)^{200}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
=> \(\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\)=> \(\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
b, Ta có: \(\left(\dfrac{1}{3}\right)^{75}=\left[\left(\dfrac{1}{3}\right)^3\right]^{25}=\left(\dfrac{1}{27}\right)^{25}\)
\(\left(\dfrac{1}{5}\right)^{50}=\left[\left(\dfrac{1}{5}\right)^2\right]^{25}\)\(=\left(\dfrac{1}{25}\right)^{25}\)
Do \(\left(\dfrac{1}{27}\right)^{25}< \left(\dfrac{1}{25}\right)^{25}=>\left(\dfrac{1}{3}\right)^{75}< \left(\dfrac{1}{5}\right)^{50}\)
Kiểm tra lại bài nhé, học tốt!!
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\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)
\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)
Bài 2:
a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)
\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)
\(=-\dfrac{3}{5}\)
b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)
\(\Leftrightarrow8x-1=5\)
\(\Leftrightarrow8x=6\)
hay \(x=\dfrac{3}{4}\)
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a: \(\left(4+\sqrt{33}\right)^2=49+8\sqrt{33}=49+2\cdot\sqrt{528}\)
\(\left(\sqrt{29}+\sqrt{14}\right)^2=43+2\cdot\sqrt{29\cdot14}=43+2\cdot\sqrt{406}\)
mà 49>43 và 528>406
nên \(\left(4+\sqrt{33}\right)^2>\left(\sqrt{29}+\sqrt{14}\right)^2\)
=>\(4+\sqrt{33}>\sqrt{29}+\sqrt{14}\)
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A=1+2+2^2+2^3+....+2^9
2A=2+2^2+2^3+....+2^10
2A-A=2^10-1
A=2^10-1/2
B=5.2^8=(2^2+1).2^8=2^10+2^8
=>B>A
2A = 2(1 + 2 + 22 + .... + 29 )
= 2 + 22 + 23 + ..... + 210
2A - A = (2 + 22 + 23 + ..... + 210) - (1 + 2 + 22 + .... + 29 )
A = 210 - 1
B = 5.28 = (22 + 1).28 = 210 + 28
210 - 1 < 210 + 28
=> A < B
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Lời giải:
$2^{299}< 2^{300}=(2^3)^{100}=8^{100}$
$3^{201}> 3^{200}=(3^2)^{100}=9^{100}$
$\Rightarrow 3^{201}> 9^{100}> 8^{100}> 2^{299}$
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a: \(\dfrac{\sqrt{81}}{\sqrt{16}}=\dfrac{9}{4}=\dfrac{36}{16}< \dfrac{81}{16}\)
b: \(\sqrt{16+25}=\sqrt{41}< 9=\sqrt{16}+\sqrt{25}\)
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\(a) 3^{200}=(3^2)^{100}=9^{100}\\2^{300}=(2^3)^{100}=8^{100}\)
Vì \(9^{100}>8^{100}\) nên \(3^{200}>2^{300}\)
\(b) 5^{40}=(5^4)^{10}=625^{10}\\3^{50}=(3^5)^{10}=243^{10}\)
Vì \(625^{10}>243^{10}\) nên \(5^{40}>3^{50}\)
#\(Toru\)
a> \(3^{200}\) và \(2^{300}\)
Ta có:\(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)
Vì 9>8 nên \(9^{100}>8^{100}\)
\(\Rightarrow\)\(3^{200}>2^{300}\)
b> \(5^{40}\) và \(3^{50}\)
Ta có:\(5^{40}=5^{4.10}=\left(5^4\right)^{10}=625^{10}\)
\(3^{50}=3^{5.10}=\left(3^5\right)^{10}=243^{10}\)
Vì 625 > 243 nên \(625^{10}>243^{10}\)
\(\Rightarrow\)\(5^{40}>3^{50}\)