so sanh
\(\left(\frac{-175}{149}\right)^{300}\)va \(\left(\frac{-952}{299}\right)^{222}\)
Những câu hỏi liên quan
Bai \(1:\)So sanh :
\(a\)) \(\left(-\frac{1}{5}\right)^{300}\) va \(\left(-\frac{1}{3}\right)^{300}\)
\(b\)) \(\left(-\frac{1}{2}\right)^{5^{1^3}}\)va \(\left(-\frac{1}{2}\right)^{3^{1^5}}\)
\(A=\left(\frac{1}{1^2}-1\right)\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right);B=-\frac{1}{2}\). hãy so sanh a va b
\(A=\left(\frac{1}{1^2}-1\right)\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right)\)
\(=0.\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right)=0>-\frac{1}{2}\)
suy ra A>B
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Cho \(A=\left(\frac{1}{^{2^2}}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)......\left(\frac{1}{2013^2}-1\right).\left(\frac{1}{2014^2}-1\right)va\)\(B=-\frac{1}{2}.\)Hay so sanh A va B
Cho A = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\) va B = \(\frac{-1}{2}\), So sanh A va B
A = \(-\frac{1.3}{2.2}.-\frac{2.4}{3.3}.\cdot\cdot\cdot-\frac{2013.2015}{2014.2014}=-\frac{\left(1.2.3...2013\right).\left(3.4.5....2015\right)}{\left(2.3....2014\right).\left(2.3....2014\right)}=-\frac{2.2015}{2014}=-\frac{4030}{2014}
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so sánh
\(\left(\frac{1}{222}\right)^{333};\left(\frac{1}{333}\right)^{222}\)
-Vì (1/222)^333=(1/222)^3.111=(3/666)^111
(1/333)^222=(1/333)^2.111=(2/666)^111
-Vì 111=111 và 3/666>2/666
=))(1/222)^333>(1/333)^222
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Ta có:
222333 = 111333.2333 = 111222.111111.(23)111 = 111222.111111.8111 = 111222.888111
333222 = 111222.3222 = 111222.(32)111 = 111222.9111
Vì 111222.888111 > 111222.9111
=> 222333 > 333222
=> \(\frac{1}{222^{333}}< \frac{1}{333^{222}}\)
hay \(\left(\frac{1}{222}\right)^{333}< \left(\frac{1}{333}\right)^{222}\)
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so sanh
a) \(\left(\frac{1}{16}\right)^{200}\)va\(\left(\frac{1}{2}\right)^{1000}\)
b) (-32)27 va (-18)39
a, Có : (1/60)^200 = [(1/2)^4]^200 = (1/2)^800
Vì 0 < 1/2 < 1 nên (1/2)^800 > (1/2)^1000
=> (1/16)^200 > (1/2)^1000
Tk mk nha
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a) \(\left(\frac{1}{16}\right)^{200}=\left(\frac{1}{2}\right)^{800}< \left(\frac{1}{2}\right)^{1000}\)
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a) \(\left(\frac{1}{16}\right)^{200}=\frac{1}{16^{200}}\)
\(\left(\frac{1}{2}\right)^{1000}=\frac{1}{2^{1000}}\)
có : \(16^{200}=\left(2^4\right)^{200}=2^{800}\)
ta thấy \(2^{800}< 2^{1000}\)
\(\Rightarrow16^{200}< 2^{1000}\)
\(\Rightarrow\frac{1}{16^{200}}>\frac{1}{2^{1000}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{200}>\left(\frac{1}{2}\right)^{100}\)
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Xem thêm câu trả lời
\(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{299\cdot400}\)chứng minh rằng \(\frac{1}{299}\left(\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right)\)=\(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{299\cdot400}\)
Cho \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{2014^2}-1\right)\)va \(B=-\frac{1}{2}\). So sanh A va B
\(b=\left(\frac{-1}{2}\right)^{10}\)\(a=\left(\frac{-1}{2}\right)^{2^5}\)So Sanh: a va b