\(\dfrac{1}{2010}và\dfrac{-7}{19}\)
so sánh
so sánh
a)\(A=\dfrac{-2015}{2015.2016}\) và \(B=\dfrac{-2014}{2014.2015}\) b)A = \(\dfrac{10^{2009}+1}{10^{2010}+1}\) và \(B=\dfrac{10^{2010}+1}{10^{2011}+1}\)
A=-2015/2015x2016
A=-1/2016
B=-2014/2014x2015
B=-1/2015
vi 2016>2015,-1/2016>-1/2015
vay A>B
b) Ta có: \(A=\dfrac{10^{2009}+1}{10^{2010}+1}\)
\(\Leftrightarrow10A=\dfrac{10^{2010}+10}{10^{2010}+1}=1+\dfrac{9}{10^{2010}+1}\)
Ta có: \(B=\dfrac{10^{2010}+1}{10^{2011}+1}\)
\(\Leftrightarrow10B=\dfrac{10^{2011}+10}{10^{2011}+1}=1+\dfrac{9}{10^{2011}+1}\)
Ta có: \(10^{2010}+1< 10^{2011}+1\)
\(\Leftrightarrow\dfrac{9}{10^{2010}+1}>\dfrac{9}{10^{2011}+1}\)
\(\Leftrightarrow\dfrac{9}{10^{2010}+1}+1>\dfrac{9}{10^{2011}+1}+1\)
\(\Leftrightarrow10A>10B\)
hay A>B
So sánh A và B biết: A= \(\dfrac{2008+2008+2010}{2009+2010+2011}\) và B= \(\dfrac{2008}{2009}\)+ \(\dfrac{2009}{2010}\)+ \(\dfrac{2010}{2011}\)
A = \(\dfrac{2008}{2009+2010+2011}+\dfrac{2009}{2009+2010+2011}+\dfrac{2010}{2009+2010+2011}\)
Ta có:
\(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)
\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)
\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)
Từ 3 điều trên suy ra : A < B
So sánh:
a) 430 và 3.2410
b) \(\dfrac{3}{1^2.2^2}\) + \(\dfrac{5}{2^2.3^2}\) + \(\dfrac{7}{3^2.4^2}\) +...+\(\dfrac{19}{9^2.10^2}\) và 1
a) \(3\cdot24^{10}=3\cdot6^{10}\cdot4^{10}=3\cdot3^{10}\cdot2^{10}\cdot2^{20}\)
\(=3^{11}\cdot2^{30}\)
\(4^{30}=2^{30}\cdot2^{30}=2^{30}\cdot4^{15}\)
Ta có \(4^{15}>3^{15}>3^{11}\) nên \(4^{15}>3^{11}\)
Khi đó \(4^{15}\cdot2^{30}>3^{11}\cdot2^{30}\) hay \(4^{30}>3\cdot24^{10}\)
b) \(\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{19}{9^2\cdot10^2}\)
\(=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+...+\dfrac{19}{81\cdot100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}< 1\)
Vậy dãy trên nhỏ hơn 1
a/
\(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}.2^{30}=\left(2^2\right)^{15}.2^{30}=4^{15}.2^{30}\)
\(3.24^{10}=3.3^{10}.\left(2^3\right)^{10}=3^{11}.2^{30}< 3^{15}.2^{30}\)
\(\Rightarrow4^{30}=4^{15}.2^{30}>3^{15}.2^{30}>3^{11}.2^{30}=3.24^{10}\)
b/
\(=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}=\)
\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}=\)
\(=1-\dfrac{1}{10^2}< 1\)
a) 4³⁰ = (2²)³⁰ = 2⁶⁰ = 2³⁰.2³⁰ = 1073741824.2³⁰
3.24¹⁰ = 3.(3.2³)¹⁰ = 3.3¹⁰.2³⁰ = 3¹¹.2³⁰ = 177147.2³⁰
Do 1073741824 > 177147
⇒ 1073741824.2³⁰ > 177147.2³⁰
Vậy 4³⁰ > 3.24¹⁰
b) 3/(1².2²) + 5/(2².3²) + ... + 19/(9².10²)
= 1/1² - 1/2² + 1/2² - 1/3² + ... + 1/9² - 1/10²
= 1 - 1/100
= 99/100
Mà 99/100 < 1
⇒ 3/(1².2²) + 5/(2².3²) + 7/(3².4²) + ... + 19/(9².10²) < 1
\(\dfrac{1}{2010}và\dfrac{-7}{19}\) \(\dfrac{497}{-499}và\dfrac{-2345}{2341}\)
\(\dfrac{-3737}{4141}và\dfrac{-37}{41}\)
So sánh các phân số sau đây bằng cách thuận tiện :
\(\dfrac{47}{95}\)và\(\dfrac{35}{69}\) \(\dfrac{53}{103}\)và \(\dfrac{71}{145}\) \(\dfrac{2009}{2010}\)và\(\dfrac{2005}{2006}\) \(\dfrac{783}{901}\)và\(\dfrac{738}{915}\)
\(\dfrac{47}{95}\) và \(\dfrac{35}{69}\)
\(\dfrac{47}{95}< \dfrac{1}{2}\) và \(\dfrac{35}{69}>\dfrac{1}{2}\)
Vậy \(\dfrac{47}{95}< \dfrac{35}{69}\)
\(\dfrac{53}{103}\) và \(\dfrac{71}{145}\)
\(\dfrac{53}{103}>\dfrac{1}{2}\) và \(\dfrac{71}{145}< \dfrac{1}{2}\)
Vậy \(\dfrac{53}{103}>\dfrac{71}{145}\)
\(\dfrac{2009}{2010}\) và \(\dfrac{2005}{2006}\)
\(1-\dfrac{2009}{2010}=\dfrac{1}{2010}\) và \(1-\dfrac{2005}{2006}=\dfrac{1}{2006}\)
Vậy \(\dfrac{2009}{2010}>\dfrac{2005}{2006}\)
\(\dfrac{783}{901}\) và \(\dfrac{738}{915}\)
\(\dfrac{738}{915}< \dfrac{783}{915}< \dfrac{783}{901}\)
Vậy \(\dfrac{783}{901}>\dfrac{738}{915}\)
nhanh giúp mình với ạ
cảm ơn bạn nhiều
So Sánh : A = \(\dfrac{2009^{2009}+1}{2009^{2010}+1}\) và B = \(\dfrac{2009^{2010}-2}{2009^{2011}-2}\)
Ta có :
\(B=\dfrac{2009^{2010}-2}{2009^{2011}-2}< 1\)
\(\Leftrightarrow B< \dfrac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\dfrac{2009^{2010}+2009}{2009^{2011}+2009}=\dfrac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\dfrac{2009^{2009}+1}{2009^{2010}+1}=A\)
\(\Leftrightarrow A>B\)
So sánh P và Q, biết: \(P=\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2013}\) và \(Q=\dfrac{2010+2011+2012}{2011+2012+2013}\)
\(Q=\dfrac{2010+2011+2012}{2011+2012+2013}=\dfrac{2010}{2011+2012+2013}+\dfrac{2011}{2011+2012+2013}+\dfrac{2012}{2011+2012+2013}\)
Ta có: \(\dfrac{2010}{2011+2012+2013}< \dfrac{2010}{2011}\)
\(\dfrac{2011}{2011+2012+2013}< \dfrac{2011}{2012}\)
\(\dfrac{2012}{2011< 2012< 2013}< \dfrac{2012}{2013}\)
\(\Rightarrow\dfrac{2010}{2011+2012+2013}+\dfrac{2011}{2011+2012+2013}+\dfrac{2012}{2011+2012+2013}\)
\(\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2013}\)
\(P>Q\)
So sánh \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{100}}\) và \(19\)
a) Đọc và thảo luận nội dung sau:
b) So sánh các phân số sau với 1:
\(\dfrac{5}{6},\dfrac{3}{2},\dfrac{9}{19},\dfrac{7}{7},\dfrac{49}{46},\dfrac{32}{71}\)
c) Viết ba phân số bé hơn 1, ba phân số lớn hơn 1, ba phân số bằng 1.
a) HS tự thực hiện
b) $\frac{5}{6}$ < 1 ; $\frac{3}{2} > 1$
$\frac{9}{{19}}$ < 1 ; $\frac{7}{7}$ = 1
$\frac{{49}}{{46}}$ > 1 ; $\frac{{32}}{{71}}$ < 1
c) Ba phân số bé hơn 1 là: $\frac{2}{7};\,\,\,\frac{{11}}{{25}};\,\,\,\frac{{37}}{{59}}$
Ba phân số lớn hơn 1 là: $\frac{7}{2};\,\,\,\frac{{15}}{7};\,\,\,\,\frac{{33}}{{12}}$
Ba phân số bằng 1 là: $\frac{9}{9};\,\,\,\,\frac{{25}}{{25}};\,\,\,\,\frac{{47}}{{47}}$