Tính nhanh
a, A=1990.32-990/1990.31+1000
b, B=2008.2009+2000/2009.2010-2018
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A=1990.32-990/1990.31+1000
A=1990.(31+1)-990/1990.31+1000
A=1990.31+1990-990/1990.31+1000
A=1990.31+1000/1990.31+1000
A=1
Trả lời:
\(a,\)\(A=\frac{1990.32-990}{1990.31+1000}\)
\(\Leftrightarrow A=\frac{\left(1000+990\right).32-990}{\left(1000+990\right).31+1000}\)
\(\Leftrightarrow A=\frac{1000.32+990.32-990}{1000.31+990.31+1000}\)
\(\Leftrightarrow A=\frac{1000.32+990.31}{1000.32+990.31}\)
\(\Leftrightarrow A=1\)
Vậy\(A=1\)
\(b,\)\(B=\frac{2008.2009+2000}{2009.2010-2018}\)
\(\Leftrightarrow B=\frac{2008.2009+2008-8}{2009.2010-2010-8}\)
\(\Leftrightarrow B=\frac{2008.\left(2009+1\right)-8}{(2009-1).2010-8}\)
\(\Leftrightarrow B=\frac{2008.2010-8}{2008.2010-8}\)
\(\Leftrightarrow B=1\)
Vậy\(B=1\)
Hok tốt!
Good girl
Ta có: 2009.2010>2008.2009
\(\frac{1}{2009\cdot2010}< \frac{1}{2008\cdot2009}\)
\(\Rightarrow E>F\)
2008/2008.2009 và 2009/2009.2010
2008/2008.2009 < 2009/2009.2010
k mk na <3
2008/ 2008 × 2009> 2009/ 2009 × 2010
Mình thề 100% CHUẨN KHÔNG CẦN CHỈNH☺
a, \(\dfrac{2009}{2010}\) và \(\dfrac{2010}{2011}\)
Ta có:
\(2009.2011=4040099\)
\(2010.2010=4040100\)
Vì \(2009.2011< 2010.2010\)
nên \(\dfrac{2009}{2010}< \dfrac{2010}{2011}\)
b, \(\dfrac{2008}{2008.2009}\) và \(\dfrac{2009}{2009.2010}\)
Ta có:
\(\dfrac{2008}{2008.2009}=\dfrac{1}{2009};\dfrac{2009}{2009.2010}=\dfrac{1}{2010}\)
Vì \(\dfrac{1}{2009}>\dfrac{1}{2010}\) nên \(\dfrac{2008}{2008.2009}>\dfrac{2009}{2009.2010}\)
Chúc bạn học tốt!!!
a)\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(\dfrac{2009}{2010}< 1\)
\(\Leftrightarrow\dfrac{2009}{2010}< \dfrac{2009+1}{2010+1}\Leftrightarrow\dfrac{2009}{2010}< \dfrac{2010}{2011}\)
b)
\(\dfrac{2008}{2008.2009}=\dfrac{1}{2009}\)
\(\dfrac{2009}{2009.2010}=\dfrac{1}{2010}\)
\(\dfrac{1}{2009}>\dfrac{1}{2010}\Leftrightarrow\dfrac{2008}{2008.2009}>\dfrac{2009}{2009.2010}\)
d)
\(\dfrac{1}{3^{400}}=\dfrac{1}{\left(3^4\right)^{100}}=\dfrac{1}{81^{100}}\)
\(\dfrac{1}{4^{300}}=\dfrac{1}{\left(4^3\right)^{100}}=\dfrac{1}{64^{100}}\)
\(81^{100}>64^{100}\Leftrightarrow\dfrac{1}{81^{100}}< \dfrac{1}{64^{100}}\)
a/ Do : 2009/2010 > 2009/2011, 2009/2011 < 2010/2011 nên 2009/2010 < 2010/2011
1 đúng
Ta có: 200/201+201/202>200+201/202 (1)
200+201/201+202<200+201/202 (2)
từ (1) và (2) suy ra 200/201+201/202>200+201/201+202
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A=1-\frac{1}{2010}\)
\(A=\frac{2009}{2010}\)
Trả lời:
\(a,\)\(A=\frac{1990.32-990}{1990.31+1000}\)
\(\Leftrightarrow A=\frac{\left(1000+990\right).32-990}{\left(1000+990\right).31+1000}\)
\(\Leftrightarrow A=\frac{1000.32+990.32-990}{1000.31+990.31+1000}\)
\(\Leftrightarrow A=\frac{1000.32+990.31}{1000.32+990.31}\)
\(\Leftrightarrow A=1\)
Vậy\(A=1\)
\(b,\)\(B=\frac{2008.2009+2000}{2009.2010-2018}\)
\(\Leftrightarrow B=\frac{2008.2009+2008-8}{2009.2010-2010-8}\)
\(\Leftrightarrow B=\frac{2008.\left(2009+1\right)-8}{(2009-1).2010-8}\)
\(\Leftrightarrow B=\frac{2008.2010-8}{2008.2010-8}\)
\(\Leftrightarrow B=1\)
Vậy\(B=1\)
Hok tốt!
Good girl
Trả lời:
\(a,\)\(A=\frac{1990.32-990}{1990.31+1000}\)
\(\Leftrightarrow A=\frac{\left(1000+990\right).32-990}{\left(1000+990\right).31+1000}\)
\(\Leftrightarrow A=\frac{1000.32+990.32-990}{1000.31+990.31+1000}\)
\(\Leftrightarrow A=\frac{1000.32+990.31}{1000.32+990.31}\)
\(\Leftrightarrow A=1\)
Vậy\(A=1\)
\(b,\)\(B=\frac{2008.2009+2000}{2009.2010-2018}\)
\(\Leftrightarrow B=\frac{2008.2009+2008-8}{2009.2010-2010-8}\)
\(\Leftrightarrow B=\frac{2008.\left(2009+1\right)-8}{(2009-1).2010-8}\)
\(\Leftrightarrow B=\frac{2008.2010-8}{2008.2010-8}\)
\(\Leftrightarrow B=1\)
Vậy\(B=1\)
Hok tốt!
Bad boy