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a: 2003.2004-1/2003.2004 < 2004.2005-1/2004.2005
b: 149/157 < 449/457
c: 1999.2000/1999.2000+1 < 2000.2001/2000.2001+1
Ta có :
+) \(\frac{2003.2004-1}{2003.2004}=\frac{2003.2004}{2003.2004}-\frac{1}{2003.2004}=1-\frac{1}{2003.2004}\)
+) \(\frac{2004.2005-1}{2004.2005}=\frac{2004.2005}{2004.2005}-\frac{1}{2004.2005}=1-\frac{1}{2004.2005}\)
ta thấy :
\(\frac{1}{2003.2004}>\frac{1}{2004.2005}\Rightarrow1-\frac{1}{2003.2004}< 1-\frac{1}{2004.2005}\)
\(\Rightarrow\frac{2003.2004-1}{2003.2004}< \frac{2004.2005-1}{2004.2005}\)
\(A=\dfrac{2003.2004-1}{2003.2004}=1-\dfrac{1}{2003.2004}\)
\(B=\dfrac{2004.2005-1}{2004.2005}=1-\dfrac{1}{2004.2005}\)
So sánh: \(\dfrac{1}{2003.2004}>\dfrac{1}{2004.2005}\)
\(\Rightarrow-\dfrac{1}{2003.2004}< -\dfrac{1}{2004.2005}\\ \Rightarrow1-\dfrac{1}{2003.2004}< 1-\dfrac{1}{2004.2005}\\ Hay.A< B\)
a) A=\(\dfrac{2003.2004-1}{2003.2004}=\dfrac{2003.2004}{2003.2004}-\dfrac{1}{2004}=1-\dfrac{1}{2003.2004}\)
B = \(\dfrac{2004.2005-1}{2004.2005}=\dfrac{2004.2005}{2004.2005}-\dfrac{1}{2004.2005}=1-\dfrac{1}{2004.2005}\)
Vì \(\dfrac{1}{2003.2004}>\dfrac{1}{2004.2005}\)
\(\Rightarrow1-\dfrac{1}{2003.2004}< 1-\dfrac{1}{2004.2005}\)
Vậy A < B
b) \(\left(3X-2^4\right).7^5=2.7^6.\dfrac{1}{2009^0}\)
\(\left(3X-2^4\right).7^5=2.7^6.1\)
\(\left(3X-2^4\right).7^5=2.7^6\)
\(\left(3X-2^4\right).=2.7^6:7^5\)
\(3X-2^4=2.7\)
\(3X-16=14\)
\(3X=16+14=30\)
\(X=30:3=10\)
Vậy X = 10
1/ \(A=\dfrac{2003.2004-1}{2003.2004}=\dfrac{2003.2004}{2003.2004}-\dfrac{1}{2003.2004}=1-\dfrac{1}{2003.2004}\)
\(B=\dfrac{2004.2005-1}{2004.2005}=\dfrac{2004.2005}{2004.2005}-\dfrac{1}{2004.2005}=1-\dfrac{1}{2004.2005}\)
Vì \(1-\dfrac{1}{2003.2004}< 1-\dfrac{1}{2004.2005}\Leftrightarrow A< B\)
2/ \(\left(3x-2^4\right).7^5=2.7^6.\dfrac{1}{2009^0}\)
\(\Leftrightarrow\left(3x-2^4\right).7^5=2.7^6.1\)
\(\Leftrightarrow3x-2^4=2.7^6:7^5\)
\(\Leftrightarrow3x-2^4=2.7\)
\(\Leftrightarrow3x-16=14\)
\(\Leftrightarrow3x=30\)
\(\Leftrightarrow x=10\left(tm\right)\)
Vậy ..
a) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)
\(\dfrac{12}{49}=\dfrac{564}{2303};\dfrac{13}{47}=\dfrac{637}{2303}\)
Vì 564<637=>\(\dfrac{564}{2303}< \dfrac{637}{2303}\)=>\(\dfrac{12}{49}< \dfrac{13}{47}\)
1, mình không ghi đề nha
A= \(\frac{1.1+1.1+1.1}{3+3.3+3.3+3}\)
A=\(\frac{1.3}{9.3}\)
A=\(\frac{1}{9}\)