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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) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)
a) Ta có: \(\dfrac{39}{-65}=\dfrac{-39}{65}=\dfrac{-39:13}{65:13}=\dfrac{-3}{5}\)
\(\dfrac{-3}{5}=\dfrac{-3}{5}\)
Do đó: \(\dfrac{-3}{5}=\dfrac{39}{-65}\)
b) Ta có: \(\dfrac{-9}{27}=\dfrac{-9:9}{27:9}=\dfrac{-1}{3}\)
\(\dfrac{-41}{123}=\dfrac{-41:41}{123:41}=\dfrac{-1}{3}\)
Do đó: \(\dfrac{-9}{27}=\dfrac{-41}{123}\)
c) Ta có: \(\dfrac{-3}{4}=\dfrac{-3\cdot5}{4\cdot5}=\dfrac{-15}{20}\)
\(\dfrac{4}{-5}=\dfrac{-4}{5}=\dfrac{-4\cdot4}{5\cdot4}=\dfrac{-16}{20}\)
mà \(\dfrac{-15}{20}>\dfrac{-16}{20}\)
nên \(\dfrac{-3}{4}>\dfrac{4}{-5}\)
d) Ta có: \(\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-2\cdot7}{3\cdot7}=\dfrac{-14}{21}\)
\(\dfrac{-5}{7}=\dfrac{-5\cdot3}{7\cdot3}=\dfrac{-15}{21}\)
mà \(\dfrac{-14}{21}>\dfrac{-15}{21}\)
nên \(\dfrac{2}{-3}>\dfrac{-5}{7}\)
Giải:
a)Ta có:
C=1957/2007=1957+50-50/2007
=2007-50/2007
=2007/2007-50/2007
=1-50/2007
D=1935/1985=1935+50-50/1985
=1985-50/1985
=1985/1985-50/1985
=1-50/1985
Vì 50/2007<50/1985 nên -50/2007>-50/1985
⇒C>D
b)Ta có:
A=20162016+2/20162016-1
A=20162016-1+3/20162016-1
A=20162016-1/20162016-1+3/20162016-1
A=1+3/20162016-1
Tương tự: B=20162016/20162016-3
B=1+3/20162016-3
Vì 20162016-1>20162016-3 nên 3/20162016-1<3/20162016-3
⇒A<B
Chúc bạn học tốt!
Làm tiếp:
c)Ta có:
M=102018+1/102019+1
10M=10.(102018+1)/202019+1
10M=102019+10/102019+1
10M=102019+1+9/102019+1
10M=102019+1/102019+1 + 9/102019+1
10M=1+9/102019+1
Tương tự:
N=102019+1/102020+1
10N=1+9/102020+1
Vì 9/102019+1>9/102020+1 nên 10M>10N
⇒M>N
Chúc bạn học tốt!
(Sửa \(cn-bm\rightarrow cn-dm\))
Ta có :
\(\left\{{}\begin{matrix}ad-bc=1\\cn-dm=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}ad=1+bc\\cn=1+dm\end{matrix}\right.\)
\(\dfrac{x}{y}=\dfrac{a}{b}.\dfrac{d}{c}=\dfrac{ad}{bc}=\dfrac{1+bc}{bc}=1+\dfrac{1}{bc}>1\left(bc>0\right)\)
\(\Rightarrow x=\dfrac{a}{b}>y=\dfrac{c}{d}\left(2\right)\)
\(\dfrac{y}{z}=\dfrac{c}{d}.\dfrac{n}{m}=\dfrac{cn}{dm}=\dfrac{1+dm}{dm}=1+\dfrac{1}{dm}>1\left(dc>0\right)\)
\(\Rightarrow y=\dfrac{c}{d}>z=\dfrac{m}{n}\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow x>y>z\)
