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Ta có:
\(A=\frac{1946}{1986}=1-\frac{40}{1986}\)
\(B=\frac{1968}{2008}=1-\frac{40}{2008}\)
Vì \(\frac{40}{1986}>\frac{40}{2008}\Rightarrow1-\frac{40}{1986}< 1-\frac{40}{2008}\)
\(\frac{1946}{1986}< \frac{1968}{2008}\Rightarrow A< B\)
ta có \(A=\frac{1946}{1986}\)
\(\Rightarrow1-A=1-\frac{1946}{1986}=\frac{1986}{1986}-\frac{1946}{1986}=\frac{40}{1986}\)
\(\Rightarrow1-B=1-\frac{1968}{2008}=\frac{2008}{2008}-\frac{1968}{2008}=\frac{40}{2008}\)
Vì \(\frac{40}{1986}>\frac{40}{2008}\Rightarrow1-\frac{1946}{1986}>1-\frac{1968}{2008}\)
nên \(\Rightarrow1-A>1-B\left(1\right)\)
từ (1) <=> A<B
1.
\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\left(\frac{1}{2^{100}}+\frac{1}{2^{100}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)
cứ làm như vậy ta được :
\(=1+1=2\)
2. Ta có :
\(\frac{2008+2009}{2009+2010}=\frac{2008}{2009+2010}+\frac{2009}{2009+2010}\)
vì \(\frac{2008}{2009}>\frac{2008}{2009+2010}\); \(\frac{2009}{2010}>\frac{2009}{2009+2010}\)
\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}>\frac{2008+2009}{2009+2010}\)
a) Xét:
\(a>b\)
\(\Rightarrow\dfrac{a}{b}>1\Rightarrow\dfrac{a+m}{b+m}>1\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{a+m}\)
\(a< b\)
\(\Rightarrow\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\)
\(a=b\)
\(\Rightarrow\dfrac{a}{b}=1\Rightarrow\dfrac{a+m}{b+m}=1\Rightarrow\dfrac{a}{b}=\dfrac{a+m}{b+m}=1\)
Mk chỉ áp dụng tính 1 câu,câu sau làm tương tự
b)
Ta có:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{10^{1993}+1}{10^{1992}+1}< 1\)
\(B< \dfrac{10^{1993}+1+9}{10^{1992}+1+9}\Rightarrow B< \dfrac{10^{1993}+10}{10^{1992}+10}\Rightarrow B< \dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\Rightarrow B< \dfrac{10^{1992}+1}{10^{1991}+1}=A\)
\(B< A\)
@@ ~ học tốt ~
tính chất trên gọi là tính chất bắc cầu, ta so sánh hai phân số với một số (phân số) thứ 3.
a) \(\dfrac{-1}{-4}\)=\(\dfrac{1}{4}>0\)
\(\dfrac{3}{-4}< 0\)
\(\Rightarrow\dfrac{1}{4}>\dfrac{3}{-4}hay\dfrac{-1}{-4}>\dfrac{3}{-4}\)
b) Ta có:
\(\dfrac{15}{17}=1-\dfrac{2}{17}\\ \)
\(\dfrac{25}{27}=1-\dfrac{2}{27}\\ \\ \)
Mà \(\dfrac{2}{17}>\dfrac{2}{27}\left(17< 27\right)\)
\(\Rightarrow1-\dfrac{2}{17}< 1-\dfrac{2}{27}\)hay \(\dfrac{15}{17}< \dfrac{25}{27}\)
\(1=\dfrac{1946}{1986}+\dfrac{40}{1986}\)
\(1=\dfrac{1968}{2008}+\dfrac{40}{2008}\)
\(\Rightarrow\)\(\dfrac{40}{1968}>\dfrac{40}{2008}\)
\(\Rightarrow\dfrac{1946}{1968}>\dfrac{1968}{2008}\)
Vì \(\left\{{}\begin{matrix}1=\dfrac{1946}{1986}+\dfrac{40}{1986}\\1=\dfrac{1968}{2008}+\dfrac{40}{2008}\end{matrix}\right.\)
Mà \(\dfrac{40}{1986}>\dfrac{40}{2008}\)
Nên \(\dfrac{1946}{1986}>\dfrac{1968}{2008}\)
Vậy \(\dfrac{1946}{1986}>\dfrac{1968}{2008}\).