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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 ~
\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}=2\times\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{240}\right)\)
\(A=2\times\left(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+....+\dfrac{1}{15\times16}\right)\)
\(A=2\times\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\times\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{3}{8}\)
b) cậu đi tìm số sốm hạng là : \(\left(2010-1\right):1+1=2010\)
\(\Rightarrow\)số cặp trong phép tính là : \(2010:2=1005\)(cặp)
\(\Rightarrow B=1-2+3-4+...+2009-2010\)(1005 cặp)
\(\Rightarrow\left(1-2\right)+\left(3-4\right)+...+\left(2009-2010\right)\)
\(\Rightarrow B=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)(1005 số -1)
\(\Rightarrow B=\left(-1\right).1005\)
\(\Rightarrow B=\left(-1005\right)\)
cậu tik cho mik nhé!!!
Ta có:
\(2007A=\dfrac{2007^{2009}+2007}{2007^{2009}+1}=1+\dfrac{2006}{2007^{2009}+1}\)\(2007B=\dfrac{2007^{2010}+10}{2007^{2010}+1}=1+\dfrac{9}{2007^{2010}+1}\)Vì \(\dfrac{2007}{2007^{2009}+1}>\dfrac{2007}{2007^{2010}+1}\)
=>2007A > 2007B
Do đó A>B
Vậy A>B
Ta có : \(B\) = \(\dfrac{2007^{2009}+1}{2007^{2010}+1}\) \(< 1\) \(\Rightarrow\dfrac{2007^{2009}+1}{2007^{2010}+1}< \dfrac{2007^{2009}+1+2006}{2007^{2010}+1+2006}\) \(=\dfrac{2007^{2009}+2007}{2007^{2010}+2007}\)
\(=\dfrac{2007\left(2007^{2008}+1\right)}{2007\left(2007^{2009}+1\right)}\) \(=\dfrac{2007^{2008}+1}{2007^{2009}+1}=A\)
Vậy \(A>B\)
Đặt \(A=\dfrac{2009^{2008}+1}{2009^{2009}+1}\) và \(B=\dfrac{2009^{2007}+1}{2009^{2008}+1}\)
Ta có:
\(2009A=\dfrac{2009.\left(2009^{2008}+1\right)}{2009^{2009}+1}=\dfrac{2009^{2009}+2009}{2009^{2009}+1}\)
\(=\dfrac{2009^{2009}+1+2008}{2009^{2009}+1}=\dfrac{2009^{2009}+1}{2009^{2009}+1}+\dfrac{2008}{2009^{2009}+1}\)
\(=1+\dfrac{1}{2009^{2009}+1}\)
\(2009B=\dfrac{2009.\left(2009^{2007}+1\right)}{2009^{2008}+1}=\dfrac{2009^{2008}+2009}{2009^{2008}+1}\)
\(=\dfrac{2008^{2008}+1+2008}{2009^{2008}+1}=\dfrac{2008^{2008}+1}{2009^{2008}+1}+\dfrac{2008}{2009^{2008}+1}\)
\(=1+\dfrac{2008}{2009^{2008}+1}\)
Vì \(1+\dfrac{2008}{2009^{2009}+1}< 1+\dfrac{2008}{2009^{2008}+1}\)
Nên \(10A< 10B\) \(\Rightarrow A< B\)
Vậy \(\dfrac{2009^{2008}+1}{2009^{2009}+1}< \dfrac{2009^{2007}+1}{2009^{2008}+1}\)
~ Học tốt ~
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2009^{2008}+1}{2009^{2009}+1}< 1\)
\(\Rightarrow A< \dfrac{2009^{2008}+1+2008}{2009^{2009}+1+2008}\Rightarrow A< \dfrac{2009^{2008}+2009}{2009^{2009}+2009}\Rightarrow A< \dfrac{2009\left(2009^{2007}+1\right)}{2009\left(2009^{2008}+1\right)}\Rightarrow A< \dfrac{2009^{2007}+1}{2009^{2008}+1}=B\)\(\Rightarrow A< B\)
a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)
\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)
\(\Leftrightarrow8x=-\frac{5}{4}\)
\(\Leftrightarrow x=-\frac{5}{32}\)
c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(\Leftrightarrow x+1=2003\)
\(\Leftrightarrow x=2002\)
Cách 2:
Ta có: \(10A=\dfrac{10^{2008}+10}{10^{2008}+1}=1+\dfrac{9}{10^{2008}+1}\)
\(10B=\dfrac{10^{2009}+10}{10^{2009}+1}=1+\dfrac{9}{10^{2009}+1}\)
Vì \(\dfrac{9}{10^{2008}+1}>\dfrac{9}{10^{2009}+1}\Rightarrow1+\dfrac{9}{10^{2008}+1}>1+\dfrac{9}{10^{2009}+1}\)
\(\Rightarrow10A>10B\Rightarrow A>B\)
Vậy A > B
A<B