Tính: A= \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{2013+\frac{2013}{2}+\frac{2012}{3}+...+\frac{1}{2014}}\)
Bài 2: Cho \(\frac{a}{b}< \frac{c}{d}\) và b;d>0
Chứng Minh: \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)
Bài 2)
Ta có \(\frac{a}{b}< \frac{c}{d}\)
\(\Rightarrow ad< bc\)
Xét \(\frac{a}{b}< \frac{a+c}{b+d}\)
\(\Rightarrow a\left(b+d\right)< b\left(a+c\right)\)
\(\Rightarrow ab+ad< ab+bc\)
\(\Rightarrow ad< bc\) ( thỏa mãn đề bài )
Vậy \(\frac{a}{b}< \frac{a+c}{b+d}\) (1)
Xét \(\frac{a+c}{b+d}< \frac{c}{d}\)
\(\Rightarrow d\left(a+c\right)< c\left(b+d\right)\)
\(\Rightarrow ad+cd< bc+cd\)
\(\Rightarrow ad< bc\) ( thỏa mãn đề bài )
Vậy \(\frac{a+c}{b+d}< \frac{c}{d}\) (2)
Từ (1) và (2)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\) (đpcm)
Cho \(A=\frac{1}{2014}+\frac{2}{2013}+\frac{3}{2012}+..+\frac{2013}{2}+\frac{2014}{1}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}+\frac{1}{2015}\)
Tính \(\frac{A}{B}\)
tính GTBT D=\(\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(D=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)+1}\)
\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}\)
\(=\frac{2013}{2014}\)
\(\frac{\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2013}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2013}}{\frac{2012}{1}+2+\frac{2012}{2}+1+\frac{2011}{3}+1+...+\frac{1}{2013}+1-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{\frac{2014}{1}+\frac{2014}{2}+...+\frac{2014}{2013}-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{2014\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}-1\right)}\)
=\(\frac{1}{2014}\)
A=\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+.....+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2014}+\frac{1}{2015}}\)=
Xét Tử số của A ta có:
\(2014+\frac{2013}{2}+\frac{2012}{3}+....+\frac{2}{2013}=1+\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+....+\left(\frac{1}{2014}+1\right)\)\(TS=\frac{2015}{2}+\frac{2015}{3}+....+\frac{2015}{2014}+\frac{2015}{2015}\)
\(TS=2015.\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)\)
\(A=\frac{2015.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)}{\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)}=2015\)
toán lớp 8 dễ quá vậy
A=2015
hình như thế
Tính nhanh
\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+\frac{2011}{4}+\frac{2010}{5}+...+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}}\)
Giải tự luận hộ mình nha!!!!!!!! Mình cảm ơn!!!
Đặt phân thức trên là D
=> D=(1+1+1+1+...+1+2013/2+2012/3+...+2/2013+1/2014)/(1/2+1/3+1/4+...+1/2014)
=> D=(1+2013/2+1+2012/3+1+2011/4+...+1+2/2013+1+1/2014+1)/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=(2015/2+2015/3+2015/4+...+2015/2013+2015/2014+1)/(1/2+1/3+1/4+...+1/2014)
=> D=[2015*(1/2+1/3+1/4+1/5+....+1/2014)]/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=2015
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đùa thôi đáp án: 2015 nha bn
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À quên nhớ FOLOW CHO TUI NHA!
Tính A=\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+\frac{2011}{4}+...+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}}\)
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\(B=\frac{1-3}{1\cdot3}+\frac{2-4}{2\cdot4}+\frac{3-5}{3\cdot5}+\frac{4-6}{4\cdot6}+............+\frac{2011-2013}{2011.2013}+\frac{2012-2014}{2012\cdot2014}-\frac{2013+2014}{2013\cdot2014}\)
giá trị biểu thức A=\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+...+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}+\frac{1}{2015}}là?\)
