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Ta có:
\(C=\dfrac{2}{60.63}+\dfrac{2}{63.66}+...+\dfrac{2}{117.120}+\dfrac{2}{2006}\)
\(C=2\left(\dfrac{1}{60.63}+\dfrac{1}{63.66}+...+\dfrac{1}{117.120}\right)+\dfrac{2}{2006}\)
\(C=2.\dfrac{1}{3}\left(\dfrac{3}{60.63}+\dfrac{3}{63.66}+...+\dfrac{3}{117.120}\right)+\dfrac{2}{2006}\)
\(C=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{66}+...+\dfrac{1}{117}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)
\(C=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)
\(C=\dfrac{2}{3}.\dfrac{1}{120}+\dfrac{2}{2006}\)
\(C=\dfrac{1}{180}+\dfrac{2}{2006}\)
Ta lại có:
\(D=\dfrac{5}{40.44}+\dfrac{5}{44.48}+...+\dfrac{5}{76.80}+\dfrac{5}{2006}\)
\(D=5\left(\dfrac{1}{40.44}+\dfrac{1}{44.48}+...+\dfrac{1}{76.80}\right)+\dfrac{5}{2006}\)
\(D=5.\dfrac{1}{4}\left(\dfrac{4}{40.44}+\dfrac{4}{44.48}+...+\dfrac{4}{76.80}\right)+\dfrac{5}{2006}\)
\(D=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{48}+...+\dfrac{1}{76}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)
\(D=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)
\(D=\dfrac{5}{4}.\dfrac{1}{80}+\dfrac{5}{2006}\)
\(D=\dfrac{1}{64}+\dfrac{5}{2006}\)
Vì \(\dfrac{1}{180}< \dfrac{1}{64}\)
\(\dfrac{2}{2006}< \dfrac{5}{2006}\)
\(\Rightarrow\dfrac{1}{180}+\dfrac{2}{2006}< \dfrac{1}{64}+\dfrac{5}{2006}\)
\(\Rightarrow C< D\)
dở ẹt nhu cu net ma ko biet lamb tao hoc lop mau giao tao cung biet tra loi dung la ngu
\(A=\frac{2}{60\cdot63}+\frac{2}{63\cdot66}+...+\frac{2}{117\cdot120}+\frac{2}{2003}\)
\(\text{Đặt }C=\frac{2}{60\cdot63}+\frac{2}{63\cdot66}+...+\frac{2}{117\cdot120}\)
\(C=\frac{2}{3}\left(\frac{3}{60\cdot63}+\frac{3}{63\cdot66}+...+\frac{3}{117\cdot120}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+...+\frac{1}{117}-\frac{1}{120}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{120}\right)\)
\(C=\frac{2}{3}\cdot\frac{1}{120}\)
\(C=\frac{1}{180}\)
\(\text{Thay }C=\frac{1}{180}\text{Ta có : }\) \(A=\frac{1}{180}+\frac{2}{2003}\)
\(B=\frac{5}{40\cdot44}+\frac{5}{44\cdot48}+...+\frac{5}{76\cdot80}+\frac{5}{2003}\)
\(\text{Đặt }D=\frac{5}{40\cdot44}+\frac{5}{44\cdot48}+...+\frac{5}{76\cdot80}\)
\(D=\frac{5}{4}\left(\frac{4}{40\cdot44}+\frac{4}{44\cdot48}+...+\frac{4}{76\cdot80}\right)\)
\(D=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+...+\frac{1}{76}-\frac{1}{80}\right)\)
\(D=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{80}\right)\)
\(D=\frac{5}{4}\cdot\frac{1}{80}\)
\(D=\frac{1}{64}\)
\(\text{Thay }D=\frac{1}{64}\text{ Ta có : }B=\frac{1}{64}+\frac{5}{2003}\)
\(\text{Vì }A=\frac{1}{180}+\frac{2}{2003}\text{ , }B=\frac{1}{64}+\frac{5}{2003}\)
\(\text{Có : }\frac{1}{180}< \frac{1}{64}\)
\(\frac{2}{2003}< \frac{5}{2003}\)
\(\Rightarrow\text{ }A< B\)
A= 2x3/3x60x63+2x3/3x63x66+...+2x3/3x117x120+2/2007
=2/3(1/60-1/63+1/63-1/66+...+1/117-1/120)+2/2007
=2/3(1/60-1/120)+2/2007=1/180+2/2007
Lam tuong tu nhu vay vs B. B nhan ca tu va mau cho 4 sua do thu 5/4 ra ngoai. Cuoi cung tinh dk A<B
a: \(=\dfrac{2}{3}\left(\dfrac{3}{60\cdot63}+\dfrac{3}{63\cdot66}+...+\dfrac{3}{117\cdot120}\right)+\dfrac{2}{2006}\)
\(=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{63}+...+\dfrac{1}{117}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)
\(=\dfrac{2}{3}\cdot\dfrac{1}{120}+\dfrac{1}{2003}=\dfrac{1}{180}+\dfrac{1}{2003}=\dfrac{2183}{180\cdot2003}\)
b: \(=\dfrac{5}{4}\left(\dfrac{4}{40\cdot44}+\dfrac{4}{44\cdot48}+...+\dfrac{4}{76\cdot80}\right)+\dfrac{5}{2006}\)
\(=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)
\(=\dfrac{5}{4}\cdot\dfrac{1}{80}+\dfrac{5}{2006}=\dfrac{1}{64}+\dfrac{5}{2006}=\dfrac{1163}{64192}\)
c: \(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+\dfrac{3}{17\cdot20}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{3}\cdot\dfrac{9}{20}=\dfrac{3}{20}\)
a) \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)
\(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x+2010=0\) ( vì 1/2003 + 1/2006 -- 1/2011 -- 1/2015 \(\ne\)0)
\(\Leftrightarrow x=-2010\)
câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<
Giải:
1) \(7^8.\left(-\dfrac{1}{7}\right)^8\)
\(=7^8.\left(\dfrac{1}{7}\right)^8\)
\(=7^8.\dfrac{1^8}{7^8}\)
\(=1\)
2) \(\left(\dfrac{4}{3}\right)^{10}.\left(-\dfrac{3}{4}\right)^{10}\)
\(=\left(\dfrac{4}{3}\right)^{10}.\left(\dfrac{3}{4}\right)^{10}\)
\(=\dfrac{4^{10}}{3^{10}}.\dfrac{3^{10}}{4^{10}}\)
\(=1\)
3) \(\left(-\dfrac{7}{2}\right)^{2006}.\left(-\dfrac{2}{7}\right)^{2006}\)
\(=\left(\dfrac{7}{2}\right)^{2006}.\left(\dfrac{2}{7}\right)^{2006}\)
\(=1\)
4) \(\left(-\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\left(\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\dfrac{5^{2007}.13^{2006}}{13^{2007}.5^{2006}}\)
\(=\dfrac{5}{13}\)
Vậy ...
C=1/2.(3/60.63+....+3/117.120)+1/1003
C=1/2.(1/60-1/63+....+1/117-1/120)+1/1003
....còn lại tự làm nha, bài còn lại cũng tương tự
Bạn ơi còn \(\frac{5}{2006}\)xử lý sao