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\(A=\dfrac{25}{18\cdot21}+\dfrac{25}{21\cdot24}+\dfrac{25}{24\cdot27}+...+\dfrac{25}{123\cdot126}\)
\(=25\left(\dfrac{1}{18\cdot21}+\dfrac{1}{21\cdot24}+\dfrac{1}{24\cdot27}+...+\dfrac{1}{123\cdot126}\right)\)
\(=\dfrac{25}{3}\left(\dfrac{3}{18\cdot21}+\dfrac{3}{21\cdot24}+\dfrac{3}{24\cdot27}+...+\dfrac{3}{123\cdot126}\right)\)
\(=\dfrac{25}{3}\left(\dfrac{1}{18}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{24}+...+\dfrac{1}{123}-\dfrac{1}{126}\right)\)
\(=\dfrac{25}{3}\left(\dfrac{1}{18}-\dfrac{1}{126}\right)\)\(=\dfrac{25}{3}\cdot\dfrac{1}{21}=\dfrac{25}{63}\)
A= \(\dfrac{25}{18}-\dfrac{25}{21}+\dfrac{25}{21}-\dfrac{25}{24}+...+\dfrac{25}{123}-\dfrac{25}{126}\)
A= \(\dfrac{25}{18}-\dfrac{25}{126}\)
A= \(\dfrac{25}{21}\)
Hoặc ngay dòng 2 bạn làm như thế này cũng được: \(25.\left(\dfrac{1}{18}-\dfrac{1}{21}+...+\dfrac{1}{123}-\dfrac{1}{126}\right)\)
1,
\(\dfrac{3}{2^2}\cdot\dfrac{8}{3^2}\cdot\dfrac{15}{4^2}...\dfrac{899}{30^2}\\ =\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}....\dfrac{29\cdot31}{30\cdot30}\\ =\left(\dfrac{1\cdot2\cdot3\cdot...\cdot29}{2\cdot3\cdot4\cdot....\cdot30}\right)\cdot\left(\dfrac{3\cdot4\cdot5\cdot....\cdot31}{2\cdot3\cdot4.....\cdot30}\right)\\ =\dfrac{1}{30}\cdot\dfrac{31}{2}\\ =\dfrac{31}{60}\)
2,
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{37\cdot38\cdot39}\\ =\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+...+\dfrac{2}{37\cdot38\cdot39}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot5}+....+\dfrac{1}{37\cdot38}-\dfrac{1}{38\cdot39}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{38\cdot39}\right)\\ =\dfrac{1}{4}-\dfrac{1}{3964}\\ =\dfrac{185}{741}\)
3, Làm tương tự, áp dụng ; \(\dfrac{n}{x\left(x+n\right)}=\dfrac{1}{x}-\dfrac{1}{x+n}\)
A= \(\frac{5}{3}\)(\(\frac{3}{18.21}+\frac{3}{21.24}+\frac{3}{24.27}+...\frac{3}{123.126}\)
A=\(\frac{5}{3}\)(\(\frac{1}{18}-\frac{1}{126}\))
A=\(\frac{5}{3}.\frac{1}{21}\)
A=\(\frac{5}{63}\)
\(A=\frac{5}{18.21}+\frac{5}{21.24}+\frac{5}{24.27}+...+\frac{4}{123.126}\)
\(=\frac{5}{3}.\left(\frac{3}{18.21}+\frac{3}{21.24}+\frac{3}{24.27}+...+\frac{3}{123.126}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{123}-\frac{1}{126}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{18}-\frac{1}{126}\right)\)
\(=\frac{5}{3}.\frac{1}{21}\)
\(=\frac{5}{63}\)
3/5*A=3/(18*21)+3/(21*24)+3/(24*27)+...+3/(123*126)
=>3/5*A=1/18-1/21+1/21-1/24+1/24-1/27+...+1/123-1/126
=>3/5*A=1/18-1/126
=>3/5*A=1/21
=>A=5/63
A=\(\frac{5}{3}\left(\frac{3}{18.21}+\frac{3}{21.24}+..........+\frac{3}{123.126}\right)\)
A=\(\frac{5}{3}\left(\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+............+\frac{1}{123}-\frac{1}{126}\right)\)
=\(\frac{5}{3}.\left(\frac{1}{18}-\frac{1}{126}\right)\)
=\(\frac{5}{3}.\frac{1}{21}\)
=\(\frac{5}{63}\)
Đây này má Ran mori
a) \(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
\(=5+\dfrac{1}{7}-3-\dfrac{3}{11}-2-\dfrac{1}{7}-1-\dfrac{8}{11}\)
\(=\left(5-3-2-1\right)+\left(\dfrac{1}{7}-\dfrac{3}{11}-\dfrac{1}{7}-\dfrac{8}{11}\right)\)
\(=-1+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=-1+0-1=-2\)
a)\(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
= \(\left(5+\dfrac{1}{7}-3+\dfrac{3}{11}\right)-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(5-\dfrac{1}{7}+3-\dfrac{3}{11}-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(\left(5-3-2-1\right)+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{8}{11}-\dfrac{3}{11}\)
= \(-1+2+\dfrac{5}{11}\)
= \(1+\dfrac{5}{11}=\dfrac{1}{1}+\dfrac{5}{11}=\dfrac{11}{11}+\dfrac{5}{11}=\dfrac{16}{11}\)
Vậy :câu a) = \(\dfrac{16}{11}\)
\(=\left(\dfrac{1}{10}+\dfrac{-1}{10}\right)+\left(-\dfrac{1}{11}+\dfrac{1}{11}\right)+\left(-\dfrac{1}{12}+\dfrac{1}{12}\right)+\left(-\dfrac{1}{13}+\dfrac{1}{13}\right)+\left(-\dfrac{1}{14}+\dfrac{1}{14}\right)+\left(-\dfrac{1}{15}+\dfrac{1}{15}\right)+\dfrac{1}{16}\\ =\dfrac{1}{16}\)
Tính nhanh :
\(\dfrac{1}{10}+\dfrac{-1}{11}+\dfrac{1}{12}+\dfrac{-1}{13}+\dfrac{1}{14}+\dfrac{-1}{15}+\dfrac{1}{16}+\dfrac{-1}{10}+\dfrac{1}{11}+\dfrac{-1}{12}+\dfrac{1}{13}+\dfrac{-1}{14}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{10}+\dfrac{-1}{10}\right)+\left(\dfrac{-1}{11}+\dfrac{1}{11}\right)+\left(\dfrac{1}{12}+\dfrac{-1}{12}\right)+\left(\dfrac{-1}{13}+\dfrac{1}{13}\right)+\left(\dfrac{1}{14}+\dfrac{-1}{14}\right)\)
\(+\left(\dfrac{-1}{15}+\dfrac{1}{15}\right)+\dfrac{1}{16}\)
\(=0+0+...+0+\dfrac{1}{16}\)
\(=\dfrac{1}{16}\)
Bài 1:
a: \(A=\dfrac{1\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}{2\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}\cdot\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{7}+\dfrac{6}{7}=\dfrac{1}{7}+\dfrac{6}{7}=1\)
b: \(B=2000:\left[\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\cdot\dfrac{-\dfrac{7}{6}+\dfrac{7}{8}-\dfrac{7}{10}}{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}\right]\)
\(=2000:\left[\dfrac{2}{7}\cdot\dfrac{-7}{2}\right]=-2000\)
c: \(C=10101\cdot\left(\dfrac{5}{111111}+\dfrac{1}{111111}-\dfrac{4}{111111}\right)\)
\(=10101\cdot\dfrac{2}{111111}=\dfrac{2}{11}\)
a: \(A=\left(\dfrac{-3}{4}+\dfrac{-2}{9}-\dfrac{1}{36}\right)+\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{3}{5}\right)+\dfrac{1}{57}\)
\(=\dfrac{-27-8-1}{36}+\dfrac{5+1+9}{15}+\dfrac{1}{57}\)
=1/57
b: \(B=\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-1}{5}-\dfrac{5}{7}-\dfrac{3}{35}\right)+\dfrac{1}{41}\)
\(=\dfrac{3+1+2}{6}+\dfrac{-7-25-3}{35}+\dfrac{1}{41}\)
=1/41
c: \(C=\left(\dfrac{-1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\dfrac{1}{107}\)
=1-1+1/107
=1/107
Đề sai. câu đầu phải là \(\dfrac{1}{18.21}\) mới đúng.
Nếu câu đầu là \(\dfrac{1}{18.21}\) thì mik có cách làm sau :
\(\dfrac{1}{18.21}+\dfrac{1}{21.24}+\dfrac{1}{24.27}+...+\dfrac{1}{123.126}\)
= \(\dfrac{1}{3}.\left(\dfrac{3}{18.21}+\dfrac{3}{21.24}+\dfrac{3}{24.27}+...+\dfrac{3}{123.126}\right)\)
= \(\dfrac{1}{3}.\left(\dfrac{1}{18}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{24}+\dfrac{1}{24}-\dfrac{1}{27}+...+\dfrac{1}{123}-\dfrac{1}{126}\right)\)
= \(\dfrac{1}{3}.\left(\dfrac{1}{18}-\dfrac{1}{126}\right)\)
= \(\dfrac{1}{3}.\dfrac{1}{21}\)
= \(\dfrac{1}{63}\)