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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: \(25.\left(\dfrac{1}{18}-\dfrac{1}{21}+...+\dfrac{1}{123}-\dfrac{1}{126}\right)\)
\(A=\frac{5}{18.21}+\frac{5}{21.24}+\frac{5}{24.27}+...+\frac{5}{123.126}\)
\(A=\frac{5}{3}\left(\frac{3}{18.21}+\frac{3}{21.24}+\frac{3}{24.27}+...+\frac{3}{123.126}\right)\)
\(A=\frac{5}{3}\left(\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+\frac{1}{24}-\frac{1}{27}+...+\frac{1}{123}-\frac{1}{126}\right)\)
\(A=\frac{5}{3}\left(\frac{1}{18}-\frac{1}{126}\right)\)
\(A=\frac{5}{3}.\frac{1}{21}\)
\(A=\frac{5}{63}\)
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 :
\(\dfrac{1}{18.21}+\dfrac{1}{21.24}+\dfrac{1}{24.27}+...+\dfrac{1}{123.126}\)
= \(\dfrac{1}{3}.\left(\dfrac{3}{18.21}+\dfrac{3}{21.24}+\dfrac{3}{24.27}+...+\dfrac{3}{123.126}\right)\)
= \(\dfrac{1}{3}.\left(\dfrac{1}{18}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{24}+\dfrac{1}{24}-\dfrac{1}{27}+...+\dfrac{1}{123}-\dfrac{1}{126}\right)\)
= \(\dfrac{1}{3}.\left(\dfrac{1}{18}-\dfrac{1}{126}\right)\)
= \(\dfrac{1}{3}.\dfrac{1}{21}\)
= \(\dfrac{1}{63}\)
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=\(\frac{5}{63}\)
\(A=\frac{5}{18.21}+\frac{5}{21.24}+\frac{5}{24.27}+...+\frac{4}{123.126}\)
\(=\frac{5}{3}.\left(\frac{3}{18.21}+\frac{3}{21.24}+\frac{3}{24.27}+...+\frac{3}{123.126}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{123}-\frac{1}{126}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{18}-\frac{1}{126}\right)\)
\(=\frac{5}{3}.\frac{1}{21}\)
\(=\frac{5}{63}\)
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 ; \(\dfrac{n}{x\left(x+n\right)}=\dfrac{1}{x}-\dfrac{1}{x+n}\)
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=\(\frac{5}{3}\left(\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+............+\frac{1}{123}-\frac{1}{126}\right)\)
=\(\frac{5}{3}.\left(\frac{1}{18}-\frac{1}{126}\right)\)
=\(\frac{5}{3}.\frac{1}{21}\)
=\(\frac{5}{63}\)
\(\dfrac{5}{12.17}+\dfrac{35}{17.18}-\dfrac{39}{18.21}+\dfrac{30}{21.72}\)
=\(\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}+\dfrac{1}{18}-\dfrac{1}{18}-\dfrac{1}{21}+\dfrac{5.6}{21.12.6}\)
= \(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{5}{21.12}\)
=\(\dfrac{21}{12.21}-\dfrac{12}{12.21}+\dfrac{5}{21.12}\)
= \(\dfrac{21-12+5}{12.21}\)
=\(\dfrac{14}{252}\)
\(=\dfrac{1}{18}\)
a, \(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
\(A=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\right)\)
\(A=5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
(do \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(A=5.\left(1-\dfrac{1}{31}\right)=5.\dfrac{30}{31}=\dfrac{150}{31}\)
b, \(B=\dfrac{6}{15.18}+\dfrac{6}{18.21}+...+\dfrac{6}{87.90}\)
\(B=2\left(\dfrac{3}{15.18}+\dfrac{3}{18.21}+...+\dfrac{13}{87.90}\right)\)
\(B=2\left(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)
(do \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(B=2\left(\dfrac{1}{15}-\dfrac{1}{90}\right)=2.\dfrac{1}{18}=\dfrac{1}{9}\)
c, \(C=\dfrac{3^2}{8.11}+\dfrac{3^2}{11.14}+...+\dfrac{3^2}{197.200}\)
\(C=3\left(\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{197.200}\right)\)
\(C=3\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
(do \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(C=3\left(\dfrac{1}{8}-\dfrac{1}{200}\right)=3.\dfrac{3}{35}=\dfrac{9}{35}\)
Chúc bạn học tốt!!!
Ta có:
`7/(-5) = (-7)/5 < -1`
`-1 < (-2)/5 < 0`
`0 < (-4)/(-5) = 4/5`
Thứ tự giảm dần là: `4/5; (-2)/5; 7/(-5)`
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