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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 :
\(\dfrac{1}{18.21}+\dfrac{1}{21.24}+\dfrac{1}{24.27}+...+\dfrac{1}{123.126}\)
= \(\dfrac{1}{3}.\left(\dfrac{3}{18.21}+\dfrac{3}{21.24}+\dfrac{3}{24.27}+...+\dfrac{3}{123.126}\right)\)
= \(\dfrac{1}{3}.\left(\dfrac{1}{18}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{24}+\dfrac{1}{24}-\dfrac{1}{27}+...+\dfrac{1}{123}-\dfrac{1}{126}\right)\)
= \(\dfrac{1}{3}.\left(\dfrac{1}{18}-\dfrac{1}{126}\right)\)
= \(\dfrac{1}{3}.\dfrac{1}{21}\)
= \(\dfrac{1}{63}\)
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=\(\frac{5}{3}\left(\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+............+\frac{1}{123}-\frac{1}{126}\right)\)
=\(\frac{5}{3}.\left(\frac{1}{18}-\frac{1}{126}\right)\)
=\(\frac{5}{3}.\frac{1}{21}\)
=\(\frac{5}{63}\)
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=\(\frac{5}{63}\)
\(A=\frac{5}{18.21}+\frac{5}{21.24}+\frac{5}{24.27}+...+\frac{4}{123.126}\)
\(=\frac{5}{3}.\left(\frac{3}{18.21}+\frac{3}{21.24}+\frac{3}{24.27}+...+\frac{3}{123.126}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{123}-\frac{1}{126}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{18}-\frac{1}{126}\right)\)
\(=\frac{5}{3}.\frac{1}{21}\)
\(=\frac{5}{63}\)
mình ghi nhầm nên các bạn cứ hết hai phân số là một câu nhé ví dụ như \(\dfrac{-5}{8}\):\(\dfrac{15}{4}\)
a)\(\dfrac{-5}{9}+\dfrac{5}{9}:\left(1\dfrac{2}{3}-2\dfrac{1}{12}\right)\)
\(=\dfrac{-5}{9}+\dfrac{5}{9}:\left(\dfrac{20}{12}-\dfrac{25}{12}\right)\)
\(=\dfrac{-5}{9}+\dfrac{5}{9}\cdot\dfrac{-12}{5}\)
\(=\dfrac{-5}{9}+\dfrac{-12}{9}\)
\(=\dfrac{-17}{9}\)
b) \(\dfrac{-7}{25}\cdot\dfrac{11}{13}+\dfrac{-7}{25}\cdot\dfrac{2}{13}-\dfrac{18}{25}\)
\(=\dfrac{-7}{25}\cdot\left(\dfrac{11}{13}+\dfrac{2}{13}\right)-\dfrac{18}{25}\)
\(=\dfrac{-7}{25}\cdot1-\dfrac{18}{25}\)
\(=\dfrac{-25}{25}\)
\(=-1\)
c) \(\left(\dfrac{5}{7}\cdot0,6-5:3\dfrac{1}{2}\right)\cdot\left(40\%-1,4\right)\cdot\left(-2\right)^3\)
\(=\left(\dfrac{3}{7}-\dfrac{10}{7}\right)\cdot\left(0,4-1,4\right)\cdot\left(-8\right)\)
\(=\left(-1\right)\cdot\left(-1\right)\cdot\left(-8\right)\)
\(=-8\)
a. \(\dfrac{-3}{5}\)
b. \(\dfrac{-2}{3}\) c. \(\dfrac{4}{39}\) d. \(\dfrac{26}{45}\)
\(A=\dfrac{25}{18\cdot21}+\dfrac{25}{21\cdot24}+\dfrac{25}{24\cdot27}+...+\dfrac{25}{123\cdot126}\)
\(=25\left(\dfrac{1}{18\cdot21}+\dfrac{1}{21\cdot24}+\dfrac{1}{24\cdot27}+...+\dfrac{1}{123\cdot126}\right)\)
\(=\dfrac{25}{3}\left(\dfrac{3}{18\cdot21}+\dfrac{3}{21\cdot24}+\dfrac{3}{24\cdot27}+...+\dfrac{3}{123\cdot126}\right)\)
\(=\dfrac{25}{3}\left(\dfrac{1}{18}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{24}+...+\dfrac{1}{123}-\dfrac{1}{126}\right)\)
\(=\dfrac{25}{3}\left(\dfrac{1}{18}-\dfrac{1}{126}\right)\)\(=\dfrac{25}{3}\cdot\dfrac{1}{21}=\dfrac{25}{63}\)
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: \(25.\left(\dfrac{1}{18}-\dfrac{1}{21}+...+\dfrac{1}{123}-\dfrac{1}{126}\right)\)