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A=\(\frac{7}{10}\)*(\(\frac{10}{3\cdot13}\)+\(\frac{10}{13\cdot23}\)+\(\frac{10}{23\cdot33}\)+\(\frac{10}{43\cdot53}\)+\(\frac{10}{53\cdot63}\))
A=\(\frac{7}{10}\)*(\(\frac{1}{3}\)-\(\frac{1}{13}\)+\(\frac{1}{13}\)-\(\frac{1}{23}\)+\(\frac{1}{23}\)-\(\frac{1}{33}\)+\(\frac{1}{43}\)-\(\frac{1}{53}\)+\(\frac{1}{53}\)-\(\frac{1}{63}\))
A=\(\frac{7}{10}\)*(\(\frac{1}{3}\)-\(\frac{1}{33}\)+\(\frac{1}{43}\)-\(\frac{1}{63}\))
A=\(\frac{7}{10}\)*(\(\frac{10}{33}\)+\(\frac{20}{2709}\))
A=\(\frac{7}{10}\)*\(\frac{9250}{29799}\)
A=\(\frac{925}{4257}\)
Lời giải:
a, \(\dfrac{12}{19}.\dfrac{7}{15}.\dfrac{-13}{17}.\dfrac{19}{12}.\dfrac{17}{13}\)
\(=\dfrac{12}{19}.\dfrac{-7}{15}.\dfrac{13}{17}.\dfrac{19}{12}.\dfrac{17}{13}\)
\(=\left(\dfrac{12}{19}.\dfrac{19}{12}\right).\left(\dfrac{13}{17}.\dfrac{17}{13}\right).\dfrac{-7}{15}\)
\(=1.1.\dfrac{-7}{15}\)
\(=\dfrac{-7}{15}\)
b, \(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{27}=\dfrac{-5}{9}\\\dfrac{-45}{b}=\dfrac{-5}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}9a=-5.27\\-5b=-45.9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}9a=-135\\-5b=-405\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=-15\\b=-81\end{matrix}\right.\)
Vậy, \(a=-15;b=-81\).
a) \(\left(\dfrac{11}{12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{5}{3}\)
\(\Rightarrow\left(\dfrac{11}{1.12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{5}{3}\)
\(\Rightarrow\left(1-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{34}+...+\dfrac{1}{89}-\dfrac{1}{100}\right)+x=\dfrac{5}{3}\)
\(\Rightarrow1-\dfrac{1}{100}+x=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{5}{3}-1+\dfrac{1}{100}\)
\(\Rightarrow x=\dfrac{500}{300}-\dfrac{300}{300}+\dfrac{3}{300}\)
\(\Rightarrow x=\dfrac{203}{300}\)
b) \(\left(\dfrac{5}{11.16}+\dfrac{5}{16.21}+...+\dfrac{5}{19.24}\right)-x+\dfrac{1}{3}=\dfrac{7}{3}\)
=>\(\left(\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{19}-\dfrac{1}{24}\right)-x=\dfrac{7}{3}-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{1}{11}-\dfrac{1}{24}-x=2\)
\(\Rightarrow-x=2-\dfrac{1}{11}+\dfrac{1}{24}\)
\(\Rightarrow-x=\dfrac{528}{264}-\dfrac{24}{264}+\dfrac{11}{264}\)
\(\Rightarrow x=\dfrac{515}{264}\)
c) Câu hỏi của Đàm Chu Hữu An - Toán lớp 6 - Học toán với OnlineMath
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(A=\dfrac{1}{2}.\dfrac{100}{101}\)
\(A=\dfrac{50}{101}\)
Bạn xem Sách bài tập toán 6 tập 2 trang 24 bài 9.3b; 9.4; 9.5; 9.6 sẽ có
Nhớ tick cho mình nha
A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+......+\dfrac{1}{99.101}\)
= \(\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+..+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{101}\right)\)
=\(\dfrac{1}{2}.\dfrac{100}{101}\)
=\(\dfrac{50}{101}\)
Câu 1:
a,\(x=\dfrac{1}{4}+\dfrac{2}{13}\)
\(x=\dfrac{13}{52}+\dfrac{8}{52}=\dfrac{21}{52}\)
Câu 2:
a,\(\dfrac{-2}{5}+\dfrac{3}{-4}+\dfrac{6}{7}+\dfrac{3}{4}+\dfrac{2}{5}\)
\(=\left(\dfrac{-2}{5}+\dfrac{2}{5}\right)+\left(\dfrac{3}{-4}+\dfrac{3}{4}\right)+\dfrac{6}{7}\)
=\(0+0+\dfrac{6}{7}=\dfrac{6}{7}\)
b,\(\dfrac{7}{15}+\dfrac{4}{-9}+\dfrac{-2}{11}+\dfrac{8}{15}+\dfrac{-5}{9}\)
=\(\left(\dfrac{7}{15}+\dfrac{8}{15}\right)+\left(\dfrac{4}{-9}+\dfrac{-5}{9}\right)+\dfrac{-2}{11}\)
=\(\dfrac{15}{15}+\dfrac{-9}{9}+\dfrac{-2}{11}=1+\left(-1\right)+\dfrac{-2}{11}\)
=\(0+\dfrac{-2}{11}=\dfrac{-2}{11}\)
c, \(\dfrac{-5}{7}+\dfrac{5}{13}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}\)
=\(\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+\dfrac{-5}{7}\)
=\(\dfrac{13}{13}+\dfrac{-41}{41}+\dfrac{-5}{7}=1+\left(-1\right)+\dfrac{-5}{7}\)
=\(0+\dfrac{-5}{7}=\dfrac{-5}{7}\)
\(\dfrac{17,8.3,7-7,8.4,8+17,8.5,7-7,8.4,6}{4,7.6,5+4,7.2,5+4,7}\)
=\(\dfrac{17,8.3,7+17,8.5,7-7,8.4,8-7,8.4,6}{4,7\left(6,5+2,5+1\right)}\)
=\(\dfrac{17,8\left(3,7+5,7\right)-\left[7,8\left(4,8+4,6\right)\right]}{47}\)
=\(\dfrac{17,8.9,4-7,8.9,4}{47}\)
=\(\dfrac{9,4\left(17,8-7,8\right)}{47}\)
=\(\dfrac{9,4.10}{47}=\dfrac{94}{47}=2\)
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