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#)Giải :
\(\left(92-\frac{1}{9}-\frac{2}{10}-\frac{3}{10}-...-\frac{92}{100}\right):\left(\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}\right)\)
\(=\left(1-\frac{1}{9}+1-\frac{2}{10}+1-\frac{3}{11}+...+1-\frac{92}{100}\right)\div\frac{1}{5}\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\)
\(=\left(\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}\right)\div\frac{1}{5}\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\)
\(=8\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\div\frac{1}{5}\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\)
\(=8\div\frac{1}{5}\)
\(=40\)
#~Will~be~Pens~#
Quên mất, bảo tối hôm đó vào làm :)). May là sang nay có ng k ms vào xem. Sorry
S=\(\frac{92-\left(1-\frac{8}{9}\right)-\left(1-\frac{8}{10}\right)-..-\left(1-\frac{8}{100}\right)}{\frac{1}{5}.\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{100}\right)}=\frac{92-92+\left(\frac{8}{9}+\frac{8}{10}+...+\frac{8}{100}\right)}{\frac{1}{5}\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{100}\right)}\)
=\(\frac{8\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{100}\right)}{\frac{1}{5}\left(\frac{1}{9}+\frac{1}{10}+....+\frac{1}{100}\right)}=\frac{8}{\frac{1}{5}}=\frac{8.5}{1}=40\)
Vậy S=40
40 nha bạn! Để hôm nào mk rảnh mk giải chi tiết ra cho! Thông cảm nha! ^_^
Câu b:
\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)
= \(\frac{63}{20}+\frac{3}{5}\)
= \(\frac{15}{4}\)
\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)
\(\frac{25}{8}:\frac{5}{6}\)
\(\frac{25}{8}.\frac{6}{5}\)
\(\frac{30}{8}\)
Ta có:
\(A=\left(x-\frac{1}{2}\right).\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\frac{9}{10}=\frac{1}{3}\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}.\frac{10}{9}\Leftrightarrow x=\frac{47}{54}\)
\(B=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}=\frac{1}{10.x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\Leftrightarrow B=\frac{1}{5}.\frac{100}{101}=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{x}=\frac{1}{10}-\frac{20}{101}=-\frac{99}{1010}\Leftrightarrow x=-\frac{1010}{99}\)
c) Sai đề nhé bạn vì không có kết quả nên không tìm được x.
d) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)
\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)
\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)
\(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}\)
\(=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\)
\(=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\right)\)
\(=1+\frac{28}{5}\)
\(=\frac{33}{5}\)
Ta có:
a) \(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{9}{15}=1+1+\frac{9}{15}=1\frac{9}{15}\)
b)\(\frac{10}{18}+\frac{4}{9}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\right)\)
\(=1+\frac{31}{128}=1\frac{31}{128}\)
a) (1,5 . 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125
=> (2,85 - x - 0,5) : 0,25 = 60
=> (2,85 - 0,5) - x = 60 . 0,25
=> 2,35 - x = 15
=> x = 2,35 - 15
=> x = -12,65
Vậy x = -12,65
b) \(1-\left(5\frac{2}{9}+x-7\frac{7}{18}\right)\div2\frac{1}{6}=0\)
\(\Rightarrow\left(5\frac{2}{9}-7\frac{7}{18}+x\right)\div2\frac{1}{6}=1-0\)
\(\Rightarrow\left(\frac{47}{9}-\frac{133}{18}+x\right)\div2\frac{1}{6}=1\)
\(\Rightarrow\frac{-13}{6}+x=2\frac{1}{6}\)
\(\Rightarrow x=2\frac{1}{6}-\frac{-13}{6}\)
\(\Rightarrow x=\frac{13}{6}+\frac{13}{6}\)
\(\Rightarrow x=\frac{26}{6}\)
\(\Rightarrow x=\frac{13}{3}\)
Vậy \(x=\frac{13}{3}\)
c) \(35\left(2\frac{1}{5}-x\right)=32\)
\(\Rightarrow2\frac{1}{5}-x=32\div35\)
\(\Rightarrow\frac{11}{5}-x=\frac{32}{35}\)
\(\Rightarrow x=\frac{11}{5}-\frac{32}{35}\)
\(\Rightarrow x=\frac{9}{7}\)
Vậy \(x=\frac{9}{7}\)
d) \(\frac{4}{3}+\left(x\div2\frac{2}{3}-0,5\right).1\frac{35}{55}=0,6\)
\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{3}{5}-\frac{4}{3}\)
\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{-11}{15}\)
\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-11}{15}\div\frac{18}{11}\)
\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-121}{270}\)
\(\Rightarrow x\div\frac{8}{3}=\frac{-121}{270}+\frac{1}{2}\)
\(\Rightarrow x\div\frac{8}{3}=\frac{7}{135}\)
\(\Rightarrow x=\frac{7}{135}.\frac{8}{3}\)
\(\Rightarrow x=\frac{56}{405}\)
Vậy \(x=\frac{56}{405}\)
e) \(1\frac{1}{3}.2\frac{2}{4}\div\frac{5}{6}.1\frac{1}{11}=11-5\div x\)
\(\Rightarrow\frac{4}{3}.\frac{5}{2}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)
\(\Rightarrow\frac{10}{3}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)
\(\Rightarrow4.\frac{12}{11}=11-5\div x\)
\(\Rightarrow11-5\div x=\frac{48}{11}\)
\(\Rightarrow5\div x=11-\frac{48}{11}\)
\(\Rightarrow5\div x=\frac{73}{11}\)
\(\Rightarrow x=5\div\frac{73}{11}\)
\(\Rightarrow x=\frac{55}{73}\)
Vậy \(x=\frac{55}{73}\)
a) (1,5 * 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125
(2,85 - x - 0,5) : 0,25 = 60
(2,85 - x - 0,5) = 60 x 0,25
(2,85 - x - 0,5) = 15
2,35 - x = 15
x = 2,35 - 15
x = -12,65