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PN
9
3 tháng 6 2018
\(\left(x-\frac{1}{2}\right).\frac{5}{2}+\frac{1}{2}=\frac{7}{4}\)
\(\left(x-\frac{1}{2}\right).\frac{5}{2}=\frac{7}{4}-\frac{1}{2}\)
\(\left(x-\frac{1}{2}\right).\frac{5}{2}=\frac{5}{4}\)
\(x-\frac{1}{2}=\frac{5}{4}:\frac{5}{2}\)
\(x-\frac{1}{2}=\frac{1}{2}\)
\(x=1\)
HT
3 tháng 7 2018
Câu b:
\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)
= \(\frac{63}{20}+\frac{3}{5}\)
= \(\frac{15}{4}\)
LT
7 tháng 7 2018
\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)
\(\frac{25}{8}:\frac{5}{6}\)
\(\frac{25}{8}.\frac{6}{5}\)
\(\frac{30}{8}\)
25 tháng 7 2015
(15/4.4-12/5.5/4)-(7/2:5/2-1/5)-1/5.x=51/5
(15-3)-(7/5-1/5)-1/5.x=51/5
12-6/5-1/5.x=51/5
54/5-1/5.x=51/5
1/5x=54/5-51/5
1/5x=3/5
x=3
3 tháng 7 2017
=) \(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{255}{256}.x=\frac{108}{7}\)
=) \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{15.17}{16.16}.x=\frac{108}{7}\)
=) \(\frac{1.2.3.4.....15}{2.3.4.....16}.\frac{3.4.5.....17}{2.3.4.....16}.x=\frac{108}{7}\)
=) \(\frac{1}{16}.\frac{17}{2}.x=\frac{108}{7}\)=) \(\frac{17}{32}.x=\frac{108}{7}\)=) \(x=\frac{108}{7}:\frac{17}{32}\)
=) \(x=\frac{3456}{119}\)
TP
16 tháng 8 2017
Ta có:
\(A=\left(x-\frac{1}{2}\right).\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\frac{9}{10}=\frac{1}{3}\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}.\frac{10}{9}\Leftrightarrow x=\frac{47}{54}\)
\(B=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}=\frac{1}{10.x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\Leftrightarrow B=\frac{1}{5}.\frac{100}{101}=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{x}=\frac{1}{10}-\frac{20}{101}=-\frac{99}{1010}\Leftrightarrow x=-\frac{1010}{99}\)
TP
16 tháng 8 2017
c) Sai đề nhé bạn vì không có kết quả nên không tìm được x.
d) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)
\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)
\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)
26 tháng 7 2018
\(\frac{11}{9}+\frac{35}{18}\)
\(=\frac{19}{6}\)
\(\frac{22}{21}:\frac{1}{2}+\frac{25}{12}:\frac{1}{2}\)
\(\frac{44}{21}+\frac{25}{6}\)
\(\frac{263}{42}\)
Ta có \(\left(x-\frac{1}{2}\right).\frac{5}{2}+\frac{1}{2}=\frac{7}{4}\)
\(\Rightarrow\left(x-\frac{1}{2}\right).\frac{5}{2}=\frac{7}{4}-\frac{1}{2}\)
\(\Rightarrow\left(x-\frac{1}{2}\right).\frac{5}{2}=\frac{5}{4}\)
\(\Rightarrow x-\frac{1}{2}=\frac{5}{4}\div\frac{5}{2}\)
\(\Rightarrow x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow x=1\)
(x-\(\frac{1}{2}\)).\(\frac{5}{2}\)+\(\frac{1}{2}\)=\(\frac{7}{4}\)
(x-\(\frac{1}{2}\)).\(\frac{5}{2}\)=\(\frac{7}{4}\)- \(\frac{1}{2}\)
(x-\(\frac{1}{2}\)).\(\frac{5}{2}\)= \(\frac{7}{4}\)- \(\frac{2}{4}\)
(x-\(\frac{1}{2}\)).\(\frac{5}{2}\)\(\frac{5}{2}\)= \(\frac{5}{4}\)
(x-\(\frac{1}{2}\)) = \(\frac{5}{4}\): \(\frac{5}{2}\)= \(\frac{5}{4}\). \(\frac{2}{5}\)= \(\frac{1}{2}\)
x = \(\frac{1}{2}\)+ \(\frac{1}{2}\)=\(\frac{1}{4}\)
vậy x\(\frac{7}{4}\)\(\frac{7}{4}\)