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\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)
\(\Rightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right):x=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(\Rightarrow\left(1-\frac{1}{16}\right):x=1-\frac{1}{5}\)
\(\Rightarrow\frac{15}{16}:x=\frac{4}{5}\)
\(\Rightarrow x=\frac{15}{16}:\frac{4}{5}=1\frac{11}{64}\)
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a) \(\frac{1}{4}+\frac{3}{4}:x=\frac{5}{8}\)
\(\frac{3}{4}:x=\frac{3}{8}\)
\(x=2\)
vậy x=2
b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)
\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2000}{2002}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1000}{2002}\)
\(\frac{1}{x+1}=\frac{1}{2002}\)
\(x+1=2002\)
\(x=2001\)
vậy x=2001
![](https://rs.olm.vn/images/avt/0.png?1311)
ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)
ta gọi B là biểu thức thứ2
\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)
\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)
\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)
\(\Rightarrow x=1\)
mk nghĩ vậy bạn ạ, mk mong nó đúng
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=>4x+(1/2+1/4+1/8+1/16)=1
<=>4x+15/16=1
=>4x=1/16
=>x=1/16:4=1/64
vậy x=1/64
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Rightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(\Rightarrow4x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right)=1\)
\(\Rightarrow4x+\left(1-\frac{1}{16}\right)=1\)
\(\Rightarrow4x+\frac{15}{16}=1\)
\(\Rightarrow4x=1-\frac{15}{16}\)
\(\Rightarrow x=\frac{1}{16}:4\)
\(\Rightarrow x=\frac{1}{64}\)
vậy \(x=\frac{1}{64}\)
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\(a,\frac{7}{8}-\frac{1}{4}.\frac{5}{2}=\frac{x}{16}\)
\(\frac{7}{8}-\frac{5}{8}=\frac{x}{16}\)
\(\frac{2}{8}=\frac{x}{16}\)
\(\frac{4}{16}=\frac{x}{16}\)
=> X=4
k nha
\(\frac{1}{2}\times+\frac{1}{4}\times+\frac{1}{8}\times=\frac{1}{16}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)\times=\frac{1}{16}\)
\(\Rightarrow\frac{7}{8}\times=\frac{1}{16}\)
\(\Rightarrow\times=\frac{1}{14}\)
\(\frac{1}{2}x+\frac{1}{4}x+\frac{1}{8}x=\frac{1}{16}\)
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)x=\frac{1}{16}\)
\(\frac{7}{8}x=\frac{1}{16}\)
\(x=\frac{1}{16}\div\frac{7}{8}\)
\(x=\frac{1}{14}\)