Tìm x:
\(\frac{x}{2}\)+ \(\frac{x}{6}\)+ \(\frac{x}{12}\)+ \(\frac{x}{20}\)+ . . . + \(\frac{x}{132}\)= 5\(\frac{1}{2}\)
Tìm cả bài giải nữa nhé !
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\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)\div x=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{32}\right)\)
\(\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)\div x=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\right)\)
\(\frac{15}{16}\div x=\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}\right)\)
\(\frac{15}{16}\div x=\left(\frac{1}{1}-\frac{1}{12}\right)\)
\(\frac{15}{16}\div x=\frac{11}{12}\)
\(x=\frac{15}{16}\div\frac{11}{12}\)
\(x=\frac{15}{16}\times\frac{12}{11}\)
\(\Rightarrow x=\frac{180}{176}=\frac{45}{44}\)
Ta có \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\)
\(\frac{15}{16}:x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\)
\(\frac{15}{16}:x=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}\)
\(\frac{15}{16}:x=1-\frac{1}{12}\)
\(\frac{15}{16}:x=\frac{11}{12}\)
\(x=\frac{15}{16}:\frac{11}{12}\)
\(x=\frac{180}{176}\)
Đúng thì like nha
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{1}{2}+\frac{1}{6}+....+\frac{1}{32}\)
\(\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right):x=\frac{1}{1\times2}+\frac{1}{2\times3}+.....+\frac{1}{11\times12}\)
\(\frac{15}{16}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{12}\)
\(\frac{15}{16}:x=1-\frac{1}{12}\)
\(\frac{15}{16}:x=\frac{11}{12}\)
\(x=\frac{15}{16}:\frac{11}{12}\)
\(x=\frac{45}{44}\)
Tính \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
2 x A = \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)
2 x A - A = A = \(1-\frac{1}{16}=\frac{15}{16}\)
Tính \(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{11\times12}\)
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}=\frac{1}{1}-\frac{1}{12}=\frac{11}{12}\)
Ta có: \(\frac{15}{16}:x=\frac{11}{12}\Rightarrow x=\frac{15}{16}:\frac{11}{12}=\frac{15}{16}\times\frac{12}{11}=\frac{45}{44}\)
Vậy...
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009
}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\frac{x+1-1}{x+1}=\frac{2008}{2009}\)
\(\frac{x}{x+1}=\frac{2008}{2009}\)
\(2009x=2008\left(x+1\right)\)
\(2009x=2008x+2008\)
\(2009x-2008x=2008\)
\(x=2008\)
Vậy x=2008
+ \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\)
=> \(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)
=> \(A=2A-A=1-\frac{1}{16}=\frac{15}{16}\)
+ \(B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{11x12}\)
\(B=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{12-11}{11x12}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(B=1-\frac{1}{12}=\frac{11}{12}\)
\(A:x=B\Rightarrow x=A:B=\frac{15}{16}:\frac{11}{12}=\frac{15}{16}x\frac{12}{11}=\frac{45}{44}=1\frac{1}{44}\)
\(x(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}=\) \(5\frac{1}{2}\)
\(x\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\right)=\frac{11}{2}\)
\(x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\right)=\frac{11}{2}\)
\(x\left(1-\frac{1}{12}\right)=\frac{11}{2}\)
\(x\cdot\frac{11}{12}=\frac{11}{2}\)
\(x=\frac{11}{2}:\frac{11}{12}\)
\(x=6\)
Vậy x = 6
\(\frac{x}{2}+\frac{x}{6}+\frac{x}{12}+\frac{x}{20}+...+\frac{x}{132}=5\frac{1}{2}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\right)=\frac{11}{2}\)
\(\Rightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\right)=\frac{11}{2}\)
\(\Rightarrow x\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}=\frac{11}{2}\right)\)
\(\Rightarrow x\left(1-\frac{1}{12}\right)=\frac{11}{2}\)
\(\Rightarrow x.\frac{11}{12}=\frac{11}{12}\)
\(\Rightarrow x=\frac{11}{12}:\frac{11}{12}=1\)
Vậy x = 1