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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
+ \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\)
=> \(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)
=> \(A=2A-A=1-\frac{1}{16}=\frac{15}{16}\)
+ \(B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{11x12}\)
\(B=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{12-11}{11x12}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(B=1-\frac{1}{12}=\frac{11}{12}\)
\(A:x=B\Rightarrow x=A:B=\frac{15}{16}:\frac{11}{12}=\frac{15}{16}x\frac{12}{11}=\frac{45}{44}=1\frac{1}{44}\)
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)\div x=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{32}\right)\)
\(\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)\div x=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\right)\)
\(\frac{15}{16}\div x=\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}\right)\)
\(\frac{15}{16}\div x=\left(\frac{1}{1}-\frac{1}{12}\right)\)
\(\frac{15}{16}\div x=\frac{11}{12}\)
\(x=\frac{15}{16}\div\frac{11}{12}\)
\(x=\frac{15}{16}\times\frac{12}{11}\)
\(\Rightarrow x=\frac{180}{176}=\frac{45}{44}\)
\(\left(y-\frac{1}{2}\right):\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(y-\frac{1}{2}\right):\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(y-\frac{1}{3}\right):\left(1-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(y-\frac{1}{2}\right):\frac{9}{10}=\frac{1}{3}\)
\(\Leftrightarrow\left(y-\frac{1}{2}\right)=\frac{3}{10}\)
\(\Leftrightarrow y=\frac{4}{5}\)
Ta có : \(\frac{1}{4}+\frac{1}{3}:\frac{1}{x}=\frac{11}{12}\)
\(\Rightarrow\frac{1}{3}:\frac{1}{x}=\frac{11}{12}-\frac{1}{4}\)
\(\frac{1}{3}:\frac{1}{x}=\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}:\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}\times\frac{3}{2}\)
\(\frac{1}{x}=\frac{1}{2}\)
=> x = 2
a) \(\frac{x\div3-16}{2}+21=38\)
\(\frac{x\div3-16}{2}=38+21\)
\(\frac{x\div3-16}{2}=59\)
\(x\div3-16=59.2\)
\(x\div3-16=118\)
\(x\div3=118+16\)
\(x\div3=134\)
\(x=134.3\)
\(x=402\)
b) \(\frac{1}{4}+\frac{1}{3}\div\frac{1}{x}=\frac{11}{12}\)
\(\frac{1}{3}\div\frac{1}{x}=\frac{11}{12}-\frac{1}{4}\)
\(\frac{1}{3}\div\frac{1}{x}=\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}\div\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{2}\)
Vậy x = ....
\(\left(1-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}\right)\cdot X=\frac{11}{6}\)
\(< =>\left(\frac{1}{2}-\frac{1}{12}-\frac{1}{60}\right)\cdot X=\frac{11}{6}\)
\(< =>\left(\frac{30}{60}-\frac{5}{60}-\frac{1}{60}\right)\cdot X=\frac{11}{6}\)
\(< =>\left(\frac{30-5-1}{60}\right)\cdot X=\frac{11}{6}\)
\(< =>\frac{2}{5}\cdot X=\frac{11}{6}\)
\(< =>X=\frac{11}{6}:\frac{2}{5}\)
\(< =>X=\frac{55}{12}\)
CHUC BAN HOC TOT >.<
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{1}{2}+\frac{1}{6}+....+\frac{1}{32}\)
\(\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right):x=\frac{1}{1\times2}+\frac{1}{2\times3}+.....+\frac{1}{11\times12}\)
\(\frac{15}{16}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{12}\)
\(\frac{15}{16}:x=1-\frac{1}{12}\)
\(\frac{15}{16}:x=\frac{11}{12}\)
\(x=\frac{15}{16}:\frac{11}{12}\)
\(x=\frac{45}{44}\)
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...