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Ta có:\(\frac{16}{25}\)+ (x+\(\frac{1}{3}\))\(^2\)=1
(x+\(\frac{1}{3}\))\(^2\)= \(\frac{9}{25}\)
x+\(\frac{1}{3}\)= \(\frac{3}{5}\)
x=\(\frac{4}{15}\)
\(\frac{16}{25}+\left(x+\frac{1}{3}\right)^2=1\)
\(\left(x+\frac{1}{3}\right)^2=1-\frac{16}{25}\)
\(\left(x+\frac{1}{3}\right)^2=\frac{9}{25}=0,36\)
\(x+\frac{1}{3}=0,6=\frac{3}{5}\)
\(x=\frac{3}{5}-\frac{1}{6}\)
\(x=\frac{13}{30}\)
Ta có: \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2011}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2012}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2012}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1006}\right)\)
\(=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}\) (ĐPCM)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}< 1\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}< 1\)
\(S=1-\frac{1}{50}< 1\)
\(S=\frac{49}{50}< 1\left(đpcm\right)\)
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(\Leftrightarrow3A=1+\frac{1}{3}+\frac{1}{3^{^2}}+...+\frac{1}{3^{98}}\)
\(\Leftrightarrow3A-A=1-\frac{1}{3^{99}}\)
\(\Leftrightarrow2A=1-\frac{1}{3^{99}}\)
\(\Leftrightarrow A=\left(1-\frac{1}{3^{99}}\right)\div2\)
B = \(\frac{1}{1+\frac{1}{2}}+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}+\frac{1}{1+\frac{1}{3}}\)
B = \(\frac{1}{\frac{2}{2}+\frac{1}{2}}+\frac{1}{1+\frac{1}{\frac{2}{2}+\frac{1}{2}}}+\frac{1}{\frac{3}{3}+\frac{1}{3}}\)
B = \(\frac{1}{\frac{3}{2}}+\frac{1}{1+\frac{1}{\frac{3}{2}}}+\frac{1}{\frac{4}{3}}\)
B = \(\frac{2}{3}+\frac{1}{1+\frac{2}{3}}+\frac{3}{4}\)
B = \(\frac{2}{3}+\frac{1}{\frac{3}{3}+\frac{2}{3}}+\frac{3}{4}\)
B = \(\frac{2}{3}+\frac{1}{\frac{5}{3}}+\frac{3}{4}\)
B = \(\frac{2}{3}+\frac{3}{5}+\frac{3}{4}\)
B = \(\frac{121}{60}\)
cái này tính từng phân số theo quy luật từ dưới lên trên rồi tính B là ra