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\(I=\dfrac{2}{1\times5}+\dfrac{2}{5\times9}+\dfrac{2}{9\times13}+...+\dfrac{2}{181\times185}\)
\(=\dfrac{1}{2}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+...+\dfrac{4}{181\times185}\right)\)
\(=\dfrac{1}{2}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{181}-\dfrac{1}{185}\right)\)
\(=\dfrac{1}{2}\times\left(1-\dfrac{1}{185}\right)=\dfrac{1}{2}\times\dfrac{184}{185}=\dfrac{92}{185}\)
\(S=\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{41\times45}\)
\(\Rightarrow4S=\dfrac{4}{5\times9}+\dfrac{4}{9\times13}+...+\dfrac{4}{41\times45}\)
\(\Rightarrow4S=\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\)
\(\Rightarrow4S=\dfrac{1}{5}-\dfrac{1}{45}\)
\(\Rightarrow4S=\dfrac{8}{45}\)
\(\Rightarrow S=\dfrac{2}{45}\)
\(=\dfrac{1}{4}\left(\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{8}{45}=\dfrac{2}{45}\)
\(\Rightarrow\dfrac{7}{x}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{29}{45}\left(x\ne0\right)\\ \Rightarrow\dfrac{7}{x}+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}+\dfrac{8}{45}=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{21}{45}\Rightarrow\dfrac{21}{3x}=\dfrac{21}{45}\Rightarrow3x=45\\ \Rightarrow x=15\)
a) \(M=\frac{2\times2}{1\times5}+\frac{2\times2}{5\times9}+\frac{2\times2}{9\times13}+...+\frac{2\times2}{45\times40}\)
\(M=\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{45\times49}\)
\(M=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{45}-\frac{1}{49}\)
\(M=1-\frac{1}{49}\)
\(M=\frac{48}{49}\)
b) \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+5+...+10}\)
= \(\frac{2}{2\times\left(1+2\right)}+\frac{2}{2\times\left(1+2+3\right)}+...+\frac{2}{2\times\left(1+2+3+...+10\right)}\)
\(=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{110}\)
\(=\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{10\times11}\)
\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(=2\times\frac{9}{22}\)
\(=\frac{9}{11}\)
Mình trả lời câu a nha M= 4/1*5+4/5*9+4/9*13+...+4/45*49 M=1-1/5+1/5-1/9+1/9-1/13+...+1/45-1/49 M=1-1/49=48/49
\(B=\dfrac{1}{4}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+\dfrac{4}{9\times13}+...+\dfrac{4}{125\times129}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{125}-\dfrac{1}{129}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{129}\right)=\dfrac{1}{4}\times\dfrac{128}{129}=\dfrac{32}{129}\)
A = 4/1 x 5 + 4/5 x 9 + 4/9 x 13 + .... + 4/91 x 95 + 4/95 x 99
A = 1 - 1/5 + 1/5 -1/9 + 1/9 - 1/13 + .... + 1/91 - 1/95 + 1/95 - 1/99
A = 1 - 1/99
A = 98/99
B = 1/6 + 1/12 + 1/20 + ... + 1/132
B = 1/2 x 3 + 1/3 x 4 + 1/4 x 5 + ... + 1/11 x 12
B = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/11 - 1/12
B = 1/2 - 1/12
B = 5/12
3/(1×5) + 3/(5×9) + 3/(9×13) + 3/(13×17) + 3/(17×21)
= 3/4 × (1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17 + 1/17 - 1/21)
= 3/4 × (1 - 1/21)
= 3/4 × 20/21
= 5/7
\(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+...+\dfrac{1}{45\times49}\)
\(=\dfrac{1}{4}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+...+\dfrac{4}{45\times49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{45}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{49}\right)=\dfrac{1}{4}\times\dfrac{48}{49}=\dfrac{12}{49}\)