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\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}-[2\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)]=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}-[2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)]=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}-[2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{5}-\frac{1}{16}\right)]=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}-[2.\frac{3}{16}]=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}=1\)
\(\Rightarrow x=2008\)
\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}=\frac{21}{45}\)
\(\Rightarrow x=15\)
\(\frac{x}{2008}-\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)
\(\frac{x}{2008}=1\)
x=2008
a) \(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{47.49}=\frac{24}{x+4}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+......+\frac{2}{47.49}\right)=\frac{24}{x+4}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{47}-\frac{1}{49}\right)=\frac{24}{x+4}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{49}\right)=\frac{24}{x+4}\)
\(\Rightarrow\frac{1}{2}.\frac{48}{49}=\frac{24}{x+4}\)
\(\Rightarrow\frac{24}{49}=\frac{24}{x+4}\)
\(\Rightarrow x+4=49\Rightarrow x=45\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}=\frac{24}{x+4}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\right)=\frac{24}{x+4}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)=\frac{24}{x+4}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{49}\right)=\frac{24}{x+4}\Leftrightarrow A=\frac{1}{2}.\frac{48}{49}=\frac{24}{x+4}\)
\(\Rightarrow A=\frac{24}{49}=\frac{24}{x+4}\Leftrightarrow x=49-4=45\)
Bài b) hình như sai đề thì phải đó bạn.
bạn gửi câu trả lời của bạn trước nhé rồi mình sẽ gửi câu trả lời cho
6666 +8888=
6666 + 8888 = 15554
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