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ở câu 1 ở mỗi phẫn số chúng ta cộng thêm 1, tổng là ta cộng thêm 5. Lấy 5 + -5=0. Rồi ta được tất cả tử là x+200,đặt chung ra ngoài,từ đó tính x=-200
đặt \(A=\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+........+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\frac{1}{2}A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+........+\frac{1}{x.\left(x+1\right)}=\frac{2}{9}.\frac{1}{2}\)
\(\frac{1}{2}A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+........+\frac{1}{x.\left(x+1\right)}=\frac{1}{9}\)
\(\frac{1}{2}A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{2}A=\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=18-1\)
\(\Rightarrow x=17\)
vậy \(x=17\)
\(A=\frac{12}{1.5}+\frac{12}{5.9}+\frac{12}{9.13}+.............+\frac{12}{101.105}\)
\(=3.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+............+\frac{4}{101.105}\right)\)
\(=3\left(1-\frac{1}{105}\right)\)
\(=3.\frac{104}{105}=\frac{312}{105}\)
Theo đầu bài ta có:
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}\right)\cdot2=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=17\)
\(b)\) Ta có: \(x-\frac{37}{45}=\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45\text{ }}\)
\(\Leftrightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)
\(\Leftrightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{45}\)
\(\Leftrightarrow x-\frac{37}{45}=1\)
\(\Leftrightarrow x=1+\frac{37}{45}\)
\(\Leftrightarrow x=\frac{82}{45}\)
Vậy \(x=\frac{82}{45}\)
Đặt \(A=\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{21.2}+\frac{1}{28.2}+\frac{1}{36.2}+...+\frac{2}{x\left(x+1\right).2}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x.\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6.4}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(\Rightarrow\) \(\frac{1}{2}A=\frac{1}{6}-\frac{1}{x+1}\)
\(\Rightarrow A=\left(\frac{1}{6}-\frac{1}{x+1}\right):\frac{1}{2}\)
Theo đề bài ta có :
\(\left(\frac{1}{6}-\frac{1}{x+1}\right):\frac{1}{2}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}.\frac{1}{2}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{2}{18}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{2}{18}\)
\(\Rightarrow\frac{1}{x+1}=\frac{3}{18}-\frac{2}{18}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=18-1\)
\(\Rightarrow x=17\)
Vậy x = 17
Sửa đề :v hình như vậy mới làm được
\(2\frac{2}{9}-x=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}\)
\(\frac{20}{9}-x=\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}\)
\(\frac{20}{9}-x=2\left[\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right]\)
\(\frac{20}{9}-x=2\left[\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right]\)
\(\frac{20}{9}-x=2\left[\frac{1}{3}-\frac{1}{9}\right]\)
\(\frac{20}{9}-x=2\cdot\frac{2}{9}\)
\(\frac{20}{9}-x=\frac{4}{9}\Leftrightarrow x=\frac{16}{9}\)
\(B=3+3^2+3^3+.....+3^{2006}\)
\(\Rightarrow3B=3^2+3^3+....+3^{2007}\)
\(\Rightarrow2B=3^{2007}-3\)
\(\Rightarrow B=\frac{3^{2007}-3}{2}\)
\(2B+3=3^x\)
\(\Rightarrow2.\frac{3^{2007}-3}{2}+3=3^x\)
\(\Rightarrow3^{2007}-3+3=3^x\Rightarrow3^{2007}=3^x\Rightarrow x=2007\)