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A=2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101
A= 2 - 1/3 + 1/3 - 1/5 + 1/5 - ... + 2/99 - 2/101
A = 2 - 2/101 = 200/101
B = 3-1/3+1/3-1/5+1/5-...+3/49-3/51
B = 3-3/51(tự tính nhé)
C = 5(5/1.6+5/6.11+5/11.16+....+5/26-5/31
C = 5(5-1/31)(tự tính)
D rút gon cho 2 rồi 3D , sau đó 5(3/.... tương tự các cách làm trên)
2E nhân lên rồi giải giống trên
3F Rồi nhân 4/77 và rút gọn thì tính được
a, A= \(\frac{1}{1}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+......+\(\frac{1}{99}\)-\(\frac{1}{100}\)
A=\(\frac{1}{1}\)-\(\frac{1}{100}\)+(-\(\frac{1}{3}\)+\(\frac{1}{3}\)-.....-\(\frac{1}{99}\)+\(\frac{1}{99}\))
A=\(\frac{1}{1}\)-\(\frac{1}{100}\)+0
A=1-\(\frac{1}{100}\)=\(\frac{100}{100}\)-\(\frac{1}{100}\)=\(\frac{99}{100}\)
=> 1/11 - 1/13 + 1/13 - 1/15 + ..... + 1/19 - 1/21 - x + 4 + 221/231 = 7/3
=> 1/11 - 1/21 - x + 4 + 221/231 = 7/3
=> 2099/420 - x = 7/3
=> x = 2099/420 - 7/3 = 373/140
Tk mk nha
Bài làm
\(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{19.21}-x+4+\frac{221}{231}=\frac{7}{3}\)
\(\Leftrightarrow2\left(\frac{1}{11.13}+\frac{1}{13.15}+...+\frac{1}{19.21}\right)-x+4+\frac{221}{231}=\frac{7}{3}\)
\(\Leftrightarrow2\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)-x+4+\frac{221}{231}=\frac{7}{3}\)
\(\Leftrightarrow2\left(\frac{1}{11}-\frac{1}{21}\right)-x+4+\frac{221}{231}=\frac{7}{3}\)
\(\Leftrightarrow2.\frac{10}{231}-x+4+\frac{221}{231}=\frac{7}{3}\)
\(\Leftrightarrow\frac{20}{231}-x+4+\frac{221}{231}=\frac{7}{3}\)
\(\Leftrightarrow\frac{20}{231}-x+\frac{924}{231}+\frac{221}{231}=\frac{539}{231}\)
\(\Leftrightarrow\frac{20}{231}-x+\frac{924}{231}=\frac{539}{231}-\frac{221}{231}\)
\(\Leftrightarrow\frac{20}{231}-x+\frac{924}{231}=\frac{318}{231}\)
\(\Leftrightarrow\frac{20}{231}-x=\frac{318}{231}-\frac{924}{231}\)
\(\Leftrightarrow\frac{20}{231}-x=-\frac{606}{231}\)
\(\Leftrightarrow x=\frac{20}{231}-\frac{606}{231}\)
\(\Leftrightarrow x=-\frac{586}{231}\)
Vậy \(\Leftrightarrow=-\frac{586}{231}\)
\(B=\frac{5}{18\cdot21}+\frac{5}{21\cdot24}+\frac{5}{24\cdot27}+...+\frac{5}{123\cdot126}\\ B=\frac{5}{3}\cdot\left(\frac{3}{18\cdot21}+\frac{3}{21\cdot24}+\frac{3}{24\cdot27}+...+\frac{3}{123\cdot126}\right)\\ B=\frac{5}{3}\cdot\left(\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+\frac{1}{24}-\frac{1}{27}+...+\frac{1}{123}-\frac{1}{126}\right)\\ B=\frac{5}{3}\cdot\left(\frac{1}{18}-\frac{1}{126}\right)\\ B=\frac{5}{3}\cdot\left(\frac{7}{126}-\frac{1}{126}\right)\\ B=\frac{5}{3}\cdot\frac{1}{21}\\ B=\frac{5}{63}\)
Sửa đề : \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Leftrightarrow\)\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}.\frac{1}{2}\)
\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow\)\(x+1=18\)
\(\Leftrightarrow\)\(x=18-1\)
\(\Leftrightarrow\)\(x=17\)
Vậy \(x=17\)
Chúc bạn học tốt ~
\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10.\frac{4}{55}=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+\frac{40}{55}=\frac{3}{11}\)
\(\Leftrightarrow\)\(x=\frac{3}{11}-\frac{40}{55}\)
\(\Leftrightarrow\)\(x=\frac{-5}{11}\)
Vậy \(x=\frac{-5}{11}\)
Chúc bạn học tốt ~
a) \(\frac{4.7}{9.32}\)=\(\frac{28}{288}\)=\(\frac{7}{72}\)
b)\(\frac{3.21}{14.15}\)=\(\frac{63}{210}\)=\(\frac{3}{10}\)
c)\(\frac{2.5.13}{26.35}\)=\(\frac{130}{910}\)=\(\frac{1}{7}\)
d)\(\frac{9.6-9.3}{18}\)=\(\frac{27}{18}\)=\(\frac{3}{2}\)
e)\(\frac{17.5-17}{3-20}\)=\(\frac{68}{-17}\)=\(-4\)
f)\(\frac{49+7.49}{49}\)=\(\frac{392}{49}\)=\(8\)
\(A=\frac{15.3^{11}+4.27^1}{9^7}\)
\(\Rightarrow A=\frac{3.5.3^{11}+4.3^{3^1}}{\left(3^2\right)^7}\)
\(\Rightarrow A=\frac{3^{12}.5+4.3^3}{3^{14}}\)
\(\Rightarrow A=\frac{3^3.\left(5.3^8+4.3^3\right)}{3^{14}}\)
\(\Rightarrow A=\frac{32805+4}{177147}\)
\(\Rightarrow A=\frac{32809}{177147}\)
=\(\frac{1}{2}\cdot\left(\frac{6}{9\cdot15}+\frac{6}{15\cdot21}+\frac{6}{21\cdot27}+...+\frac{6}{99\cdot105}\right)\)
=\(\frac{1}{2}\cdot\left(\frac{1}{9}-\frac{1}{15}+\frac{1}{15}-\frac{1}{21}+\frac{1}{21}-\frac{1}{27}+...+\frac{1}{99}-\frac{1}{105}\right)\)
=\(\frac{1}{2}\cdot\left(\frac{1}{9}-\frac{1}{105}\right)\)
=\(\frac{1}{2}\cdot\left(\frac{35}{315}-\frac{3}{315}\right)\)
=\(\frac{1}{2}\cdot\left(\frac{32}{315}\right)\)
=\(\frac{1\cdot16}{1\cdot315}\)
=\(\frac{16}{315}\)