2\3x-780\11:[13\2.(1\3.5+1\5.7+1\7.9+1\9.11]=-5

2\3x-780\11:[13\2.(1\3-1\5+1\5-1\7+....+1\9-1\11)]=-5

2\3x-780\11:[13\2.(1\3-1\11)]=-5

2\3x-780\11:[13\2.8\33]=-5

2\3x-780\11:52\33=-5

2\3x-525\13=-5

2\3x=-5+525\13

2\3x=460\13

x=460\13:2\3

x=690\13

1/2(2/3.5+2/5.7+2/7.9+...+2/(2x+1)(2x+3))=15/93

1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/2x+1-1/2x+3)=15/93

1/2(1/3-1/2x+3)=15/93

=>1/3-1/2x+3=10/31

=>1/2x+3=1/93

=>2x+3=93

2x=93-3=90

=>x=45

Đặt \(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

\(\Rightarrow2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{93}\)

\(\Rightarrow2x+3=93\)

\(2x=90\)

\(x=45\)

Vậy \(x=45\).

\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

=> \(\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{5}{31}\)

=> \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{5}{31}\)

=> \(\frac{1}{3}-\frac{1}{2x+3}=\frac{5}{31}:\frac{1}{2}=\frac{5}{31}\cdot2=\frac{10}{31}\)

=> \(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}=\frac{1}{93}\)

=> 2x +3  = 93

=> 2x = 90 => x = 45

2/3.5+2/5.7+2/7.9+...+2/(2x+1)(2x+3)=2.15/93

1/3-1/5+1/5-1/7+...+1/2x+1-1/2x+3=10/31

1/3-1/2x+3=10/31

1/(2x+3)=1/93

2x+3=93

2x=90

x=45

A =(1/2 +1)×(1/3 +1)×(1/4 +1)×....×(1/99 +1)

=3/2x4/3x...............x100/99

=2-1/99

=197/99

A= \(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot.....\cdot\frac{100}{99}\)

A=\(\frac{\left(3\cdot4\cdot5\cdot....\cdot99\right)\cdot100}{2\cdot\left(3\cdot4\cdot5\cdot...\cdot99\right)}\)

A=\(\frac{100}{2}=50\)

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

=> \(\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)>\(\frac{32}{100}\)=32%

Câu đầu tiên:

\(A=\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot...\cdot\left(\frac{1}{99}+1\right)\)

\(A=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{100}{99}=\frac{3\cdot4\cdot5\cdot...\cdot99\cdot100}{3\cdot4\cdot5\cdot...\cdot99\cdot2}=\frac{100}{2}=50\)

Câu thứ 2:

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}>\frac{32}{100}\)

\(a,\frac{x-1}{21}=\frac{3}{x+1}\)

\(\Leftrightarrow\left[x-1\right]\left[x+1\right]=63\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

\(\Leftrightarrow x^2=8^2\)

\(\Leftrightarrow x=\pm8\)

\(b,\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left[\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\right]=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right]=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{45}\right]=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}=\frac{21}{45}\)

\(\Leftrightarrow\frac{7}{x}=\frac{7}{15}\)

\(\Leftrightarrow x=15\)

Vậy x = 15

Bài cuối tương tự

a) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.........+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)

\(\Leftrightarrow2.\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{x\left(x+1\right)}\right]=\frac{1998}{2000}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{x\left(x+1\right)}=\frac{999}{2000}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{999}{2000}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{999}{2000}\)\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2000}\)

\(\Leftrightarrow x+1=2000\)\(\Leftrightarrow x=1999\)

Vậy \(x=1999\)

b) \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

\(\Leftrightarrow\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{15.2}{93}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+......+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\Leftrightarrow\frac{1}{2x+3}=\frac{1}{93}\)\(\Leftrightarrow2x+3=93\)

\(\Leftrightarrow2x=90\)\(\Leftrightarrow x=45\)

Vậy \(x=45\)

a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)

(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)

(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)

x=\(\frac{1}{3}:\frac{13}{12}\)

x=\(\frac{4}{13}\)

b)( \(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{9}-\frac{2}{11}_{ }\)):x =\(\frac{2}{3}\)

Giống câu a

\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{31}{32}.\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{31}{32}\)

\(\Leftrightarrow\frac{1}{32}=\frac{1}{x+2}\)

\(\Leftrightarrow x+2=32\Rightarrow x=30\)

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