Đặt bt bằng A

Ta có 2A= 2(2/3 + 2/6 + 2/12 +2/24 + 2/48 + 2/96 + 2/192)

2A= 4/3 +2/3 + 2/6 + 2/12 + 2/24 + 2/48 + 2/96

A= 2A-A= (4/3 +2/3 + 2/6 + 2/12 + 2/24 + 2/48 + 2/96) - (2/3 + 2/6 + 2/12 +2/24 + 2/48 + 2/96  2/192)

A=4/3 +2/3 + 2/6 + 2/12 + 2/24 + 2/48 + 2/96 - 2/3 - 2/6 - 2/12 - 2/24 - 2/48 - 2/96 - 2/192

A=(2/3 - 2/3) + (2/6 - 2/6) + ( 2/12 -  2/12) + (2/24 - 2/24) + (2/48 - 2/48) + ( 2/96 -  2/96) + (4/3 - 2/192)

A=0+0+0+0+0+0+ (256/192 - 2/192)

A=254/192

A=127/96(rút gọn phân số) 

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{192}\)

\(=2\times\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{192}\right)\)

\(=2\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{96}-\frac{1}{192}\right)\)

\(=2\times\left(1-\frac{1}{192}\right)\)

\(=2\times\frac{191}{192}\)

\(=\frac{382}{192}=\frac{191}{96}\)

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)

\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+...+\frac{1}{96}-\frac{1}{192}\right)\)

\(=2.\left(1-\frac{1}{192}\right)\)

\(=2.\frac{191}{192}\)

\(=\frac{382}{192}=\frac{191}{96}\)

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+...+\frac{2}{192}.\)

\(=2\times\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{192}\right)\)

\(=2\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{96}-\frac{1}{192}\right)\)

\(=2\times\left(1-\frac{1}{192}\right)\)

\(=2\times\frac{191}{192}=\frac{191}{68}\)

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+...+\frac{2}{192}\)

\(=\frac{1}{3.1}+\frac{1}{3.2}+\frac{1}{3.2^2}+...+\frac{1}{3.2^6}\)

\(=\frac{1}{3}.\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)

\(=\frac{1}{3}.A\)với \(A=\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)

\(\Rightarrow2A=2.\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)

\(\Rightarrow2A=2+\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2^5}\)

\(\Rightarrow2A-A=\left(2+\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2^5}\right)-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)

\(\Rightarrow A=2-\frac{1}{2^6}=2-\frac{1}{64}=\frac{127}{64}\)

\(\Rightarrow\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+...+\frac{2}{192}=\frac{1}{3}.\frac{127}{64}=\frac{127}{192}\)

Câu 1

Ta có \(\frac{119x83-183}{120x83-266}=\frac{119x83-183}{119x83+83-266}=\frac{119x83-183}{119x83-183}=1\) 

\(E=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}...\frac{9^2}{8.10}=\frac{\left(2.3.4...9\right)^2}{1.2.\left(3.4...8\right)^2.9.10}=\frac{2^2.9^2}{1.2.9.10}=\frac{18}{10}=\frac{9}{5}\)

\(H=\frac{8}{1^2\cdot3^2}+\frac{16}{3^2\cdot5^2}+...+\frac{48}{11^2\cdot13^2}\)

\(H=\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{5^2}+...+\frac{1}{11^2}-\frac{1}{13^2}\)

\(H=1-\frac{1}{13^2}\)

\(H=\frac{168}{169}\)

Phương thiếu bước nhé 

\(H=\frac{8}{1^2.3^2}+\frac{16}{3^2.5^2}+\frac{24}{5^2.7^2}+...+\frac{48}{11^2.13^2}\)

\(H=\frac{3^2-1^2}{1^2.3^2}+\frac{5^2-3^2}{3^2.5^2}+\frac{7^2-5^2}{5^2.7^2}+...+\frac{13^2-11^2}{11^2.13^2}\)

\(H=\frac{3^2}{1^2.3^2}-\frac{1^2}{1^2.3^2}+\frac{5^2}{3^2.5^2}-\frac{3^2}{3^2.5^2}+\frac{7^2}{5^2.7^2}-\frac{5^2}{5^2.7^2}+...+\frac{13^2}{11^2.13^2}-\frac{11^2}{11^2.13^2}\)

\(H=\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{5^2}+\frac{1}{5^2}-\frac{1}{7^2}+...+\frac{1}{11^2}-\frac{1}{13^2}\)

\(H=1-\frac{1}{13^2}=1-\frac{1}{169}=\frac{168}{169}\)

Chúc bạn học tốt ~ 

\(\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+...+\frac{2}{40400}\)

\(=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{200.202}\)

\(=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{202-200}{200.202}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{200}-\frac{1}{202}\)

\(=\frac{1}{2}-\frac{1}{202}\)

\(=\frac{50}{101}\)

A = 2/2*4+2/4*6+2/6*8+...+2/200*202

=1/2-1/4+1/4-1/6+1/6-1/8+...+1/200-1/202

=1/2-1/202

=50/101

Đặt A=2/8+2/24+2/48+..+2/40400
Ta có:A=2/8+2/24+2/48+2/40400
=>A=2/2.4+2/4.6+2/6.8+..+2/200.202
=>A=1/2-1/4+1/4-1/6+1/6-1/8+..+1/200-1/202
=>A=1/2-1/202
=>A=100/202=50/101
Vậy A=50/101

   \(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}\)+\(\frac{2}{96}\)

=\(2\)x (\(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}\)\(+\frac{1}{96}\))

=\(2\)x (\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+...\)\(+\frac{1}{48}-\frac{1}{96}\))

=\(2\)x (\(1-\frac{1}{96}\))

=\(2\)x \(\frac{95}{96}\)

=\(\frac{190}{96}=\frac{95}{48}\)

cái a bằng 1962

cái b bằng 127/192

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