Ta có :

\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)

\(=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}\)

Giá trị không đổi khi cả tử và mẫu cùng nhân với 2, ta được :

\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}+\frac{2}{16.19}\)

\(=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{19}\right)=\frac{2}{3}.\frac{18}{19}=\frac{12}{19}\)

\(A=\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}=\frac{1}{2}.\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}+\frac{1}{304}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{19}\right)=\frac{9}{19}\)

Đặng Minh Triều làm sai. \(\frac{1}{2}.\frac{1}{14}\ne\frac{1}{28}\)

A=27

BẠN NHÉ

chúc bạn học giỏi

tk mình nha

thank you very much

a = 27

k mình nha

Chúc bạn học giỏi

Mình cảm ơn bạn nhiều

cảm ơn hai bạn rất nhiều !

x=100

Ta sẽ có: 1-1+1+1-1+1-1+1=0

\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)

\(\Rightarrow\frac{x-1}{99}-1-\frac{x+1}{101}+1+\frac{x-2}{98}-1-\frac{x+2}{102}+1+\frac{x-3}{97}-1-\frac{x+3}{103}+1+\frac{x-4}{96}-1-\frac{x+4}{104}+1=0\)

\(\Rightarrow\frac{x-100}{99}-\frac{x-100}{101}+\frac{x-100}{98}-\frac{x-100}{102}+\frac{x-100}{97}-\frac{x-100}{103}+\frac{x-100}{96}-\frac{x-100}{104}=0\)

\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\right)=0\)

Vì \(\frac{1}{99}>\frac{1}{101};\frac{1}{98}>\frac{1}{102};\frac{1}{97}>\frac{1}{103};\frac{1}{96}>\frac{1}{104}\)

\(\Rightarrow\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\ne0\)

\(\Rightarrow x-100=0\)

\(\Rightarrow x=100\)

Vậy \(x=100\)

x thuoc R

\(\dfrac{x-1}{99}-\dfrac{x+1}{101}+\dfrac{x-2}{98}-\dfrac{x+2}{102}+\dfrac{x-3}{97}-\dfrac{x+3}{103}+\dfrac{x-4}{96}-\dfrac{x+4}{104}=0\)
<=> \(\dfrac{x-1}{99}-1-\dfrac{x+1}{101}-1+\dfrac{x-2}{98}-1-\dfrac{x-2}{102}-1+\dfrac{x-3}{97}-1-\dfrac{x+3}{103}-1+\dfrac{x-4}{96}-1-\dfrac{x+4}{104}=0\)

Không biết

cái a bằng 1962

cái b bằng 127/192

à quên mình chưa rút gọn phân số đấy đâu bạn ạ

ban rút gọn phân số đấy hộ mình nha

bạn giải từng bước ra giúp mình nhé

A = \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{7.8}\)

  \(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{7}-\frac{1}{8}\)

  \(=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)

B = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)

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

  \(=1-\frac{1}{13}=\frac{12}{13}\)

\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{7.8}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)

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

\(=1-\frac{1}{13}=\frac{12}{13}\)

A=\(\frac{1}{6}\)+\(\frac{1}{12}\)+......+\(\frac{1}{56}\)

A=\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{7.8}\)

A=\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+....+\(\frac{1}{7}\)-\(\frac{1}{8}\)

A=\(\frac{1}{2}\)-\(\frac{1}{8}\)=\(\frac{3}{8}\)

B=\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+...+\(\frac{2}{11.13}\)

B=1-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+....+\(\frac{1}{11}\)-\(\frac{1}{13}\)

B=1-\(\frac{1}{13}\)=\(\frac{12}{13}\)

C=\(\frac{3}{4}\)+\(\frac{3}{28}\)+....+\(\frac{3}{304}\)

C=\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+..+\(\frac{3}{16.19}\)

Rồi bạn cũng tách ra như câu A và câu B