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\(A=2-\left(\frac{2^3}{25}+\frac{2^3}{63}+...+\frac{2^3}{255}+\frac{2^3}{323}\right)\)
\(=2-4.\left(\frac{2}{35}+\frac{2}{63}+...+\frac{2}{255}+\frac{2}{323}\right)\)
\(=2-4.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{15.17}+\frac{2}{17.19}\right)\)
\(=2-4.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}\right)\)
\(=2-4.\left(\frac{1}{5}-\frac{1}{19}\right)\)
\(=2-4.\frac{14}{95}=2-\frac{56}{95}=\frac{134}{95}\)
Câu này tính hợp lý nha bạn !
-5/7 . 2/11 + -5/7 . 9/11 + 1 5/7 = -5/7 . (2/11+9/11) + 1 5/7
= -5/7 . 1 + 1 5/7
= -5/7 + 1 5/7
= -5/7 + (1 + 5/7)
= -5/7 + 1 + 5/7
= (-5/7 + 5/7) + 1
= 0 + 1 = 1
Lưu ý : dấu / nghĩa là phân số nhé bạn, còn 1 5/7 nghĩ là hỗn số nhé !
Có gì không hiểu bạn cứ hỏi mình ! Chúc bạn học tốt ! 
-5/7*2/11+-5/7*9/11+1/5/7
=-5/7*(2/11+9/11)+12/7
=-5/7*1+12/7
=-5/7+12/7
=1
Câu 1 :\(P=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{98}{100}=\frac{1}{100}\)
\(\frac{24\cdot47-23}{24+47\cdot23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
\(=\frac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\frac{3\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}\)
\(=\frac{24^2+24\cdot23-23}{24+24\cdot23+23^2}\cdot\frac{3}{9}\) \(=\frac{24^2+23\cdot\left(24-1\right)}{\left(23+1\right)\cdot24\cdot23^2}\cdot\frac{1}{3}=1\cdot\frac{1}{3}=\frac{1}{3}\)
a) ta có:
\(\frac{-1}{2}-1\le x\le\frac{1}{2}.3\)
hay \(-1,5\le x\le1,5\)
vì x\(\in Z\) nên ta chọn x=-1,0,1
ta có:
3S=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
3S-S=\(\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^9}\right)\)
2S=1-\(\frac{1}{3^9}\)
s=\(\left(1-\frac{1}{3^9}\right):2\)
S = \(3+3.\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
Đặt A = \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=> 2A = \(1+\frac{1}{2}+...+\frac{1}{2^8}\)
=> 2A - A = A = \(\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)=1-\frac{1}{2^9}\)
=> S = 3 + 3 . A = \(3+3.\left(1-\frac{1}{2^9}\right)=3+3-\frac{3}{2^9}=6-\frac{3}{2^9}\)