\(a.\)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x-1\right).\left(2x+1\right)}\right)=\frac{49}{99}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}-\frac{1}{2x+1}\right)=\frac{49}{99}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{2x+1}\right)=\frac{49}{99}\)

\(\Rightarrow\frac{x}{2x+1}=\frac{49}{99}\)

\(\Rightarrow99x=49.\left(2x+1\right)\)

\(\Rightarrow99x=98x+49\)

\(\Rightarrow x=49\)

Vậy : \(x=49\)

\(b.\)

\(1-3+3^2-3^3+...+\left(-3^x\right)=\frac{1-9^{1006}}{4}\)

Đặt \(A=1-3+3^2-3^3+...+\left(-3^x\right)\)

\(\Rightarrow3A=3-3^2+3^3-3^4+...+\left(-3^{x+1}\right)\)

\(\Rightarrow3A+A=1+\left(-3^{x+1}\right)\)

\(\Rightarrow4A=1+\left(-3^{x+1}\right)\)

\(\Rightarrow A=\frac{1+\left(-3^{x+1}\right)}{4}\)

\(\Rightarrow\frac{1+\left(-3^{x+1}\right)}{4}=\frac{1-9^{1006}}{4}\)

\(\Rightarrow-3^{x+1}=-9^{1006}\)

\(\Rightarrow-3^{x+1}=-3^{2012}\)

\(\Rightarrow x+1=2012\)

\(\Rightarrow x=2012-1\)

\(\Rightarrow x=2011\)

Vậy : \(x=2011\)

a) A = \(\frac{5}{1.4}+\frac{29}{4.7}+\frac{71}{7.10}+....+\frac{10301}{100.103}\) (có 34 số hạng)

A = \(\frac{4+1}{1.4}+\frac{4.7+1}{4.7}+\frac{7.10+1}{7.10}+....+\frac{100.103+1}{103.100}\)

A = \(1+\frac{1}{1.4}+1+\frac{1}{4.7}+1+\frac{1}{7.10}+....+1+\frac{1}{100.103}\)

A = \(1.34+\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)

A = \(34+\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

A = \(34+\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

A = \(34+\frac{1}{3}\cdot\frac{102}{103}\)

A = \(34+\frac{34}{103}=\frac{3536}{103}\)

bạn làm hộ mik câu B với

a)Ta có:

\(A=4\frac{25}{16}+25\left(\frac{9}{16}:\frac{125}{64}\right):\frac{-27}{8}\)

\(\Rightarrow A=\frac{89}{16}+25.\frac{36}{125}:\frac{-27}{8}\)

\(\Rightarrow A=\frac{89}{16}+\frac{36}{5}:\frac{-27}{8}\)

\(\Rightarrow A=\frac{89}{16}+\frac{-32}{15}\)

\(\Rightarrow A=\frac{823}{240}\)

Vậy A=.....

b)Ta có:

\(C=\frac{2^3}{3.5}+\frac{2^3}{5.7}+\frac{2^3}{7.9}+...+\frac{2^3}{101.103}\)

\(\Rightarrow C=\frac{2^2.2}{3.5}+\frac{2^2.2}{5.7}+\frac{2^2.2}{7.9}+...+\frac{2^2.2}{101.103}\)

\(\Rightarrow C=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)

\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{101}-\frac{1}{103}\right)\)

\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{103}\right)\)

\(\Rightarrow C=4.\frac{100}{309}\)

\(\Rightarrow C=\frac{400}{309}\)

Vậy C=.....

B, C=2^3/3.5 + 2^3/5.7+......+2^3/101.103

C= 2^3(1/3-1/5+1/5-1/7+....+1/101-1/103)

C=8(1/3-1/103)

C=8.100/309

C=800/309

VẬY C= 800/309

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test

\(S=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+..+\frac{5}{97.99}\)

\(=\frac{5}{2}.\left(5+\frac{5}{3}+\frac{5}{5}+\frac{5}{7}+...+\frac{5}{97}+\frac{5}{99}\right)\)

\(=\frac{5}{2}.\left(5+\frac{5}{99}\right)\)

\(=\frac{5}{2}.\frac{500}{99}\)

\(=\frac{1250}{99}\)(có gì sai sót xin bỏ qua cho T^T)

5a.

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{19.21}\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}.\dfrac{20}{21}=\dfrac{10}{21}\)

b.

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}.1=\dfrac{1}{2}\)

theo công thức, ta tính đc: 

A = 1- 1/3 + 1/3 - 1/5 + 1/5 -1/7 +..... + 1/49 - 1/51

=> A bằng 1- 1/51 ( các cặp phân số đối nhau thì lược bỏ như - 1/3 và + 1/3 )

1-1/3=2/3 chứ ko phải 1-1/3=1/3 đâu nha bạn

Đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\)

\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{99}-\frac{1}{101}\)

\(2A=\frac{100}{101}\)

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

b) \(\frac{2^{10}+3^{31}+2^{40}+3^6}{2^{11}\cdot3^{31}+2^{41}\cdot3^6}=\frac{2^{10}+2^{40}}{2^{11}+2^{41}}\)

\(\frac{2^{10}+2^{40}}{2^{11}+2^{41}}=\frac{1}{2}\)

=1/2x(1/1.3+1/3.5+...+1/99.101)

=1/2.(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)

=1/2.(1-1/101)

=1/2.100/101

=50/101

chúc bạn học tốt

D = \(\frac{1}{54}-\frac{3}{1.3}-\frac{3}{3.5}-\frac{3}{5.7}-...-\frac{1}{79.81}\)

\(=\frac{1}{54}-\left(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{79.81}\right)\)

\(=\frac{1}{54}-\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{79.81}\right)\)

\(=\frac{1}{54}-\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{79}-\frac{1}{81}\right)\)

\(=\frac{1}{54}-\frac{3}{2}.\left(1-\frac{1}{81}\right)\)

\(=\frac{1}{54}-\frac{3}{2}.\frac{80}{81}\)

\(=\frac{1}{54}-\frac{40}{27}\)

\(=\frac{1}{54}-\frac{80}{54}\)

\(=\frac{79}{54}\)