=> x- (\(\frac{20}{11.13}\) + \(\frac{20}{13.15}\) +...+ \(\frac{20}{53.55}\)) = \(\frac{3}{11}\)

=> x - 10.(\(\frac{2}{11.13}\) + \(\frac{2}{13.15}\) +...+ \(\frac{2}{53.55}\)) = \(\frac{3}{11}\)

=> x - 10.( \(\frac{1}{11}\) - \(\frac{1}{13}\) + \(\frac{1}{13}\) -  \(\frac{1}{15}\) +...+ \(\frac{1}{53}\) - \(\frac{1}{55}\)) = \(\frac{3}{11}\)

=> x - 10. (\(\frac{1}{11}\) - \(\frac{1}{55}\)) = \(\frac{3}{11}\)

=> x - 10.\(\frac{4}{55}\) = \(\frac{3}{11}\) 

=> x - \(\frac{8}{11}\) = \(\frac{3}{11}\)=> x=1 Vậy x=1

giải được xin cảm ơn

thank you

Bạn xem kỹ lại đề có đúng không?

1.

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{99}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x-1}=\frac{98}{99}\)

\(1-\frac{1}{x-1}=\frac{98}{99}\)

\(\frac{1}{x-1}=1-\frac{98}{99}\)

\(\frac{1}{x-1}=\frac{1}{99}\)

\(\Rightarrow x-1=99\)

\(\Rightarrow x=99+1=100\)

b) \(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)

\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}\right)+10.\left(\frac{1}{13}-\frac{1}{15}\right)+10.\left(\frac{1}{15}-\frac{1}{17}\right)+...+10.\left(\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-10.\frac{4}{55}=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\)

\(\Rightarrow x=\frac{3}{11}+\frac{8}{11}=1\)

c) 5^x + 2 . 5^x + 2³ = 83

5^x . ( 1 + 2 ) + 8 = 83

5^x . 3 = 83 - 8

5^x . 3 = 75

5^x = 75 : 3

5^x = 25

\(\Rightarrow\)5^x = 5²

\(\Rightarrow\)x = 2

2.

Ta thấy \(2016^{2016}>2016^{2016}-3\)

\(\Rightarrow B=\frac{2016^{2016}}{2016^{2016}-3}>\frac{2016^{2016}+2}{2016^{2016}-3+2}=\frac{2016^{2016}+2}{2016^{2016}-1}=A\)

\(\Rightarrow A< B\)

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{99}\)

Ta có \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{99}\)

      = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{98}{99}\)(áp dụng công thức)

      = \(1-\frac{1}{x+1}=\frac{98}{99}\)

      = \(\frac{1}{x+1}=1-\frac{98}{99}\)(quy tắc tìm số trừ)

      = \(\frac{1}{x+1}=\frac{1}{99}\Rightarrow\frac{1}{x+1}=\frac{1}{98+1}\Rightarrow x=98\)

Vậy x = 98 :)

Còn nữa, công thức mà mình áp dụng là: \(\frac{a}{b.c}=\frac{1}{b}-\frac{1}{c}\)nếu \(a=c-b\)

cảm ơn nhé các bạn

\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)

\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(x-\left[10.\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)\right]=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-\left[\frac{10}{11}-\frac{10}{55}\right]=\frac{3}{11}\)

\(x-\left[\frac{10}{11}-\frac{2}{11}\right]=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\)

\(x=\frac{3}{11}+\frac{8}{11}=1\)

1 , ủng hộ mk nha

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