1) Có nhận xét sau:

\(\frac{1}{a\sqrt{a+1}+\left(a+1\right)\sqrt{a}}=\frac{1}{\sqrt{a^2+a}\left(\sqrt{a}+\sqrt{a+1}\right)}=\frac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a^2+a}}\)

\(=\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a+1}}.\)Do đó biểu thức có giá trị bằng: \(\frac{1}{1}-\frac{1}{\sqrt{2}}+..-\frac{1}{\sqrt{1999}}=1-\frac{1}{\sqrt{1999}}.\)

2) Có nhận xét sau:

\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a+1}-\sqrt{a}}{\left(\sqrt{a}+\sqrt{a+1}\right)\left(\sqrt{a+1}-\sqrt{a}\right)}=\sqrt{a+1}-\sqrt{a}.\) Thay vào biểu thức ta được biểu thức

có giá trị bằng: \(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{1999}-\sqrt{1998}=\sqrt{1999}-1.\)

Bạn áp dụng \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)với n = 1, 2 , 3 , ... , 1999

Xét mẫu số của P ta có:

MS = 1999/ 1 + 1998/ 2+...+ 1/1999

Xét dãy số: 1; 2; ...; 1999 dãy số trên có 1999 số hạng

Tách 1999 thành tổng của 1999 số 1 khi đó ta có:

MS = (1 + 1998/2) + (1 + 1997/3)+..+(1+1/1999)+2000/2000

MS = 2000/2 +2000/3+...+2000/2000

MS = 2000.(1/2 + 1/3 + ...+ 1/2000)

Vậy P = 1/2000

Gọi 1/4 số a là 0,25 . Ta có :

                   a . 3 - a . 0,25 = 147,07

                   a . (3 - 0,25) = 147,07 ( 1 số nhân 1 hiệu )

                      a . 2,75 = 147,07

                         a = 147,07 : 2,75

                          a = 53,48

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)

Ta có : \(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)

          \(2A=2+\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+...+\frac{2}{2^{2017}}\)

          \(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)

    \(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)\)

          \(A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{2016}}-\frac{1}{2^{2017}}\)

         \(A=2-\frac{1}{2^{2017}}=\frac{2^{2018}-1}{2^{2017}}\)

Vậy   \(A=\frac{2^{2018}-1}{2^{2017}}\)

A=đã cho.

2A=1+1/2+1/2^2+1/2^3+...+1/2^2016.

2A-A=1-1/2^2017(khử).

A=1-1/2^2017.

           \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)

\(\Leftrightarrow\)\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\)

\(\Leftrightarrow\)\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)

\(\Leftrightarrow\)\(A=1-\frac{1}{2^{2016}}\)

\(A=\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{2016}}\)

\(2A=1+\frac{1}{2}+.........+\frac{1}{2^{2015}}\)

\(2A-A=\left(1+\frac{1}{2}+.....+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2016}}\right)\)

\(A=1-\frac{1}{2^{2016}}\)

A=1/2+1/2^2+1/2^3+...+1/2^2016     (1)

2A=1+1/2+1/2^2+...+1/2^2015         (2)

Lấy (2)-(1) được:

A=1+1/2+1/2^2+...+1/2^2015-(1/2+1/2^2+1/2^3+...+1/2^2016)

A=1+1/2+1/2^2+...+1/2^2015-1/2-1/2^2-1/2^3-...-1/2^2016

A=1-1/2^2016

Vậy A=1-1/2^2016

TẦM NHƯ HƠI CĂNG

\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+....+\frac{1}{1999}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{1+\left(\frac{1998}{2}+1\right)+\left(\frac{1997}{3}+1\right)+....+\left(\frac{1}{1999}+1\right)}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{2000}{2}+\frac{2000}{3}+\frac{2000}{4}+....+\frac{2000}{2000}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{2000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}\)

\(=\frac{1}{2000}\)