\(S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{100}\)

\(\Rightarrow2S=2+1+\frac{1}{2}+\frac{1}{2^2}...+\frac{1}{99}\)

\(2S-S=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)

\(\Leftrightarrow2S-S=S=2-\frac{1}{2^{100}}=\frac{2^{101}}{2^{100}}-\frac{1}{2^{100}}=\frac{2^{101}-1}{2^{100}}\)

Ta có:

\(S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2S=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2S-S=2-\frac{1}{2^{100}}\)

\(\Rightarrow S=2-\frac{1}{2^{100}}\)

Với a , b , c là số hữu tỉ t/m a = b + c ta luôn có \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right|\)

Thật vậy : \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2-2\left(\frac{1}{bc}-\frac{1}{ac}-\frac{1}{ab}\right)}\)

                                                       \(=\sqrt{\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2-\frac{2.abc\left(a-b-c\right)}{a^2b^2c^2}}\)(quy đồng lên )

                                                         \(=\sqrt{\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2}\left(\text{do a-b-c=0}\right)\)

                                                          \(=\left|\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right|\)

Áp dụng ta được \(S=\left|\frac{1}{2}-\frac{1}{1}-1\right|+\left|\frac{1}{3}-\frac{1}{2}-1\right|+...+\left|\frac{1}{100}-\frac{1}{99}-1\right|\)

                               \(=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{99}-\frac{1}{100}\)

                                \(=\left(1+1+1+...+1\right)+\left(1+\frac{1}{2}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{100}\right)\)

                                                    (có 99 số 1) 

                                 \(=99+1-\frac{1}{100}\)            

                                 \(=100-\frac{1}{100}=\frac{9999}{100}\)

Ôi, trang wed không tự nhận diện được công thức latex. Mình đăng lại bài giải:

a) Ta có

\(4T=\frac{4}{1+\sqrt{5}}+\frac{4}{\sqrt{5}+\sqrt{9}}+...+\frac{4}{\sqrt{2013}+\sqrt{2017}}\)

\(=\frac{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}{\sqrt{5}+1}+...+\frac{\left(\sqrt{2017}+\sqrt{2013}\right)\left(\sqrt{2017}-\sqrt{2013}\right)}{\sqrt{2017}+\sqrt{2013}}\)

\(=\sqrt{5}-1+\sqrt{9}-\sqrt{5}+\sqrt{13}-\sqrt{9}+...+\sqrt{2017}-\sqrt{2013}\)

\(=\sqrt{2017}-1\)

\(\Rightarrow T=\frac{\sqrt{2017}-1}{4}\)

b) Ta có

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

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

\(=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}\sqrt{1}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)

Tương tự ta có

\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)

......................

\(\frac{1}{100\sqrt{99}+99\sqrt{100}}=\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)

Suy ra

\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)

\(=1-\frac{1}{10}=\frac{9}{10}\)

a)\[\begin{array}{l}
4T = \frac{4}{{1 + \sqrt 5 }} + \frac{4}{{\sqrt 5  + \sqrt 9 }} + ... + \frac{4}{{\sqrt {2013}  + \sqrt {2017} }}\\
 = \frac{{(\sqrt 5  + 1)(\sqrt 5  - 1)}}{{1 + \sqrt 5 }} + ... + \frac{{(\sqrt {2017}  + \sqrt {2013} )(\sqrt {2017}  - \sqrt {2013} )}}{{\sqrt {2013}  + \sqrt {2017} }}\\
 = \sqrt 5  - 1 + \sqrt 9  - \sqrt 5  + ... + \sqrt {2017}  - \sqrt {2013} \\
 = 1 + \sqrt 5  - \sqrt 5  + \sqrt 9  - \sqrt 9  + ... + \sqrt {2013}  - \sqrt {2013}  + \sqrt {2017} \\
 = 1 + \sqrt {2017} \\
 \Rightarrow T = \frac{{1 + \sqrt {2017} }}{4}
\end{array}\]

Cảm ơn Tran Tu, kết bn đê

Câu a) Mik chữa lại một chút 

Ta có: \(\frac{1}{2^2}< \frac{1}{1\cdot2}\); \(\frac{1}{3^2}< \frac{1}{2\cdot3}\);.......; \(\frac{1}{100^2}< \frac{1}{99\cdot100}\)

Suy ra: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)

Suy ra: \(VT< \frac{1}{1}-\frac{1}{100}=\frac{99}{100}< 1\)

Vậy : \(VT+1< 1+1=2\)

1 : 29 x ( 19 -13 ) - 19 x ( 29 - 13 )

= 29 x 6 - 19 x 16

= 174 - 304

=  - 130

2 : 1 - \(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

⁼ 1 - \(\frac{1}{100}\)

= \(\frac{99}{100}\)

\(-4\frac{1}{2}\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le\frac{-2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

<=>  \(-1,5\le x\le\frac{11}{18}\)

đến đây tự làm

mk k biết điều kiện của x  nên giúp đến đó

\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^{3.}}+.............+\frac{1}{2^{100}}\)

\(2B=1+\frac{1}{2}+\frac{1}{2^2}+.................+\frac{1}{2^{99}}\)

\(2B-B=1-\frac{1}{2^{100}}\)

\(B=1-\frac{1}{2^{100}}\)

\( C=\frac{1}{2}-\frac{1}{2^2}+.................+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)

\(2 C=1-\frac{1}{2}+......................+\frac{1}{2^{98}}-\frac{1}{2^{99}}\)

\(2 C+C=1-\frac{1}{2^{100}}\)

\(C=\left(1-\frac{1}{2^{100}}\right):3\)