1/2+1/3+1/4+...+1/63>1/31+1/31+...+1/31(62 số hạng 1/31)

hay 1/2+1/3+1/4+...+1/63>62 x 1/31

nên 1/2+1/3+1/4+...+1/63>2(dpcm)

Này là bài lớp 4 rồi 

A = \(\frac{8}{\sqrt{5}-1}\)  - (\(2\sqrt{5}-1\) ) ( chúng ta cần trục căn thức lên để khử mẫu )                                    

= \(\frac{8\left(\sqrt{5}+1\right)}{5-1}\)- \(\left(2\sqrt{5}-1\right)\)

= \(2\sqrt{5}\)+ 2 - \(2\sqrt{5}\)+1

= 3

B = \(\frac{\left(1-\sqrt{x}\right)^2+4\sqrt{x}}{1+\sqrt{x}}\)( x \(\ge\)0 )

= \(\frac{1-2\sqrt{x}+x+4\sqrt{x}}{1+\sqrt{x}}\)

= \(\frac{1+2\sqrt{x}+x}{1+\sqrt{x}}\)

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

= 1 +\(\sqrt{x}\)

#mã mã#

\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{20}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{19}{20}\)

\(=\frac{1.2.3.....19}{2.3.4.....20}\)

\(=\frac{1}{20}\)

\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{20}\right)\)

\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{18}{19}.\frac{19}{20}\)

\(B=\frac{1}{20}\)

Hok tốt

\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{18}{19}.\frac{19}{20}\)

\(B=\frac{1}{20}\)

D=115/132/103/132=115/103

\(D=\frac{\frac{88}{132}-\frac{33}{132}+\frac{60}{132}}{\frac{55}{132}+\frac{132}{132}-\frac{84}{132}}\)

\(D=\frac{\frac{115}{132}}{\frac{103}{132}}\)

\(D=\frac{115}{103}\)

D = \(\frac{115}{103}\)

A= 1+ 1/2 + 1/2² + ... + 1/2²⁰¹² 

﴾1/2﴿A= 1/2+1/2²+...+1/2²⁰¹³

A‐﴾1/2﴿A= ﴾1+ 1/2 + 1/2² + ... + 1/2²⁰¹² ﴿ ‐ ﴾ 1/2+1/2²+...+1/2²⁰¹³ ﴿

﴾1/2﴿A = 1 ‐ 1/2²⁰¹³ 

A= ﴾1‐ 1/2²⁰¹³ ﴿ : 1/2

A= 2 ‐ 1/2²⁰¹²

\(A=2-\frac{1}{2^{2012}}\)

A= 1+ 1/2 + 1/2² + ... + 1/2²⁰¹² 

﴾1/2﴿A= 1/2+1/2²+...+1/2²⁰¹³

A‐﴾1/2﴿A= ﴾1+ 1/2 + 1/2² + ... + 1/2²⁰¹² ﴿ ‐ ﴾ 1/2+1/2²+...+1/2²⁰¹³ ﴿

﴾1/2﴿A = 1 ‐ 1/2²⁰¹³ 

A= ﴾1‐ 1/2²⁰¹³ ﴿ : 1/2

A= 2 ‐ 1/2²⁰¹²