Nguyen svtkvtm Khôi Bùi Nguyễn Việt Lâm Lê Anh Duy Nguyễn Thành Trương DƯƠNG PHAN KHÁNH DƯƠNG An Võ (leo) Ribi Nkok Ngok Bonking ...

sửa đề câu 1 :

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

\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)

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

\(=1-\frac{1}{100!}< 1\)

sửa đề câu 2

\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)

\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)

\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)

\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)

khi cộng cac số có tử bé hơn mẫu thì tổng sẽ <1 nha 

Ta có:

\(S=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)\(=\frac{1}{4}-\frac{1}{100}=\frac{24}{100}< \frac{50}{100}=\frac{1}{2}\)

Ta có : \(\frac{1}{5^2}=\frac{1}{5.5}< \frac{1}{4.5}\)

\(\frac{1}{6^2}=\frac{1}{6.6}< \frac{1}{5.6}\)

\(\frac{1}{7^2}=\frac{1}{7.7}< \frac{1}{6.7}\)

...

\(\frac{1}{100^2}=\frac{1}{100.100}< \frac{1}{99.100}\)

\(\Rightarrow S< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)

\(S< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)

\(S< \frac{1}{4}-\frac{1}{100}=\frac{6}{25}=\frac{24}{100}\)

Mà \(\frac{24}{100}< \frac{50}{100}=\frac{1}{2}\)

\(\Rightarrow S< \frac{1}{2}\)

Vậy S<\(\frac{1}{2}\).

Ta sẽ chứng minh :

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) với x, y > 0

Thật vậy : \(x+y+z\ge3\sqrt[3]{xyz}\)( bđt Cô - si )

Và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{abc}}\) ( bđt Cô - si )

\(\Rightarrow x+y+z\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) ( Dấu " = " \(\Leftrightarrow x=y=z\) )

Ta có :

\(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)

\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)

( Dấu " = " xay ra khi a=b)

Tương tự ta cũng có :

\(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c}\le\frac{1}{9}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\) ( Dấu " = " xảy ra khi b=c)

\(\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\le\frac{1}{9}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\) ( Dấu " = " xay ra khi c = a )

\(VT=\sum_{cyc}\frac{1}{\sqrt{5a^2+2ab+b^2}}\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)

\(\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)

Dấu " = " xay ra khi \(a=b=c=\frac{2}{3}\)

Chúc bạn học tốt !!

\(\frac{1}{\sqrt{4a^2+2ab+b^2+a^2+b^2}}\le\frac{1}{\sqrt{4a^2+2ab+b^2+2ab}}=\frac{1}{\sqrt{\left(2a+b\right)^2}}=\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)

\(\Rightarrow VT\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}+\frac{2}{b}+\frac{1}{c}+\frac{2}{c}+\frac{1}{a}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{2}{3}\)

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

\(\Rightarrow100=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}+1+\frac{1}{2}+...+\frac{1}{100}\)

\(\Rightarrow100=1+\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{2}{3}\right)+...+\left(\frac{99}{100}+\frac{1}{100}\right)\)

\(\Rightarrow100=1+1+1+...+1\) (100 chữ số 1)

\(\Rightarrow100=100\)

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

Ta có :\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

=\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}=\)\(\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)\)\(+...+\left(1-\frac{1}{100}\right)\)

=\(\left(1+1+1+....+1\right)\)\(-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=             \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=  \(100-1-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=\(100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)= vế trên (đpcm)

\(S=100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1+1+...+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(S=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(\RightarrowĐPCM\)

cunasai

đầu bài sai bạn nhá, lớn hơn 1/100. Ta đi cm tổng những phân số có dấu âm > 1-1/100

Có: \(2^2>1.2 \Rightarrow\frac{1}{2^2}< \frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)

mấy cái kia tương tự suy ra tổng các p/s trong ngoặc < 1-1/100
=> vế trái>1-(1-1/100)=1/100

cam on ban