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

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

\(=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4!}+...+\frac{1}{98!}-\frac{1}{100!}\)

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

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

Ta có:

\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(......................\)

\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

Cộng theo vế ta có: \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{99}{10}\)

\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{99}{10}+\frac{1}{10}=\frac{100}{10}=10\)

A = 1/2.2 + 1/3.3 +......+ 1/50.50

A < 1/1.2 + 1/2.3 +......+ 1/49.50

A < 1 - 1/2 + 1/2 - 1/3 +.....+ 1/49 - 1/50

A < 1 - 1/50

A < 49/50 < 3/4 

=> A < 3/4 (đpcm)

Hình như bạn Killua giải sai thì phải.. 49/50 > 3/4 chứ

Theo mình thì bài này nên giữ nguyên phân số 1/2^2( vì nó bằng 1/4)

Xét : B = 1/3^2 + 1/4^2 +...+ 1/50^2

       => B < 1/2.3 + 1/3.4 +...+ 1/49.50

       => B<  1/2-1/3+1/3-1/4+...+1/49-1/50

       => B < 1/2-1/50 < 1/2

Suy ra A < 1/2^2 + 1/2 = 3/4 

Vậy A< 3/4

1) \(\frac{100}{\sqrt{100}}=\frac{100}{10}=10\)

=> \(\frac{1}{\sqrt{1}}+\frac{2}{\sqrt{2}}+....+\frac{100}{\sqrt{100}}>\frac{100}{\sqrt{100}}=10\)

2) Xét hiệu: \(\frac{a+b}{2}-\sqrt{ab}=\frac{a+b-2\sqrt{ab}}{2}=\frac{a-\sqrt{ab}-\sqrt{ab}+b}{2}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2}\ge0\)

=> \(\frac{a+b}{2}\ge\sqrt{ab}\) 

Vậy....

a) đặt A=vế trái ,sau khi rút gọc tử và mẫu của từng số hạng ta đc

A=\(\sqrt{1}\)+\(\sqrt{2}\)+\(\sqrt{3}\)+...+\(\sqrt{100}\)

vì \(\sqrt{100}\)=10

\(\sqrt{1}\)>0

\(\sqrt{2}\)>0

...

\(\sqrt{99}\)>0

cộng lại ta sẽ đc A>0

b) 

(a-b)²>=0

=>a²+b²>=2ab

=>a²+2ab+b²>=2ab+2ab

=>(a+b)²>=4ab

=>a+b>=2.\(\sqrt{ab}\) với mọi a,b>0

=>dpcm (chia cả 2 vế cho 2)

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 

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)

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

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