tính
A=\(\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{199}}{\frac{1}{1.199}+\frac{1}{3.197}+...+\frac{1}{197.3}+\frac{1}{199.1}}\)
CMR:
\(\frac{1}{\sqrt{1.199}}+\frac{1}{\sqrt{2.198}}+\frac{1}{\sqrt{3.197}}+...+\frac{1}{\sqrt{199.1}}>1,99\)
\(VT=2.\left(\frac{1}{\sqrt{1.199}}+\frac{1}{\sqrt{2.198}}+...+\frac{1}{\sqrt{99.101}}+\frac{1}{\sqrt{100.100}}\right)\)
\(=2\left(\frac{1}{\sqrt{1.199}}+...+\frac{1}{\sqrt{n\left(200-n\right)}}+...+\frac{1}{\sqrt{99.101}}+\frac{1}{100}\right)\)\(\left(1\le n\le99\right)\)
Ta chứng minh \(\sqrt{n\left(200-n\right)}\le100\text{ }\left(\text{*}\right)\)
\(\left(\text{*}\right)\Leftrightarrow200n-n^2\le100^2\Leftrightarrow n^2-2.100n+100^2\ge0\)
\(\Leftrightarrow\left(100-n\right)^2\ge0\)
Do bất đẳng thức cuối đúng nên (*) là đúng, do đó ta có:
\(A\ge2\left(\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}\right)\text{ }\left(\text{100 số }\frac{1}{100}\right)\)
\(=2>1,99\)
A=\(\frac{1}{\sqrt{1.199}}+\frac{1}{\sqrt{2.198}}+\frac{1}{\sqrt{3.197}}+...+\frac{1}{\sqrt{199.1}}\)
Chứng minh rằng: A > 1,99
Áp dụng BĐT sau : \(\frac{1}{\sqrt{a.b}}>\frac{2}{a+b}\) với \(a\ne b\) (bạn tự chứng minh) , ta được :
\(A=\frac{1}{\sqrt{1.199}}+\frac{1}{\sqrt{2.198}}+\frac{1}{\sqrt{3.197}}+...+\frac{1}{\sqrt{199.1}}\)
\(>2.\left(\frac{1}{1+199}+\frac{1}{2+198}+\frac{1}{3+197}+...+\frac{1}{199+1}\right)\)
\(=2.\frac{199}{200}=1,99\)
Vậy A > 1,99
mi tích tau tau tích mi xong tau trả lời nka
việt nam nói là làm
Chứng minh BĐT đó dễ thôi , suy ra từ BĐT Cauchy: \(a+b\ge2\sqrt{ab}\Rightarrow\frac{1}{\sqrt{ab}}\ge\frac{2}{a+b}\)
A=\(\frac{1}{\sqrt{1.199}}\) +\(\frac{1}{\sqrt{2.198}}\) +\(\frac{1}{\sqrt{3.197}}\)+...+\(\frac{1}{\sqrt{198.2}}\)+\(\frac{1}{\sqrt{199.1}}\)
Chứng minh A>1,99
Ta có với a,b là hai số dương và khác nhau thì \(\sqrt{ab}< \frac{a+b}{2}\Leftrightarrow\frac{1}{\sqrt{ab}}>\frac{2}{a+b}\)
Áp dụng điều trên , ta có :
\(A=\frac{1}{\sqrt{1.199}}+\frac{1}{\sqrt{2.198}}+\frac{1}{\sqrt{3.197}}+...+\frac{1}{\sqrt{198.2}}+\frac{1}{\sqrt{199.1}}\)
\(>2\left(\frac{1}{1+199}+\frac{1}{2+198}+\frac{1}{3+197}+...+\frac{1}{198+2}+\frac{1}{199+1}\right)\)
\(\Rightarrow A>2.\frac{199}{200}=1,99\)
A= \(\frac{1}{\sqrt{1.199}}+\frac{1}{\sqrt{2.198}}+...+\frac{1}{\sqrt{199.1}}\)
so sánh A với 1
Áp dụng bđt \(\frac{1}{\sqrt{ab}}>\frac{2}{a+b}\) với a > 0; b > 0; a \(\ne\) b ta có:
\(A=\frac{1}{\sqrt{1.199}}+\frac{1}{\sqrt{2.198}}+...+\frac{1}{\sqrt{199.1}}>\frac{2}{1+199}+\frac{2}{2+198}+...+\frac{2}{199+1}\)
\(A>\frac{2}{200}+\frac{2}{200}+...+\frac{2}{200}\) (199 số \(\frac{2}{200}\))
\(A>\frac{2}{200}.199\)
\(A>\frac{1}{100}.199=1,99>1\)
=> A > 1
Tính bằng cách hợp lí:
a) A=\(\left(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+....+\frac{1}{101.400}\right):\left(\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\right)\)
b) B=\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{200}\right):\left(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+....\frac{198}{2}+\frac{199}{1}\right)\)
S= 1/199 + 2/198 + ... + 198/2 + 199/1
S= (1/199 + 1) + (2/198 + 1)+ ... + (198/2 + 1) +1
S= 200/200 + 200/199 + 200/198 + ... + 200/2
S= 200.(1/200 + 1/199 + ... + 1/2)
Suy ra , B=(1/2 + 1/3 + ... +1/200) : 200.(1/2 + 1/3 + ... + 1/200)
B=1 : 200 = 1/200
1)\(A=\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{8}+...+\frac{197}{198}-\frac{199}{200}\)
\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
tính B:A
2)\(A=1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{4026}\)
\(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{4025}\)
So sánh \(\frac{A}{B}và1\frac{2013}{2014}\)
1. Tính nhanh
a, \(\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{21}{39}+\frac{49}{91}.\frac{8}{15}\)
b, \(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
c,\(\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{192}{2}+\frac{199}{1}}\)
Tính nhanh giúp mình nha ! NHANH HẾT MỨC CÓ THỂ NHÉ
a)Tính\(\frac{\left(17\frac{8}{19}-16\frac{9}{18}\right)\left(17,5+16\frac{17}{51}-32\frac{15}{22}\right)}{\frac{7}{3.13}+\frac{7}{13.23}+\frac{7}{23.33}}\)
b) Chứng tò rằng:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
phần a dễ bạn tự làm đi tử thì bạn tính như bình thường còn mẫu thì:7.(\(\frac{1}{3.13}\)+\(\frac{1}{13.23}\)+\(\frac{1}{23.33}\))
\(\frac{7}{10}\).(\(\frac{1}{3}\)-\(\frac{1}{33}\))=\(\frac{7}{33}\)
b)(1+1/3+1/5+..+1/199)-(1/2+1/4+...+1/200)
(1+1/2+1/3+...+1/199+1/200)-(1/2+1/2+1/4+1/4+...+1/200+1/200)
=1+1/2+1/3+...+1/199+1/200-(1+1/2+1/3+...+1/100)
=1/101+1/102+...+1/200
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