Chứng minh (1-1/3). (1-1/6)....(1-1/253)<2/5
Ai nhanh mình tick
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Tích trên có thừa số 1 - 253 = -252 còn các thừa số kia trong tích đều dương. Vậy tích trên âm.
Mà 2/5 dương nên đpcm
(1−13)(1−16)...(1−1253)
=23⋅56⋅...⋅252253=46⋅1012⋅...⋅504506
=1⋅42⋅3⋅2⋅53⋅4⋅...⋅21⋅2422⋅23
=1⋅2⋅3⋅42⋅52⋅...⋅212⋅22⋅23⋅242⋅32⋅42⋅...⋅222⋅23=1⋅243⋅22=2466<25
\(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)...\left(1-\dfrac{1}{253}\right)\\=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{252}{253}\\ =\dfrac{4}{6}\cdot\dfrac{10}{12}\cdot...\cdot\dfrac{504}{506}\\ =\dfrac{1\cdot4}{2\cdot3}\cdot\dfrac{2\cdot5}{3\cdot4}\cdot...\cdot\dfrac{21\cdot24}{22\cdot23}\\ =\dfrac{1\cdot2\cdot3\cdot4^2\cdot5^2\cdot...\cdot21^2\cdot22\cdot23\cdot24}{2\cdot3^2\cdot4^2\cdot...\cdot22^2\cdot23}\\ =\dfrac{1\cdot24}{3\cdot22}=\dfrac{24}{66}< \dfrac{2}{5}\)
a>
\(\frac{1}{2^2}+\frac{1}{100^2}\)=1/4+1/10000
ta có 1/4<1/2(vì 2 đề bài muốn chứng minh tổng đó nhỏ 1 thì chúng ta phải xét xem có bao nhiêu lũy thừa hoặc sht thì ta sẽ lấy 1 : cho số số hạng )
1/100^2<1/2
=>A<1
Ta có : 1-1/2+1/3-1/4+1/5-1/6
= 1+1/2+1/3+1/4+1/5+1/6-2.(1/2+1/4+1/6)
= 1+1/2+1/3+1/4+1/5+1/6-(1+1/2+1/3)
=1+1/2+1/3+1/4+1/5+1/6-1-1/2-1/3
=1/4+1/5+1/6
1) \(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\dfrac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-\sqrt{x}}\)
\(A=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\cdot\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b) Ta có:
\(A\cdot\sqrt{x}=25\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\cdot\sqrt{x}=25\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=25\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=5^2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=5\\\sqrt{x}+1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=16\\\sqrt{x}=-6\text{(vô lý)}\end{matrix}\right.\)
c) Ta xét hiệu:
\(A-4=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}-4\)
\(A-4=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}-\dfrac{4\sqrt{x}}{\sqrt{x}}\)
\(A-4=\dfrac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}}\)
\(A-4=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}\)
\(A-4=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Với \(x>0\) thì \(\left(\sqrt{x}-1\right)>0\) và \(\sqrt{x}>0\)
\(\Rightarrow\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
Nên A > 4 (đpcm)
1: \(A=\dfrac{x-1}{\sqrt{x}}:\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1+1-\sqrt{x}}\)
\(=\dfrac{\left(x-1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
2: A*căn x=25
=>(căn x+1)^2=25
=>căn x+1=5
=>x=16
3: \(A-4=\dfrac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
=>A>4