CMR : A = \(\frac{1}{\sqrt{1}}\)+\(\frac{1}{\sqrt{2}}\)+\(\frac{1}{\sqrt{3}}\)+\(\frac{1}{\sqrt{4}}\)+...+\(\frac{1}{\sqrt{n}}\)>\(\sqrt{n+1}\)với n > 2
giúp mình với nhé
1) CMR \(\frac{1}{\sqrt{1.1999}}+\frac{1}{\sqrt{2.1998}}+\frac{1}{\sqrt{3.1997}}+...+\frac{1}{\sqrt{1999.1}}\ge1,999\)
2) CMR \(\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{95\sqrt{94}+94\sqrt{95}}< 1\)
3) CMR \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
4) CMR \(\sqrt{n}< \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}< 2\sqrt{n}\)
a) CMR: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}\) với \(n\in N\)*
b) tính \(B=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+......+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
a/ Quy đồng vế phải, hình như lộn mẫu cuối là căn 2 của (n+1) mới đúng
\(VP=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\frac{\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}}{\sqrt{n+1}.\sqrt{n}}\)
\(=\frac{1}{\left(\sqrt{n+1}+\sqrt{n}\right).\sqrt{n+1}.\sqrt{n}}=\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=VT\)
\(B=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
\(=1-\frac{1}{\sqrt{25}}\)
CMR: Với n thuộc N* thì:
\(\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{x+1}}\right)\)
Từ đó suy ra tổng sau k là số nguyên tố:
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2008\sqrt{2007}}\)
Các bạn giúp mk với nhé! Mk cần gấp
\(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}=\frac{\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}}{\sqrt{n}.\sqrt{n+1}}=\frac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(>\frac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n+1}\right)}=\frac{1}{2\left(n+1\right).\sqrt{n}}\)
Suy ra \(\text{Tổng }=...< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{199}}-\frac{1}{\sqrt{200}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{200}}\right)< 2\)
Một số < 2 thì hiển nhiên ko phải là một số nguyên tố (SNT nhỏ nhất là 2)
Bài 1: CMR
Bài 2: CMR
Các bạn ơi giải giúp mình với nha :
Rút gọn biểu thức:
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{90}+\sqrt{100}}\)
Chứng minh đẳng thức :
\(\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\)với n là số tự nhiên
Chứng minh các đại thức :
\(\left(\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{6-4\sqrt{2}}}\right)^2=\sqrt{8}\)
Giúp mình với nhé!
a)= \(\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{100}-\sqrt{99}}{100-99}\)
=\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)
= \(-1+\sqrt{100}\)
= -1 +10
=9
b)Ta có\(\left(\sqrt{n+1}-\sqrt{n}\right)\cdot\left(\sqrt{n+1}+\sqrt{n}\right)\)=n+1-n=1 (1)
Lại có:\(\frac{1}{\sqrt{n+1}+1}\cdot\left(\sqrt{n+1}+1\right)=1\)(2)
Từ (1) và (2)=>\(\left(\sqrt{n+1}-1\right)=\frac{1}{\sqrt{n+1}+1}\)
c)\(\left(\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{6-4\sqrt{2}}}\right)^2\)
=\(\left(\frac{\left(2+\sqrt{2}\right)^2}{\sqrt{2}+\sqrt{\left(2+\sqrt{2}\right)^2}}+\frac{\left(2-\sqrt{2}\right)^2}{\sqrt{2}-\sqrt{\left(2-\sqrt{2}\right)^2}}\right)^2\)
=\(\left(\frac{\left(2+\sqrt{2}\right)^2}{2+2\sqrt{2}}+\frac{\left(2-\sqrt{2}\right)^2}{-2+2\sqrt{2}}\right)^2\)
=\(\left(\frac{\left(2+\sqrt{2}\right)^2\cdot\left(2\sqrt{2}-2\right)}{\left(2\sqrt{2}+2\right)\cdot\left(2\sqrt{2}-2\right)}+\frac{\left(2-\sqrt{2}\right)^2\cdot\left(2\sqrt{2}+2\right)}{\left(2\sqrt{2}-2\right)\left(2\sqrt{2}+2\right)}\right)^2\)
=\(\left(\frac{\left(2+\sqrt{2}\right)^2\cdot\left(2\sqrt{2}-2\right)+\left(2-\sqrt{2}\right)^2\cdot\left(2\sqrt{2}+2\right)}{4}\right)^2\)
=\(\left(\frac{12\sqrt{2}-12+16-8\sqrt{2}+12\sqrt{2}+12-16-8\sqrt{2}}{4}\right)^2\)
=\(\left(\frac{8\sqrt{2}}{4}\right)^2=8\)
CMR:
Với n thuộc N*
\(a)1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}>\sqrt{n}\\ b)\frac{1}{\sqrt{n}}>2\left(\sqrt{n-1}-\sqrt{n}\right)\)
CMR n\(\in\)N, n>3
a,\(\frac{1}{2\sqrt{1} }+\frac{1}{3\sqrt{2} } +\frac{1}{4\sqrt{3} }+...+\frac{1}{(n+1)\sqrt{n} }<2 \)
b,S=\(\frac{1}{3(1+\sqrt{2}) }+\frac{1}{5(\sqrt{2}+\sqrt{3} }+...+\frac{1}{(2n+1)(\sqrt{n}+\sqrt{n+1}) } \)
a, Chắc xét hàm số tổng quát!
Xét hàm số tổng quát:
\(\dfrac{1}{\left(k+1\right)\sqrt{k}}=\dfrac{\sqrt{k}}{k\left(k+1\right)}=\sqrt{k}\left(\dfrac{1}{k\left(k+1\right)}\right)\)
\(=\sqrt{k}\left[\sqrt{\dfrac{1}{k}}^2-\sqrt{\dfrac{1}{k+1}}^2\right]\)
\(=\sqrt{k}\left(\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}}\right)\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\)
\(=\left(1+\dfrac{\sqrt{k}}{\sqrt{k+1}}\right)\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\)
Vì \(\dfrac{\sqrt{k}}{\sqrt{k+1}}< 1\Rightarrow1+\dfrac{\sqrt{k}}{\sqrt{k+1}}< 2\)
Do đó \(\left(1+\dfrac{\sqrt{k}}{\sqrt{k+1}}\right)\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)< 2.\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\)
\(\Rightarrow\dfrac{1}{\left(k+1\right)\sqrt{k}}< 2\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\) (1)
Áp dụng điểu (1) ta được:
\(\dfrac{1}{2}< 2\left(\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}\right)\)
\(\dfrac{1}{3\sqrt{2}}< 2\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\right)\)
...................................
\(\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
\(\Rightarrow\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+....+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
\(\Rightarrow\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(1-\dfrac{1}{\sqrt{n+1}}\right)\)
Với mọi giá trị của \(n>0\) ta luôn có: \(\sqrt{n+1}>0\)
Do đó \(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\) (đpcm)
CMR\(\frac{1}{2\sqrt{n+1}}< \sqrt{n+1}-\sqrt{n}\)với n thuộc N
Áp dụng CMR \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2500}}< 100\)
Ta có :
\(\hept{\begin{cases}\frac{1}{2\sqrt{n+1}}< \frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\\\sqrt{n+1}-\sqrt{n}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\end{cases}}\forall n\in N\)
Suy ra : \(\frac{1}{2\sqrt{n+1}}< \sqrt{n+1}-\sqrt{n}\)
Đặt \(M=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2499}}+\frac{1}{\sqrt{2500}}\)
\(\Leftrightarrow\frac{1}{2}M=\frac{1}{2\sqrt{2500}}+\frac{1}{2\sqrt{2499}}+...+\frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt{2}}+\frac{1}{2}\)
Áp dụng BĐT , ta có :
\(\frac{1}{2}M< \sqrt{2500}-\sqrt{2499}+\sqrt{2499}-\sqrt{2498}+...+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}+\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}M< \sqrt{2500}-\sqrt{1}+\frac{1}{2}=50-\frac{1}{2}< 50\)
\(\Rightarrow M< 100\)
CMR:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}>\sqrt{n}\)
(Với n\(\in\)N và n>1).
Đặt A =\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{n}}\)
=> A > \(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}+.....+\frac{1}{\sqrt{n}}\)
=> A > \(\frac{1}{\sqrt{n}}.n\)
=> A > \(\sqrt{n}\)
=> \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{n}}>\sqrt{n}\)(Đpcm)