Tính giá trị biểu thức
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2015\sqrt{2014}+2014\sqrt{2015}}\)
Tính
\(M=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+.......+\dfrac{1}{2015\sqrt{2014}+2014\sqrt{2015}}\)
Cho M=\(\dfrac{\sqrt{2}-\sqrt{1}}{1+2}+\dfrac{\sqrt{3}-\sqrt{2}}{2+3}+\dfrac{\sqrt{4}-\sqrt{3}}{3+4}+...+\dfrac{\sqrt{2015}-\sqrt{2014}}{2014+2015}\)
Hãy so sánh M với \(\dfrac{1}{2}\)
cm rằng
\(2\left(\sqrt{a}-\sqrt{b}\right)< \dfrac{1}{\sqrt{b}}< 2\left(\sqrt{b}-\sqrt{c}\right)\)
biết a,b,c là 3 số thực thỏa mãn đk a=b+1=c+2 , c>0
b)biểu thức \(B=\sqrt{1+2014^2+\dfrac{2014^2}{2015^2}}+\dfrac{2014}{2015}\) có giá trị là 1 số nguyên
b,\(B=\sqrt{1+2014^2+\dfrac{2014^2}{2015^2}}+\dfrac{2014}{2015}\)
Ta có :\(\left(2014+1\right)^2=2014^2+1+2.2014\)
\(\Rightarrow2014^2+1=2015^2-2.2014\)
\(\Rightarrow B=\sqrt{2015^2-2.2014+\left(\dfrac{2014}{2015}\right)^2}+\dfrac{2014}{2015}\)
\(=\sqrt{\left(2015-\dfrac{2014}{2015}\right)^2}+\dfrac{2014}{2015}\)
\(=2015-\dfrac{2014}{2015}+\dfrac{2014}{2015}\)
\(=2015\)
Vậy B=2015
Tính giá trị biểu thức : \(S=\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+.....+\frac{1}{2014\sqrt{2015}+2015\sqrt{2014}}\)
1. Tính
a. \(\left(3\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-3\sqrt{2}\right)\)
b. \(\dfrac{1}{\sqrt{2013}-\sqrt{2014}}-\dfrac{1}{\sqrt{2014}-\sqrt{2015}}\)
c. \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}\)
d. \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}\)
c: Ta có: \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}\)
\(=4+\sqrt{10}-4+\sqrt{10}\)
\(=2\sqrt{10}\)
d: Ta có: \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}\)
\(=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1\)
\(=2\sqrt{2}\)
a) \(=\left(2\sqrt{3}\right)^2-\left(3\sqrt{2}\right)^2=12-18=-6\)
b) \(=\dfrac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\dfrac{\sqrt{2014}+\sqrt{2015}}{2014-2015}=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}-\sqrt{2015}=-\sqrt{2013}-\sqrt{2015}\)
c) \(=4+\sqrt{10}-4+\sqrt{10}=2\sqrt{10}\)
d) \(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1=2\sqrt{2}\)
Tính gía trị biểu thức:
\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+....+\frac{1}{2014\sqrt{2013}+2013\sqrt{2014}}+\frac{1}{2015\sqrt{2014}+2014\sqrt{2015}}\)
Chứng minh \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\) rồi áp dụng với n = 1,2,....,2014
Chứng tỏ :
\(\dfrac{1}{\sqrt{x+2014}+\sqrt{y+2014}}-\dfrac{1}{\sqrt{2015-x}+\sqrt{2015-y}}+\dfrac{1}{\sqrt{2014-x}+\sqrt{2014-y}}\ne0\)
CMR: b) Biểu thức B=\(\sqrt{1+2014^2+\dfrac{2014^2}{2015^2}}+\dfrac{2014}{2015}\) có giá trị là một số nguyên
Với \(\forall a\in N\left(a\ne0\right)\cdot\),ta có:\(\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}=\sqrt{\dfrac{\left(a^2+1\right)\left(a^2+2a+1\right)+a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}=\sqrt{\dfrac{\left(a^2+1\right)^2+2a\left(a^2+1\right)+a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}=\sqrt{\dfrac{\left(a^2+a+1\right)^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}=\dfrac{a^2+a+1}{a+1}+\dfrac{a}{a+1}=\dfrac{\left(a+1\right)^2}{a+1}=a+1\in Z\)(Vì a là số tự nhiên)
Thay a=2014 vào thì ta có: B=2014+1=2015 là số nguyên
Cho biểu thức P= \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{2\sqrt{x}}{x+1-x\sqrt{x}-\sqrt{x}}\right):\left(1-\dfrac{2\sqrt{x}}{x+1}\right)\)
a, Rút gọn P
b, Tính giá trị của P khi x=2015-2\(\sqrt{2014}\)
c, Tính gia strij của x sao cho P ≥ 1
a: \(P=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{2\sqrt{x}}{\left(x+1\right)\left(1-\sqrt{x}\right)}\right):\dfrac{x+1-2\sqrt{x}}{x+1}\)
\(=\dfrac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x-2\sqrt{x}+1}=\dfrac{1}{\sqrt{x}-1}\)
b: Khi x=2015-2 căn 2014 thì
\(P=\dfrac{1}{\sqrt{2014}-1-1}=\dfrac{1}{\sqrt{2014}-2}=\dfrac{\sqrt{2014}+2}{2010}\)
c: Để P>=1 thì P-1>=0
=>(1-căn x+1)/căn x-1>=0
=>(căn x-2)/(căn x-1)<=0
=>1<căn x<=2
=>1<x<=4