#include <bits/stdc++.h>
using namespace std;

int main()
{
double i,n,tong,t;
cin>>t;
for(i=1;i<=t;i++)
{
cin>>n;
tong=0;
for(i=1;i<=n;i++) tong=sqrt(tong+i);
cout<<fixed<<setprecision(10)<<tong<<'\n';
}
return 0;
}

Nguyễn Minh Lệ

a, Để A nhận giá trị dương thì \(A>0\)hay \(x-1>0\Leftrightarrow x>1\)

b, \(B=2\sqrt{2^2.5}-3\sqrt{3^2.5}+4\sqrt{4^2.5}\)

\(=4\sqrt{5}-9\sqrt{5}+16\sqrt{5}=\left(4-9+16\right)\sqrt{5}=11\sqrt{5}\)

( theo công thức \(A\sqrt{B}=\sqrt{A^2B}\))

c, Với \(a\ge0;a\ne1\)

\(C=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)

\(=\left(\sqrt{a}+1\right)^2.\frac{1}{\left(\sqrt{a}+1\right)^2}=1\)

a) \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right).\sqrt{\dfrac{8-2\sqrt{15}}{2}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{25.6}-\sqrt{9.10}\right).\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right).\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}\)

\(=\left(\sqrt{10}+\sqrt{6}\right).\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}=\sqrt{2}.\left(\sqrt{5}+\sqrt{3}\right).\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)

\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)

a) Ta có: \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\sqrt{8-2\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(\sqrt{5}-\sqrt{3}\right)^2\cdot\left(4+\sqrt{15}\right)\)

\(=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)

\(=32+8\sqrt{15}-8\sqrt{15}-30\)

=2

ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)

1) Ta có: \(N=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\cdot\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)

\(=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\cdot\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)

\(=1-a\)

2) Để N=-2016 thì 1-a=-2016

\(\Leftrightarrow1-a+2016=0\)

\(\Leftrightarrow2017-a=0\)

hay a=2017(thỏa ĐK)

Vậy: Để N=-2016 thì a=2017

a: Ta có: \(N=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)

\(=x-\sqrt{x}+1\)

Đặt \(A=\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]+\left[\sqrt{4}\right]+...+\left[\sqrt{212041}\right]\)

\(=\left(\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]\right)+\left(\left[\sqrt{4}\right]+...+\left[\sqrt{8}\right]\right)+\left(\left[\sqrt{9}\right]+...+\left[\sqrt{15}\right]\right)+...+\left(\left[\sqrt{210681}\right]+...+\left[\sqrt{211599}\right]\right)+\left(\left[\sqrt{211600}\right]+\left[\sqrt{212041}\right]\right)\)

Theo cách chia nhóm như trên, nhóm 1 có 3 số, nhóm 2 có 5 số, nhóm 3 có 7 số, nhóm 4 có 9 số, ..., nhóm 459 có 919 số, nhóm cuối cùng có 442 số. Các số thuộc nhóm 1 bằng 1, các số thuộc nhóm 2 bằng 2, các số thuộc nhóm 3 bằng 3, ..., các số thuộc nhóm 459 bằng 459, Các số thuộc nhóm cuối cùng bằng 460.

Do đó \(A=1.3+2.5+3.7+...+459.919+460.442\)

            \(=1\left(1.2+1\right)+2.\left(2.2+1\right)+3.\left(3.2+1\right)+...+459.\left(459.2+1\right)+203320\)

            \(=\left(2.1^2+1\right)+\left(2.2^2+1\right)+\left(2.3^2+1\right)+...+\left(2.459^2+1\right)+203320\)

            \(=2.\left(1^2+2^2+3^2+...+459^2\right)+\left(1+2+3+...+459\right)+203320\)

            \(=2.\frac{1}{6}.459.460.919+105570+203320=64988110\)

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