cho bt: P=\(\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right).\dfrac{1}{a\sqrt{a}}\) với a>0; a\(\ne\)1
a, rút gọn P
b, tính giá trị tại a=\(\sqrt{9+4\sqrt{2}}\)
giải với ạ e cần gấp!
1\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\cdot\left(\sqrt{a}\cdot\dfrac{4}{\sqrt{a}}\right)\) (với a>0,a≠4)
2\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\cdot\left(1-\dfrac{1}{\sqrt{a}}\right)\) (với a>0,a≠1)
3\(\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\cdot\left(1-\dfrac{2}{a+1}\right)^2\) (với a>0,a≠1)
Rút gọn các biểru thức sau
1. \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}.\dfrac{4}{\sqrt{a}}\right)=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{-64\sqrt{a}}{a-4}\)Nếu nhân tu thứ 2 của phép tính là \(\sqrt{a}-\dfrac{4}{\sqrt{a}}\) thì kết quả của phép tính là -16 nha bạn
2.\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right).\left(1-\dfrac{1}{\sqrt{a}}\right)=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{-\left(1-\sqrt{a}\right)}{\sqrt{a}}=\dfrac{-2\sqrt{a}}{\left(1+\sqrt{a}\right)\sqrt{a}}=\dfrac{-2}{1+\sqrt{a}}\)\(\left(a>0,a\ne1\right)\)
1. cho biểu thức
A=\(\left(\dfrac{\sqrt{a}}{\sqrt{a}+2}-\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{4\sqrt{a}-1}{a-4}\right):\dfrac{1}{\sqrt{a}+2}\)
a, rút gọn bt a
Lời giải:
ĐKXĐ: $a\geq 0; a\neq 4$
\(A=\left[\frac{\sqrt{a}(\sqrt{a}-2)-\sqrt{a}(\sqrt{a}+2)}{(\sqrt{a}+2)(\sqrt{a}-2)}+\frac{4\sqrt{a}-1}{(\sqrt{a}-2)(\sqrt{a}+2)}\right].(\sqrt{a}+2)\)
\(=\frac{-4\sqrt{a}+4\sqrt{a}-1}{(\sqrt{a}-2)(\sqrt{a}+2)}.(\sqrt{a}+2)=\frac{-1}{(\sqrt{a}-2)(\sqrt{a}+2)}.(\sqrt{a}+2)=\frac{1}{2-\sqrt{a}}\)
A= \(\dfrac{7\sqrt{a}}{a-9}-\left(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{\sqrt{a}-1}{\sqrt{a}+3}\right)\) ĐK:(a≥0, a≠9)
B= \(\left(\dfrac{1}{\sqrt{a}-3}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}\right)\) ĐK:(a≥0, a≠9)
C= \(\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\sqrt{a}-a}\right).\left(\dfrac{1}{a}-2\right)\) ĐK:(a>0, a≠1)
D= \(\dfrac{a\sqrt{a}+1}{a-1}-\dfrac{a-1}{\sqrt{a}+1}\) ĐK:(a≥0, a≠1)
E= \(\dfrac{a}{a-4}+\dfrac{1}{\sqrt{a}-2}+\dfrac{1}{\sqrt{a}+2}\) ĐK:(a≥0, a≠4)
Giúp mìnk với nha !!!
\(A=\dfrac{7\sqrt{a}}{a-9}-\left(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{\sqrt{a}-1}{\sqrt{a}+3}\right)=\dfrac{7\sqrt{a}}{a-9}-\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\dfrac{7\sqrt{a}}{a-9}-\dfrac{a+3\sqrt{a}-a+3\sqrt{a}+\sqrt{a}-3}{a-9}=\dfrac{3}{a-9}\)\(B=\left(\dfrac{1}{\sqrt{a}-3}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}\right)=\dfrac{\sqrt{a}-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-3\right)}:\dfrac{a-9-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\dfrac{3}{\sqrt{a}\left(\sqrt{a}-3\right)}.\dfrac{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}{-5}=\dfrac{3\sqrt{a}-6}{-5\sqrt{a}}\)
\(C=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\sqrt{a}-a}\right).\left(\dfrac{1}{a}-2\right)=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\left(\sqrt{a}-1\right)}\right).\dfrac{1-2a}{a}=\dfrac{a\sqrt{a}-a}{\sqrt{a}-1}.\dfrac{1-2a}{a}=\dfrac{a\left(\sqrt{a}-1\right)}{\sqrt{a}-1}.\dfrac{1-2a}{a}=1-2a\)\(D=\dfrac{a\sqrt{a}+1}{a-1}-\dfrac{a-1}{\sqrt{a}+1}=\dfrac{a\sqrt{a}+1-\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1}=\dfrac{a\sqrt{a}+1-a\sqrt{a}+a+\sqrt{a}-1}{a-1}=\dfrac{a+\sqrt{a}}{a-1}=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-1}\)
\(E=\dfrac{a}{a-4}+\dfrac{1}{\sqrt{a}-2}+\dfrac{1}{\sqrt{a}+2}=\dfrac{a+\sqrt{a}+2+\sqrt{a}-2}{a-4}=\dfrac{a+2\sqrt{a}}{a-4}=\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-2}\)
rút gọn
\(\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right):\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\) với a>0; akhacs 1
\(\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right):\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(a>0;a\ne1\right)\\ =\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4\sqrt{a}\left(a-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\dfrac{a-1}{\sqrt{a}}\\ =\dfrac{4a\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\\ =\dfrac{4a^2}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)^2}\)
A= \(\left(\dfrac{a-1}{\sqrt{a}-1}-2\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+1\right)\) ĐK: (a≥0, a≠1)
B= \(\left(\dfrac{a\sqrt{a}-a}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\) ĐK: (a>0, a≠0, a≠2)
C= \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{a}{a-1}\right):\left(\sqrt{a}-\dfrac{\sqrt{a}}{\sqrt{a}+1}\right)\) ĐK: (a>0, a≠1)
D= \(\dfrac{a+\sqrt{a}}{\sqrt{a}}+\dfrac{a+4}{\sqrt{a}+2}\) ĐK: (a>0)
E= \(\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}+\dfrac{1-\sqrt{a}}{a+\sqrt{a}}\right)\) ĐK: (a>0, a≠1)
A = \(\left(\dfrac{a-1}{\sqrt{a}-1}-2\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+1\right)=\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-2\right)\left(\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}+1\right)=\left(\sqrt{a}+1-2\right)\left(\sqrt{a}+1\right)=\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)=a-1\)
\(B=\left(\dfrac{a\sqrt{a}-a}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}=\left(\dfrac{a\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}=\left(\dfrac{a}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\right)\cdot\dfrac{a-2}{a+2}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}=\dfrac{\left(\sqrt{a}-1\right)\left(a-2\right)}{\sqrt{a}\left(a+2\right)}\)
\(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{a}{a-1}\right):\left(\sqrt{a}-\dfrac{\sqrt{a}}{\sqrt{a}+1}\right)=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{a-1}-\dfrac{a}{a-1}\right):\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)-\sqrt{a}}{\sqrt{a}+1}\right)=\dfrac{\sqrt{a}}{a-1}:\dfrac{a}{\sqrt{a}+1}=\dfrac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}+1}{a}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\)
\(D=\dfrac{a+\sqrt{a}}{\sqrt{a}}+\dfrac{a+4}{\sqrt{a}+2}=\sqrt{a}+1+\dfrac{a+4}{\sqrt{a}+2}=\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)+\sqrt{a}+2+a+4}{\sqrt{a}+2}=\dfrac{a+2\sqrt{a}+\sqrt{a}+2+a+4}{\sqrt{a}+2}=\dfrac{2a+3\sqrt{a}+6}{\sqrt{a}+2}\)
\(E=\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}+\dfrac{1-\sqrt{a}}{a+\sqrt{a}}\right)=\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)+1-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\cdot\dfrac{a-1+1-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}=\dfrac{\left(\sqrt{a}-1\right)\cdot\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}\cdot\sqrt{a}}=\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}\)
Rút gọn
a) với x>0 , x\(\ne\)1
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}+2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\)
b) với a>0,a\(\ne\)4
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)
c)\(\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\) với a>0 ,a\(\ne\)1
d)\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\) với x>1
a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
cho M=\(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{ }a}\right)\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)với a>0
\(M=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right)\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)
\(=\left[\dfrac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]\left[\dfrac{\sqrt{a}-1}{a-1}+\dfrac{2}{a-1}\right]\)
\(=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\sqrt{a}-1}{a-1}\)
\(=\dfrac{1}{\sqrt{a}}\)
\(M=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right)\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)
\(=\left[\dfrac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]\left[\dfrac{\sqrt{a}-1}{a-1}+\dfrac{2}{a-1}\right]\)
\(=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\sqrt{a}+1}{a-1}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\)
1. Tìm x để bt có nghĩa
A=\(\dfrac{\sqrt{2x+3}}{\sqrt{x-3}}\)
B=\(\sqrt{\dfrac{2x+3}{x-3}}\)
C=\(\sqrt{-\dfrac{5}{x+2}}\)
D=\(\sqrt{-x}+\dfrac{1}{x+3}\)
2. Rút gọn bt
A=\(\sqrt{\dfrac{a+\sqrt{a^2-1}}{2}}-\sqrt{\dfrac{a-\sqrt{a^2-1}}{2}};\left(a>1\right)\)
B=\(\sqrt{\dfrac{a+\sqrt{a^2-1}}{2}}-\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}};\left(a\ge\sqrt{b};b\ge0\right)\)
C=\(\left(1+\dfrac{a+\sqrt{a}}{a+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}+1}\right);\left(a\ge0,a\ne1\right)\)
D=\(\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}};\left(x>0\right)\)
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
Rút gọn:
A = \(\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
B = \(\dfrac{3\sqrt{2}+\sqrt{11}}{\sqrt{2}+\sqrt{6+\sqrt{11}}}+\dfrac{3\sqrt{2}-\sqrt{11}}{\sqrt{2}-\sqrt{6-\sqrt{11}}}+18\)
C = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n+3}}\)với n thuộc N*
D = \(\left(\sqrt{3}+1\right)\left(\sqrt{5}-1\right)\left(\sqrt{15}-1\right)\left(7-2\sqrt{3}+\sqrt{5}\right)\)
E=\(\dfrac{\left(4+\sqrt{3}\right)}{\sqrt[]{1}+\sqrt{3}}+\dfrac{\left(8+\sqrt{15}\right)}{\sqrt{3}+\sqrt{5}}+...+\dfrac{2k+\sqrt{k^2-1}}{\sqrt{k-1}+\sqrt{k+1}}+...+\dfrac{240+\sqrt{14399}}{\sqrt{119}+\sqrt{121}}\)
F = \(\left(\dfrac{2a+1}{a\sqrt{a}-1}-\dfrac{\sqrt{a}}{a+\sqrt{a}+1}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\) với a >= 0 và a khác 1