ĐKXĐ : \(0\le x\ne1\)

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

\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{x-1}{2}\) = \(\dfrac{2\sqrt{x}}{x-1}.\dfrac{x-1}{2}=\sqrt{x}\)

Vậy ... 

ĐKXĐ : \(x;y\ne0;x\ne-y\)

Đặt s = x + y \(\ne0\) ; p = xy \(\ne0\) ( \(s^2\ge4p\) ) 

 HPT trở thành : \(\left\{{}\begin{matrix}s-\dfrac{4s}{p}=3\left(1\right)\\s+\dfrac{6}{s}=-5\left(2\right)\end{matrix}\right.\)

(2) \(\Leftrightarrow s^2+6+5s=0\)  \(\Leftrightarrow\left(s+2\right)\left(s+3\right)=0\)  \(\Leftrightarrow\left[{}\begin{matrix}s=-2\\s=-3\end{matrix}\right.\)

Với s = -2 thay vào (1) ; ta có : \(p=\dfrac{8}{5}\)  (L)

Với s = -3 thay vào (1) ; ta có : p = 2 ( t/m)

Khi đó : x là no của p/t : x^2 + 3x + 2 = 0 \(\Leftrightarrow\left(x+2\right)\left(x+1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-1\end{matrix}\right.\)

x = -2 => y = -1 ; x = -1 => y = -2

Vậy ... 

\(2\sqrt{9}-\sqrt{4}=2.3-2=4\)

\(\left\{{}\begin{matrix}x^3+y^3=1\\x^2+y^5=x^5+y^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)\left(x^2+y^2-xy\right)=1\left(1\right)\\x^2-y^2-\left(x^5-y^5\right)=0\left(2\right)\end{matrix}\right.\)

Đặt s = x + y ; p = xy \(\left(s^2\ge4p\right)\)

(1) \(\Leftrightarrow\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]=1\)  hay \(s.\left(s^2-3p\right)=1\)  

(2) \(\Leftrightarrow\left(x-y\right)\left[x+y-\left(x^4+x^3y+yx^3+y^4+x^2y^2\right)\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\\left(x+y\right)-\left[\left(x^2+y^2\right)^2-x^2y^2+xy\left(x^2+y^2\right)\right]=0\left(3\right)\end{matrix}\right.\)

Với x = y ; ta có : \(2x^3=1\Leftrightarrow x^3=\dfrac{1}{2}\)  \(\Leftrightarrow x=\sqrt[3]{\dfrac{1}{2}}=y\)  (t/m)

(3) \(\Leftrightarrow s-\left[\left(s^2-2p\right)^2-p^2+p\left(s^2-2p\right)\right]=0\)

\(\Leftrightarrow s-\left[s^4-3ps^2+p^2\right]=0\)  \(\Leftrightarrow s-\left[s^2\left(s^2-3p\right)+p^2\right]=0\)

Nếu s = 0 thì x = -y thay vào (1) ko t/m . Do đó : s \(\ne0\)

Ta có : \(s^2-3p=\dfrac{1}{s}\)  . Khi đó : \(-p^2=0\Leftrightarrow p=0\)  hay xy = 0 \(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

x = 0 thay vào (1) thì y = 1

y = 0 thay vào (1) thì x = 1

Vậy ... 

... =  \(\left[a^3\left(a+b\right)+b^2\left(a^2+ab+b^2\right)\right]\left(a-b\right)\) 

=  \(a^3\left(a^2-b^2\right)+b^2\left(a^3-b^3\right)=a^5-b^5\)

Đề câu a sai rồi 

\(\dfrac{\sqrt{12,1}}{22,5}=\dfrac{\sqrt{10}.\sqrt{12,1.10}}{225}=\dfrac{\sqrt{121}.\sqrt{10}}{225}=\dfrac{11\sqrt{10}}{225}\)

\(\dfrac{\sqrt{12^3}}{\sqrt{3^3}.2^2}=\dfrac{12\sqrt{12}}{3\sqrt{3}.4}=\sqrt{4}=2\)

Dạng tổng quát với n = 2;3;...;99

\(1-\dfrac{2}{n\left(n+1\right)}=\dfrac{n^2+n-2}{n\left(n+1\right)}=\dfrac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

Ta có : \(B=\dfrac{4.1}{2.3}.\dfrac{5.2}{3.4}.\dfrac{6.3}{4.5}...\dfrac{100.97}{98.99}.\dfrac{101.98}{99.100}\)

\(=\dfrac{\left(4.5.6...101\right).\left(1.2.3...98\right)}{\left(2.3.4...99\right)\left(3.4.5...100\right)}\) = \(\dfrac{100.101}{2.3}.\dfrac{2}{99.100}\)

= \(\dfrac{101}{3.99}=\dfrac{101}{297}\)

\(4x^2-\dfrac{4}{5}x+1=25\)  \(\Leftrightarrow20x^2-4x+5=125\)  \(\Leftrightarrow20x^2-4x-120=0\)

\(\Leftrightarrow5x^2-x-30=0\)   

\(\Delta=1-4.5.\left(-30\right)=601\) > 0 . P/t có 2 no : 

\(x_1=\dfrac{1+\sqrt{601}}{10};x_2=\dfrac{1-\sqrt{601}}{10}\)

\(\left(a-b\right)^2=a^2+b^2-2ab\) 

Vì a ; b \(\in N\) * \(\Rightarrow ab>0\) 

Do đó : \(a^2+b^2>\left(a-b\right)^2\)