\(=\dfrac{\sqrt{a}+2+\sqrt{a}-2}{a-4}:\dfrac{\sqrt{a}+2-2}{\sqrt{a}+2}\)

\(=\dfrac{2\sqrt{a}}{a-4}\cdot\dfrac{\sqrt{a}+2}{\sqrt{a}}=\dfrac{2}{\sqrt{a}-2}\)

1: ĐKXĐ: x+3>=0

=>x>=-3

\(\sqrt{x+3}>2\)

=>x+3>4

=>x>4-3=1

2: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)

\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}< 1\)

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-1< 0\)

=>\(\dfrac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)

=>\(\dfrac{3}{\sqrt{x}-2}< 0\)

=>\(\sqrt{x}-2< 0\)

=>\(\sqrt{x}< 2\)

=>0<=x<4

3: ĐKXĐ: x>=0

\(\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)-5=\sqrt{x}\left(\sqrt{x}+2\right)-5\)

=>\(x-4\sqrt{x}+3-5=x+2\sqrt{x}-5\)

=>\(x-4\sqrt{x}-2-x-2\sqrt{x}+5=0\)

=>\(-6\sqrt{x}+3=0\)

=>\(-6\sqrt{x}=-3\)

=>\(\sqrt{x}=\dfrac{1}{2}\)

=>x=1/4(nhận)

b, \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)

\(=\dfrac{2\left(\sqrt{5}-2\right)}{5-4}-\dfrac{2\left(\sqrt{5}+2\right)}{5-4}\)

\(=2\sqrt{5}-4-2\sqrt{5}-4=-8\)

a, \(\sqrt{2}\left(\sqrt{8}+\sqrt{32}-\sqrt{98}\right)\)

\(=\sqrt{2}\left(2\sqrt{2}+4\sqrt{2}-7\sqrt{2}\right)\)

\(=\sqrt{2}.\left(-\sqrt{2}\right)=-2\)

c, \(\left(2+\sqrt{3}\right)\sqrt{11-6\sqrt{2}}\)

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

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

\(=6-2\sqrt{2}+3\sqrt{3}-\sqrt{6}\)

Tiếc quá 

mình chưa học đến

bik thì giúp cho

ĐK \(x\ge-\frac{1}{2}\)

Đặt như trên... (\(a\ge\sqrt{\frac{1}{2}};b\ge0\)) ta có hệ:

\(\hept{\begin{cases}2a^2b=a+b^3\\2a^2-b^2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(b^2+1\right)b=a+b^3\\2a^2=b^2+1\end{cases}}\)

Xét pt trình đầu của hệ \(\Leftrightarrow a=b\). Thay b bởi a ở pt dưới ta được:

\(2a^2-a^2-1=0\Leftrightarrow\orbr{\begin{cases}a=1\left(TM\right)\\a=-\frac{1}{2}\left(KTM\right)\end{cases}}\). Với a = 1 thì ta có:

\(\sqrt{1+x}=1\Leftrightarrow x=0\) (TM)

Vậy...

\(u_1=\dfrac{1}{\sqrt{2}};q=\dfrac{1}{\sqrt{2}}\)

\(S_{99}=\dfrac{\dfrac{1}{\sqrt{2}}\cdot\left(\dfrac{1}{\sqrt{2}}^{99}-1\right)}{\dfrac{1}{\sqrt{2}}-1}=\dfrac{1}{\sqrt{2}}\cdot\left(\dfrac{1-2^{49}\cdot\sqrt{2}}{2^{49}\cdot\sqrt{2}}\right):\dfrac{1-\sqrt{2}}{\sqrt{2}}\)

\(=\dfrac{1}{1-\sqrt{2}}\cdot\dfrac{1-2^{49}\cdot\sqrt{2}}{2^{49}\cdot\sqrt{2}}\)

Bài 20:

a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)

b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)

\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)

c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)

\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)

\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)

=2

d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)

\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)

\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)

\(=8+4\sqrt{3}-4\sqrt{3}-6\)

=2

a) \(\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\)

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

\(=\left(9-5\right).\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\)

\(=4.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\)

\(=4.\sqrt{5-2\sqrt{5}+1}.\left(\sqrt{5}+1\right)\)

\(=4.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\)

\(=4.\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=4.\left(5-1\right)=16\)

b) \(2\sqrt{4+\sqrt{6-2\sqrt{5}}}.\left(\sqrt{10}-\sqrt{2}\right)\)

\(=2\sqrt{4+\sqrt{5-2\sqrt{5}+1}}.\left(\sqrt{10}-\sqrt{2}\right)\)

\(=2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}.\left(\sqrt{10}-\sqrt{2}\right)\)

\(=2\sqrt{3+\sqrt{5}}.\sqrt{2}.\left(\sqrt{5}-\sqrt{1}\right)\)

\(=2\sqrt{6+2\sqrt{5}}.\left(\sqrt{5}-1\right)\)

\(=2\sqrt{5+2\sqrt{5}+1}.\left(\sqrt{5}-1\right)\)

\(=2\sqrt{\left(\sqrt{5}+1\right)^2}.\left(\sqrt{5}-1\right)=2.\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\)

\(=2.\left(5-1\right)=2.4=8\)