\(=\sqrt{\dfrac{2\left(3-\sqrt{5}\right)}{2\left(2-\sqrt{3}\right)}}=\sqrt{\dfrac{6-2\sqrt{5}}{4-2\sqrt{3}}}=\sqrt{\dfrac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{3}-1\right)^2}}=\dfrac{\sqrt{5}-1}{\sqrt{3}-1}\)

13) \(\sqrt{x^2}=\left|x\right|\)

14) \(\sqrt{x^4}=x^2\)

15) \(\sqrt{\left(3x-2\right)^2}=\left|3x-2\right|\) 

16) \(\sqrt{\left(2x+1\right)^2}=\left|2x+1\right|\)

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

18) \(\sqrt{4a^2-4a+1}=\left|2a-1\right|\)

19) \(\sqrt{x^2+6x+9}=\left|x+3\right|\)

20) \(\sqrt{a^2-a+\dfrac{1}{4}}=\left|a-\dfrac{1}{2}\right|\)

21) \(\sqrt{4x^2-4xy+y^2}=\left|2x-y\right|\)

`sqrt{8-4sqrt3}-sqrt{14+8sqrt3}`

`=sqrt{2(4-2sqrt3)}-sqrt{2(7+4sqrt3)}`

`=sqrt{2(3-2sqrt3+1)}-sqrt{2(4+2.2.sqrt3+3)}`

`=sqrt{2(sqrt3-1)^2}-sqrt{2(2+sqrt3)^2}`

`=sqrt2(sqrt3-1)-sqrt2(2+sqrt3)`

`=sqrt6-sqrt2-2sqrt2-sqrt6`

`=-3sqrt2`

Sửa đề: a = b => x = y

\(P=\left(1-\dfrac{\sqrt{2xy}}{\sqrt{x^2+y^2}}\right)\left(1+\dfrac{\sqrt{2xy}}{\sqrt{x^2+y^2}}\right)\) (ĐK: \(x,y>0\))

\(=1-\left(\dfrac{\sqrt{2xy}}{\sqrt{x^2+y^2}}\right)^2\)

\(=1-\dfrac{\left(\sqrt{2xy}\right)^2}{\left(\sqrt{x^2+y^2}\right)^2}\)

\(=1-\dfrac{2xy}{x^2+y^2}\)

\(=\dfrac{x^2+y^2-2xy}{x^2+y^2}\)

\(=\dfrac{\left(x-y\right)^2}{x^2+y^2}\)

Khi x = y, ta được: \(P=\dfrac{\left(x-y\right)^2}{x^2+y^2}=\dfrac{\left(x-x\right)^2}{x^2+y^2}=0\)

#Urushi

`sqrt{3-sqrt5}-sqrt{3+sqrt5}`

`=sqrt{(6-2sqrt5)/2}-sqrt{(6+2sqrt5)/2}`

`=sqrt{(sqrt5-1)^2/2}-sqrt{(sqrt5+1)^2/2}`

`=(sqrt5-1)/sqrt2-(sqrt5+1)/sqrt2`

`=(sqrt5-1-sqrt5-1)/sqrt2`

`=(-2)/sqrt2=-sqrt2`

Cảm ơn nha :^

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

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

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

Lời giải:

ĐKXĐ: $x>0; x\neq 1; x\neq 9$

\(A=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{(\sqrt{x}+1)(\sqrt{x}-1)+(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}-1)}\)

\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{x-1-(x-9)}{(\sqrt{x}-3)(\sqrt{x}-1)}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{8}{(\sqrt{x}-1)(\sqrt{x}-3)}\)

\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-1)(\sqrt{x}-3)}{8}=\frac{\sqrt{x}-3}{8\sqrt{x}}\)

Để $A<0\Leftrightarrow \frac{\sqrt{x}-3}{8\sqrt{x}}<0$

$\Leftrightarrow \sqrt{x}-3<0$ (do $8\sqrt{x}>0$)

$\Leftrightarrow \sqrt{x}<3$

$\Leftrightarrow 0\leq x< 9$

Kết hợp với đkxđ suy ra $0< x< 9; x\neq 1$

Khi $x=3-2\sqrt{2}=(\sqrt{2}-1)^2$

$\Rightarrow \sqrt{x}=\sqrt{2}-1$

Khi đó: $A=\frac{\sqrt{x}-3}{8\sqrt{x}}=\frac{\sqrt{2}-4}{8(\sqrt{2}-1)}=\frac{-2-3\sqrt{2}}{8}$

sao chỗ x−1−(x−9) lại là trừ ạ đáng lẽ nó phải là (x−1)+(x−9) chứ ạ

\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\right)\) (ĐK: \(x>0;x\ne1;x\ne9\))

\(=\left[\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\right]\)

\(=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1+x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{2x-10}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\)

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

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

\(=\dfrac{\sqrt{x}-3}{2x\sqrt{x}-10\sqrt{x}}\)

\(A>0\) khi

\(\dfrac{\sqrt{x}-3}{2x\sqrt{x}-10\sqrt{x}}>0\)

TH1: 

\(\sqrt{x}-3>0\) và \(2x\sqrt{x}-10\sqrt{x}>0\)

\(\Leftrightarrow\sqrt{x}>3\) và \(2\sqrt{x}\left(x-5\right)>0\)

\(\Leftrightarrow x>9\) và \(x>5\)

\(\Leftrightarrow x>9\)

TH2: 

\(\sqrt{x}-3< 0\) và \(2x\sqrt{x}-10\sqrt{x}< 0\)

\(\Leftrightarrow\sqrt{x}< 3\) và \(2\sqrt{x}\left(x-5\right)< 0\)

\(\Leftrightarrow x< 9\) và \(x< 5\)

\(\Leftrightarrow x< 5\)

Vậy A > 0 khi \(\left[{}\begin{matrix}x>9\\x< 5\end{matrix}\right.\) 

Ta có: 

\(x=3-2\sqrt{2}=\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2=\left(\sqrt{2}-1\right)^2\)

\(A=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}-3}{2\cdot\left(\sqrt{2}-1\right)^2\cdot\sqrt{\left(\sqrt{2}-1\right)^2}-10\cdot\sqrt{\left(\sqrt{2}-1\right)^2}}\)

\(A=\dfrac{\left|\sqrt{2}-1\right|-3}{2\cdot\left(3-2\sqrt{2}\right)\cdot\left|\sqrt{2}-1\right|-10\cdot\left|\sqrt{2}-1\right|}\)

\(A=\dfrac{\sqrt{2}-1-3}{\left(6-4\sqrt{2}\right)\left(\sqrt{2}-1\right)-10\left(\sqrt{2}-1\right)}\)

\(A=\dfrac{\sqrt{2}-4}{6\sqrt{2}-6-8+4\sqrt{2}-10\sqrt{2}+10}\)

\(A=\dfrac{\sqrt{2}-4}{-4}\)

\(A=\dfrac{4-\sqrt{2}}{4}\)

\(3\sqrt{144}-5\sqrt{49}+\dfrac{1}{2}\sqrt{36}\)

\(=3.12-5.7+\dfrac{1}{2}.6\)

\(=36-35+3=4\)