ĐKXĐ: \(\sqrt{6}\cdot4-4x>=0\)

=>x<=0

`a)->` ĐKXĐ : `x>=0;x\ne1`

`b)` Ta có :

`P=(\sqrtx)/(\sqrtx-1)-(2\sqrtx)/(\sqrtx+1)+(x-3)/(x-1)`

`P=(\sqrtx(\sqrtx+1)-2\sqrtx(\sqrtx-1)+x-3)/(x-1)`

`P=(x+\sqrtx-2x+2\sqrtx+x-3)/(x-1)`

`P=(3\sqrtx-3)/(x-1)`

`P=(3(\sqrtx-1))/((\sqrtx-1)(\sqrtx+1))`

`P=3/(\sqrtx+1)`

Vậy `P=3/(\sqrtx+1)` khi `x>=0;x\ne1`

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

\(=\dfrac{3}{\sqrt{x}+1}\)

Bổ sung \(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne1\)

\(\sqrt{\sqrt{9}-\dfrac{4}{x-3}}=\sqrt{3-\dfrac{4}{x-3}}\)

\(=\sqrt{\dfrac{3x-9-4}{x-3}}=\sqrt{\dfrac{3x-13}{x-3}}\)

để biểu thức có nghĩa thì (3x-13)/(x-3)>=0

=>x>=13/3 hoặc x<3

\(a,P=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}\right)\cdot\left(\dfrac{\sqrt{x}+2}{2}\right)^2\left(x\ge0;x\ne4\right)\\ P=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{4}\\ P=\dfrac{4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{4}=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)

\(b,\)Ta có \(x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)

Thay vào \(P\), ta được:

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

\(c,\)Để \(P< 1\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-2}< 1\)

\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-1< 0\\ \Leftrightarrow\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\sqrt{x}-2}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-2}< 0\\ \Leftrightarrow\sqrt{x}-2< 0\left(4>0\right)\\ \Leftrightarrow\sqrt{x}< 2\\ \Leftrightarrow x< 4\)

Vậy để \(P< 1\) thì \(x< 4\)

Tick nha

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

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

\(=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{4}\)

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

b: Thay \(x=6-2\sqrt{5}\) vào P, ta được:

\(P=\dfrac{\sqrt{5}+1+2}{\sqrt{5}+1-2}=\dfrac{3+\sqrt{5}}{\sqrt{5}+1}=\dfrac{1+\sqrt{5}}{2}\)

a)x≥-2

b)x≥1/6

c)x≥0,x≥3/2

\(\sqrt[]{\dfrac{1+3x}{5}}\) xác định \(\Leftrightarrow\dfrac{1+3x}{5}\ge0\)

\(\Leftrightarrow1+3x\ge0\)

\(\Leftrightarrow3x\ge-1\)

\(\Leftrightarrow x\ge-\dfrac{1}{3}\)

\(\Rightarrow D=[-\dfrac{1}{3};+\infty)\)

giúp mình vớii