`a)A=\sqrt{4+2sqrt3}`

`=\sqrt{3+2sqrt3+1}`

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

`=sqrt3+1`

`B)1/(2-sqrt3)+1/(2+sqrt3)`

`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`

`=2+sqrt3+2-sqrt3`

`=4`

`\sqrt{4x-12}+sqrtx{x-3}-1/3sqrt{9x-27}=8`

`đk:x>=3`

`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`

`<=>2sqrt{x-3}=8`

`<=>sqrt{x-3}=4`

`<=>x-3=16`

`<=>x=19`

Vậy `S={19}`

`a)A=\sqrt{4+2sqrt3}`

`=\sqrt{3+2sqrt3+1}`

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

`=sqrt3+1`

`B)1/(2-sqrt3)+1/(2+sqrt3)`

`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`

`=2+sqrt3+2-sqrt3`

`=4`

`\sqrt{4x-12}+sqrt{x-3}-1/3sqrt{9x-27}=8`

`đk:x>=3`

`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`

`<=>2sqrt{x-3}=8`

`<=>sqrt{x-3}=4`

`<=>x-3=16`

`<=>x=19`

Vậy `S={19}`

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

\(=x+\sqrt{x}+1+\dfrac{1}{\sqrt{x}}\ge2017+\sqrt{2017}\Leftrightarrow x+\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2016+\sqrt{2017}\Leftrightarrow x+\sqrt{x}+\dfrac{1}{\sqrt{x}}-2016-\sqrt{2017}\ge0\)

Bài toán sắp hoàn thành rồi đấy cậu giải tiếp nhé! =))

2

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

b

\(y=\dfrac{18}{4+\sqrt{7}}=\dfrac{18\left(4-\sqrt{7}\right)}{16-7}=\dfrac{72-18\sqrt{7}}{9}=\dfrac{72}{9}-\dfrac{18\sqrt{7}}{9}=8-2\sqrt{7}\\ =7-2\sqrt{7}.1+1=\left(\sqrt{7}-1\right)^2\)

Thế x = 2 và y = \(\left(\sqrt{7}-1\right)^2\) vào M được:

\(M=2\left(\sqrt{7}-1\right)^2-3.2.\sqrt{\left(\sqrt{7}-1\right)^2}+2^2\\ =2\left(8-2\sqrt{7}\right)-6.\left(\sqrt{7}-1\right)+4\\ =16-4\sqrt{7}-6\sqrt{7}+6+4\\ =26-10\sqrt{7}\)

1:

a: =>2x-2căn x+3căn x-3-5=2x-4

=>căn x-8=-4

=>căn x=4

=>x=16

b: \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)-3\sqrt{x}\left(\sqrt{x}-2\right)=0\)

=>(căn x-2)(x-căn x+4)=0

=>căn x-2=0

=>x=4

ĐKXĐ \(x\ge1\)

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

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

\(P=\dfrac{2x-2\sqrt{x}}{x-1}\)

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

\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)

Giải phương trình ???

x > 1 

.-.

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

ptr thiếu 1 vế rồi. hay là rút gọn nhỉ?

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

làm lại nhé :((( 

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

ĐKXĐ: \(\left\{{}\begin{matrix}-1\le x\le3\\x\ne1\end{matrix}\right.\)

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

\(\Leftrightarrow\dfrac{x+1+\sqrt{-x^2+2x+3}}{x-1}>2x-1\)

- TH1: Với \(x>1\) BPT tương đương:

\(x+1+\sqrt{-x^2+2x+3}>\left(2x-1\right)\left(x-1\right)\)

\(\Leftrightarrow\sqrt{-x^2+2x+3}>2x^2-4x\)

Đặt \(\sqrt{-x^2+2x+3}=t\ge0\Rightarrow2x^2-4x=-2t^2+6\)

BPt trở thành: \(t>-2t^2+6\Leftrightarrow2t^2+t-6>0\)

\(\Rightarrow t>\dfrac{3}{2}\Rightarrow-x^2+2x+3>\dfrac{9}{4}\Rightarrow1< x< \dfrac{2+\sqrt{7}}{2}\)

TH2: với \(x< 1\) BPT tương đương:

\(x+1+\sqrt{-x^2+2x+3}< \left(2x-1\right)\left(x-1\right)\)

\(\Leftrightarrow\sqrt{-x^2+2x+3}< 2x^2-4x\)

Tương tự như trên, đặt  \(t=\sqrt{-x^2+2x+3}\ge0\) ta được \(0\le t< \dfrac{3}{2}\)

\(\Rightarrow-x^2+2x+3< \dfrac{9}{4}\) \(\Rightarrow-1\le x< \dfrac{2-\sqrt{7}}{2}\)

Vậy nghiệm của BPT là: \(\left[{}\begin{matrix}-1\le x< \dfrac{2-\sqrt{7}}{2}\\1< x< \dfrac{2+\sqrt{7}}{2}\end{matrix}\right.\)

a) Ta có: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}}{x-1}\)

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

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