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

\(B=\dfrac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\dfrac{1}{x+\sqrt{5}}\)

a) -17√3/3                                                  b) 11√6 

c) 21                                                            d) 11

a)  326+236−426 và làm tiếp.
136x+6x):6x và làm tiếp

a) \(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\)

\(=\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-\frac{4}{2}\sqrt{6}\)

\(=\left(\frac{3}{2}+\frac{2}{3}-\frac{4}{2}\right)\sqrt{6}\)

\(=\frac{1}{6}\cdot\sqrt{6}=\frac{\sqrt{6}}{6}\left(đpcm\right)\)

b) \(\left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}\)

\(=\left(\sqrt{6x}+\frac{1}{3}\sqrt{6x}+\sqrt{6x}\right):\sqrt{6x}\)

\(=\left[\left(1+\frac{1}{3}+1\right)\sqrt{6x}\right]:\sqrt{6x}\)

\(=\frac{7}{3}\sqrt{6x}:\sqrt{6x}=\frac{7}{3}=2\frac{1}{3}\left(đpcm\right)\)

Lời giải: 
ĐKXĐ: $x\geq 0; x\neq 1$
a.

\(A=\left[\frac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\right].\frac{2}{\sqrt{x}-1}\)

\(=\frac{x+2+x-\sqrt{x}-(x+\sqrt{x}+1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\frac{2}{\sqrt{x}-1}\)

\(=\frac{2(x-2\sqrt{x}+1)}{(\sqrt{x}-1)^2(x+\sqrt{x}+1)}=\frac{2(\sqrt{x}-1)^2}{(\sqrt{x}-1)^2(x+\sqrt{x}+1)}=\frac{2}{x+\sqrt{x}+1}\)

b.

Ta thấy với $x\geq 0 ; x\neq 1$ thì $x+\sqrt{x}+1\geq 1$

$\Rightarrow A=\frac{2}{x+\sqrt{x}+1}\leq 2$

Vậy $A$ đạt max bằng $2$ khi $x=0$

Bo thi:>

+ đk x > 0 , x khác 1

ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

\(A=\left(\dfrac{3\sqrt{x}}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{x+6\sqrt{x}+9}\)

\(=\left(\dfrac{3}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{\left(\sqrt{x}+3\right)^2}\)

\(=\dfrac{3\sqrt{x}+9-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}\)

\(=\dfrac{\sqrt{x}+13}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}\)

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

Vậy...

Bài 2 

b, `\sqrt{3x^2}=x+2`          ĐKXĐ : `x>=0`

`=>(\sqrt{3x^2})^2=(x+2)^2`

`=>3x^2=x^2+4x+4`

`=>3x^2-x^2-4x-4=0`

`=>2x^2-4x-4=0`

`=>x^2-2x-2=0`

`=>(x^2-2x+1)-3=0`

`=>(x-1)^2=3`

`=>(x-1)^2=(\pm \sqrt{3})^2`

`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$

Vậy `S={1+\sqrt{3};1-\sqrt{3}}`

Bài 1 

a, `3x-7\sqrt{x}+4=0`            ĐKXĐ : `x>=0`

`<=>3x-3\sqrt{x}-4\sqrt{x}+4=0`

`<=>3\sqrt{x}(\sqrt{x}-1)-4(\sqrt{x}-1)=0`

`<=>(3\sqrt{x}-4)(\sqrt{x}-1)=0`

TH1 :

`3\sqrt{x}-4=0`

`<=>\sqrt{x}=4/3`

`<=>x=16/9` ( tm )

TH2

`\sqrt{x}-1=0`

`<=>\sqrt{x}=1` (tm)

Vậy `S={16/9;1}`

b, `1/2\sqrt{x-1}-9/2\sqrt{x-1}+3\sqrt{x-1}=-17`     ĐKXĐ : `x>=1`

`<=>(1/2-9/2+3)\sqrt{x-1}=-17`

`<=>-\sqrt{x-1}=-17`

`<=>\sqrt{x-1}=17`

`<=>x-1=289`

`<=>x=290` ( tm )

Vậy `S={290}`

Bài 1: 

a) Ta có: \(3x-7\sqrt{x}+4=0\)

\(\Leftrightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)

b) Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}\cdot\left(-1\right)=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b) Ta có: \(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)

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

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

Thay x=3 vào B, ta được:

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

a, \(\sqrt{\left(2x-1\right)^2}=3\Leftrightarrow\left|2x-1\right|=3\)

Với \(x\ge\frac{1}{2}\)pt có dạng : \(2x-1=3\Leftrightarrow x=2\)( tm )

Với \(x< \frac{1}{2}\)pt có dạng : \(-2x+1=3\Leftrightarrow x=-1\)( tm ) 

Vậy tập nghiệm của pt là S = { -1 ; 2 } 

b, \(\frac{5}{3}\sqrt{15x}-\sqrt{15x}-2=\frac{1}{3}\sqrt{15x}\)ĐK : \(x\ge0\)

\(\Leftrightarrow\frac{2}{3}\sqrt{15x}-2=\frac{1}{3}\sqrt{15x}\Leftrightarrow\frac{1}{3}\sqrt{15x}=2\)

\(\Leftrightarrow\sqrt{15x}=6\)bình phương 2 vế : \(\Leftrightarrow15x=36\Leftrightarrow x=\frac{36}{15}=\frac{12}{5}\)( tm ) 

Vậy tập nghiệm của pt là S = { 12/5 } 

a) \sqrt{(2x-1)^{2}}=3(2x−1)2​=3

$\Rightarrow$ $|2x-1|=3$

$\iff$ $2x-1=\pm 3$

+) TH1: $2x-1=3$

$\Rightarrow$ $2x=4$

$\Rightarrow$ $x=2$
+) TH2: $2x-1=-3$

$\Rightarrow$ $2x=-2$
$\Rightarrow$ $x=-1$
Vậy  $x=-1$ ; $x=2$ .
b) Điều kiện: $x\ge 0$

\dfrac{5}{3} \sqrt{15 x}-\sqrt{15x}-2=\dfrac{1}{3} \sqrt{15 x}35​15x​−15x​−2=31​15x​

$\iff$ \dfrac{5}{3} \sqrt{15 x}-\sqrt{15 x}-\dfrac{1}{3} \sqrt{15 x}=235​15x​−15x​−31​15x​=2

$\iff$ \left(\dfrac{5}{3}-1-\dfrac{1}{3}\right) \sqrt{15} x=2(35​−1−31​)15​x=2

$\iff$ \dfrac{1}{3} \sqrt{15 x}=231​15x​=2

$\iff$ \sqrt{15 x}=615x​=6

$\iff$ $15x=36$

$\iff$ x=\dfrac{12}{5}x=512​

Vậy x=\dfrac{12}{5}x=512​ .

Bài 1:
1) A = \(10\sqrt{\dfrac{1}{5}}\) - \(3\sqrt{\left(2-\sqrt{5}\right)^2}\) + \(\sqrt{5}\)

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

       = \(3\sqrt{5}\) - \(3\sqrt{5}\) + 6 + \(\sqrt{5}\)

        = 6

     B = \(\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\) \(\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\) (ĐKXĐ: x > 0; x ≠ 4)

        = \(\dfrac{4x-8\sqrt{x}-8x}{x-4}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{x-2\sqrt{x}}\)

        = \(\dfrac{-4x-8\sqrt{x}}{x-4}\cdot\dfrac{x-2\sqrt{x}}{3-\sqrt{x}}\)

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

        = \(-\dfrac{4x}{3-\sqrt{x}}\)

        = \(\dfrac{4x}{\sqrt{x}-3}\)

Bài 2: \(\sqrt{4x^2-4x+1}=\sqrt[3]{27}\)

     ⇔ \(\sqrt{\left(2x-1\right)^2}=3\)

     ⇔ \(\left|2x-1\right|=3\)

     ⇔ \(\left\{{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)

     ⇔ \(\left\{{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)

     ⇔ \(\left\{{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

Vậy S = {2; -1}

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