\(B=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}\cdot\dfrac{\sqrt{x}-2+3}{3}\)

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

a: \(x=4+\sqrt{3}+4-\sqrt{3}=8\)

Khi x=8 thì \(A=\dfrac{2-5\cdot2\sqrt{2}}{2\sqrt{2}+1}=\dfrac{2-10\sqrt{2}}{2\sqrt{2}+1}=-6+2\sqrt{2}\)

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 } 

) √ ( 2 x − 1 ) 2 = 3 ⇒ | 2 x − 1 | = 3 ⇔ 2 x − 1 = ± 3 +) TH1: 2 x − 1 = 3 ⇒ 2 x = 4 ⇒ x = 2 +) TH2: 2 x − 1 = − 3 ⇒ 2 x = − 2 ⇒ x = − 1 Vậy x = − 1 ; x = 2 . b) Điều kiện: x ≥ 0 5 3 √ 15 x − √ 15 x − 2 = 1 3 √ 15 x ⇔ 5 3 √ 15 x − √ 15 x − 1 3 √ 15 x = 2 ⇔ ( 5 3 − 1 − 1 3 ) √ 15 x = 2 ⇔ 1 3 √ 15 x = 2 ⇔ √ 15 x = 6 ⇔ 15 x = 36 ⇔ x = 12 5 Vậy x = 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​ .

....

ĐKXĐ: x>0

\(\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

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

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

a: \(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)

\(\Leftrightarrow\sqrt{x-2}=4\)

=>x-2=16

hay x=18

b: \(\Leftrightarrow\left|3x+2\right|=4x\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)

c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)

\(\Leftrightarrow4\sqrt{x-2}=40\)

=>x-2=100

hay x=102

d: =>5x-6=9

hay x=3

\(a,\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\left(dk:x\ge2\right)\)

\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)

\(\Leftrightarrow\sqrt{x-2}=4\)

\(\Leftrightarrow x-2=16\)

\(\Leftrightarrow x=18\left(tmdk\right)\)

b,\(\sqrt{9x^2-12x+4=3x\left(dk:x\ge0\right)}\)

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

\(\Leftrightarrow\left|3x-2\right|=3x\)

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

\(\Rightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=\dfrac{1}{3}\left(tmdk\right)\end{matrix}\right.\)

Các câu còn lại làm tương tự nhé 

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: x≥2)

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)

\(\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)

\(-\sqrt{x-2}=-4\)

\(\sqrt{x-2}=4\)

\(\left|x-2\right|=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)

a: \(=\left|x-4\right|-\left|x-2\right|\)

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

\(=5-3\sqrt{2}-\left(3\sqrt{2}-3\right)=-6\sqrt{2}+8\)

b: \(=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)

\(=\left|\sqrt{7}-1+1\right|+\left|\sqrt{7}-1-1\right|\)

\(=\sqrt{7}+4-\sqrt{7}=4\)

\(\Leftrightarrow\left(3-x\right)\sqrt{x-1}+\sqrt{5-2x}=\sqrt{\left[\left(x-3\right)^2+1\right]\left(4-x\right)}\)

đặt 3-x=a;\(\sqrt{x-1}=b;\sqrt{5-2x}=c\Rightarrow b^2+c^2=4-x\)

\(\Leftrightarrow ab+c=\sqrt{\left(a^2+1\right)\left(b^2+c^2\right)}\)

<=>a²b²+2abc+c²=a²b²+b²+a²c²+c²

<=>b²-2abc+a²c²=0

<=>(b-ac)²=0

<=>b=ac

đến đây thì dễ rồi