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. 

PT \(\Leftrightarrow \left\{\begin{matrix} 2x-2\geq 0\\ x^2-2x+4=(2x-2)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 3x^2-6x=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 3x(x-2)=0\end{matrix}\right.\Leftrightarrow x=2\)

b. ĐK: $x\geq 1$
PT $\Leftrightarrow \sqrt{(x-1)+2\sqrt{x-1}+1}=2$

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

$\Leftrightarrow |\sqrt{x-1}+1|=2$

$\Leftrightarrow \sqrt{x-1}+1=2$
$\Leftrightarrow \sqrt{x-1}=1$

$\Leftrightarrow x=2$ (tm)

c. 

PT \(\Leftrightarrow \left\{\begin{matrix} 2x-1\geq 0\\ 2x^2-2x+1=(2x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x+1=4x^2-4x+1\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x=2x(x-1)=0\end{matrix}\right.\Leftrightarrow x=1\) (tm)

d.

ĐKXĐ: $x\geq 4$

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

$\Leftrightarrow \sqrt{(\sqrt{x-4}+2)^2}=2$
$\Leftrightarrow |\sqrt{x-4}+2|=2$

$\Leftrightarrow \sqrt{x-4}+2=2$

$\Leftrightarrow \sqrt{x-4}=0$

$\Leftrightarrow x=4$ (tm)

a: Ta có: \(\sqrt{x^2-2x+4}=2x-2\)

\(\Leftrightarrow x^2-2x+4=4x^2-8x+4\)

\(\Leftrightarrow-3x^2+6x=0\)

\(\Leftrightarrow-3x\left(x-2\right)=0\)

\(\Leftrightarrow x=2\)

b: Ta có: \(\sqrt{x+2\sqrt{x-1}}=2\)

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

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

\(\Leftrightarrow x-1=1\)

hay x=2

c: Ta có: \(\sqrt{2x^2-2x+1}=2x-1\)

\(\Leftrightarrow2x^2-2x+1=4x^2-4x+1\)

\(\Leftrightarrow-2x^2+2x=0\)

\(\Leftrightarrow-2x\left(x-1\right)=0\)

hay x=1

a) \(\sqrt{\left(x-3\right)^2}=2\Rightarrow\left|x-3\right|=2\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Rightarrow\sqrt{9\left(x+2\right)}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25\left(x+2\right)}=6\)

\(\Rightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Rightarrow2\sqrt{x+2}=6\Rightarrow\sqrt{x+2}=3\Rightarrow x+2=9\Rightarrow x=7\)

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

Ta có: \(x-2\sqrt{x}+3=x-2\sqrt{x}+1+2=\left(\sqrt{x}-1\right)^2+2\ge2\)

\(\Rightarrow\dfrac{1}{x-2\sqrt{x}+3}\le2\Rightarrow Q_{max}=2\) khi \(x=1\)

a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

Vậy...

b)Đk: \(x\ge-1\)

Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)

\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)

Vậy...

\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)

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

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

b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\) 

Vậy \(A_{min}=-\dfrac{1}{4}\)

a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)

\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)

\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)

a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)

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

\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)

b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)

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

\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)

✱ giải pt:

a.\(\sqrt{x^2-4x+4}\)\(=5\)

⇔\(\sqrt{\left(x-2\right)^2}=5\)

⇒\(\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

vậy....

b.\(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)

⇔ \(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

⇔ \(4\sqrt{x+1}=16\)

⇔ \(\sqrt{x+1}=16\)

⇒ \(x+1=256\)

⇔ \(x=255\)

vậy.....

a) Ta có: \(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)

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

\(\Leftrightarrow\sqrt{x-1}=\sqrt{x-1}+1+1\)(Vô lý)

Vậy: \(S=\varnothing\)

b) Ta có: \(\sqrt{x^4+2x^2+1}=\sqrt{x^2+10x+25}-10x+22\)

\(\Leftrightarrow x^2+1=\left|x+5\right|-10x+22\)

\(\Leftrightarrow\left|x+5\right|=x^2+1+10x-22=x^2+10x-21\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2+10x-21\left(x\ge-5\right)\\-x-5=x^2+10x-21\left(x< -5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+10x-21-x-5=0\\x^2+10x-21+x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+9x-26=0\\x^2+11x-16=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9+\sqrt{185}}{2}\\x=\dfrac{-11-\sqrt{185}}{2}\end{matrix}\right.\)

a) \(\sqrt{x^2-x-4}=\sqrt{x-1}\)

\(x^2-x-4=x-1\)

\(x^2-x-4-x+1=0\)

\(x^2-2x-5=0\)

\(\left(x^2-2.x.1+1^2\right)-6=0\)

\(\left(x-1\right)^2=6\)

⇒\(\left\{{}\begin{matrix}x-1=6\\x-1=-6\end{matrix}\right.\)         ⇒\(\left\{{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)

a: Ta có: \(\sqrt{4-3x}=8\)

\(\Leftrightarrow4-3x=64\)

\(\Leftrightarrow3x=-60\)

hay x=-20

b: ta có: \(\sqrt{4x-8}-12\sqrt{\dfrac{x-2}{9}}=-1\)

\(\Leftrightarrow2\sqrt{x-2}-12\cdot\dfrac{\sqrt{x-2}}{3}=-1\)

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

hay \(x=\dfrac{9}{4}\)

\(\left\{{}\begin{matrix}8>0\left(luondung\right)\\4-3x=64\end{matrix}\right.\) \(\Leftrightarrow x=-20\left(ktm\right)\)

a) Ta có: \(\sqrt{\left(x+1\right)^2}=3\)

\(\Leftrightarrow\left|x+1\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

b) Ta có: \(3\sqrt{4x+4}-\sqrt{9x-9}-8\sqrt{\dfrac{x+1}{16}}=5\)

\(\Leftrightarrow6\sqrt{x+1}-3\sqrt{x-3}-2\sqrt{x+1}=5\)

\(\Leftrightarrow4\sqrt{x+1}=5+3\sqrt{x-3}\)

\(\Leftrightarrow16\left(x+1\right)=25+30\sqrt{x-3}+9\left(x-3\right)\)

\(\Leftrightarrow16x+16=25+9x-27+30\sqrt{x-3}\)

\(\Leftrightarrow30\sqrt{x-3}=16x+16+2-9x\)

\(\Leftrightarrow30\sqrt{x-3}=7x+18\)

\(\Leftrightarrow x-3=\left(\dfrac{7x+18}{30}\right)^2\)

\(\Leftrightarrow x-3=\dfrac{49x^2}{900}+\dfrac{7}{25}x+\dfrac{9}{25}\)

\(\Leftrightarrow\dfrac{49}{900}x^2-\dfrac{18}{25}x+\dfrac{84}{25}=0\)

\(\Delta=\left(-\dfrac{18}{25}\right)^2-4\cdot\dfrac{49}{900}\cdot\dfrac{84}{25}=-\dfrac{16}{75}< 0\)

Vậy: Phương trình vô nghiệm

a)Pt\(\Leftrightarrow\left|x+1\right|=3\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

b)Đk:\(x\ge-1\)

Sửa đề: \(3\sqrt{4x+4}-\sqrt{9x+9}-8\sqrt{\dfrac{x+1}{16}}=5\)

Pt \(\Leftrightarrow6\sqrt{x+1}-3\sqrt{x+1}-2\sqrt{x+1}=5\)

\(\Leftrightarrow\sqrt{x+1}=5\)

\(\Leftrightarrow x=24\left(tm\right)\)

a.  \(\sqrt{\left(x+1\right)^2}\)  \(=3\)

⇔   \(\left|x+1\right|=3\)

⇔    \(\left|x\right|=2\)

⇒     \(x=2\) và  \(x=-2\)