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

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

b) \(\Leftrightarrow\sqrt{\left(x-3\right)^2}=7\)

\(\Leftrightarrow\left|x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=7\\x-3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\)

c) \(\Leftrightarrow3\left|x-2\right|=45\)

\(\Leftrightarrow\left|x-2\right|=15\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=15\\x-2=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-13\end{matrix}\right.\)

\(a,PT\Leftrightarrow\sqrt{3}\left(x-1\right)=1-x\\ \Leftrightarrow\sqrt{3}\left(x-1\right)+\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(\sqrt{3}+1\right)=0\\ \Leftrightarrow x=1\left(\sqrt{3}+1\ne0\right)\\ b,ĐK:x\in R\\ PT\Leftrightarrow\left|x-3\right|=7\Leftrightarrow\left[{}\begin{matrix}x-3=7\\3-x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\\ c,ĐK:x\in R\\ PT\Leftrightarrow3\left|x-2\right|=45\Leftrightarrow\left|x-2\right|=15\\ \Leftrightarrow\left[{}\begin{matrix}x-2=15\\2-x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-13\end{matrix}\right.\)

1.

ĐKXĐ: \(x\ge3\)

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

Pt trở thành:

\(t^2+3-7t-9=0\)

\(\Leftrightarrow t^2-7t-6=0\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{7-\sqrt{73}}{2}< 0\left(loại\right)\\t=\dfrac{7+\sqrt{73}}{2}\end{matrix}\right.\)

\(\Rightarrow\sqrt{x-3}=\dfrac{7+\sqrt{73}}{2}\)

\(\Rightarrow x=\dfrac{67+7\sqrt{73}}{2}\)

Nghiệm xấu quá, em nói giáo viên ra đề kiểm tra lại đề là \(x-7\sqrt{x-3}-9=0\) hay \(x-7\sqrt{x-3}+9=0\) nhé

2.

ĐKXĐ: \(x\ge2\)

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

\(\Leftrightarrow2x+1+2\sqrt{\left(x+3\right)\left(x-2\right)}=25\)

\(\Leftrightarrow\sqrt{x^2+x-6}=12-x\) (\(x\le12\))

\(\Rightarrow x^2+x-6=\left(12-x\right)^2\)

\(\Leftrightarrow x^2+x-6=144-24x+x^2\)

\(\Rightarrow x=6\)

Cách 2:

\(\Leftrightarrow\sqrt{x+3}-3+\sqrt{x-2}-2=0\)

\(\Leftrightarrow\dfrac{x-6}{\sqrt{x+3}+3}+\dfrac{x-6}{\sqrt{x-2}+2}=0\)

\(\Leftrightarrow\left(x-6\right)\left(\dfrac{1}{\sqrt{x+3}+3}+\dfrac{1}{\sqrt{x-2}+2}\right)=0\)

\(\Leftrightarrow x=6\)

3.

ĐKXĐ: \(x\ge8+8\sqrt{2}\)

Đặt \(\sqrt{x+4}=t>0\) \(\Rightarrow x=t^2-4\)

Pt trở thành:

\(\sqrt{t^2-4-4t}=3\)

\(\Leftrightarrow t^2-4t-4=9\)

\(\Leftrightarrow t^2-4t-13=0\)

\(\Rightarrow\left[{}\begin{matrix}t=2+\sqrt{17}\\t=2-\sqrt{17}< 0\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x+4}=2+\sqrt{17}\)

\(\Leftrightarrow x=17+4\sqrt{17}\)

Như câu 1, em nhờ giáo viên ra đề kiểm tra lại là \(\sqrt{x-4\sqrt{x+4}}=3\) hay \(\sqrt{x-4\sqrt{x-4}}=3\)

2. ĐKXĐ: $x\geq 2$

PT \(\Rightarrow x+3=(5-\sqrt{x-2})^2\)

\(\Leftrightarrow x+3=25+x-2-10\sqrt{x-2}\)

\(\Leftrightarrow 20=10\sqrt{x-2}\Leftrightarrow x-2=4\Leftrightarrow x=6\)

Thử lại thấy thỏa mãn

Vậy $x=6$

3. ĐKXĐ: $x\geq -4$

PT $\Leftrightarrow \sqrt{(x+4)-4\sqrt{x+4}+4}=3$

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

$\Leftrightarrow |\sqrt{x+4}-2|=3$

$\Leftrightarrow \sqrt{x+4}-2=\pm 3$. TH $\sqrt{x+4}-2=-3$ loại vì $\sqrt{x+4}-2\geq -2> -3$

Do đó: $\sqrt{x+4}-2=3$

$\Leftrightarrow \sqrt{x+4}=5$

$\Leftrightarrow x+4=25$

$\Leftrightarrow x=21$ (thỏa mãn)

Vậy $x=21$

** Lần sau bạn chú ý ghi đầy đủ yêu cầu của đề.

Lời giải:

1. ĐKXĐ: $x\geq 3$

PT $\Leftrightarrow x-9=7\sqrt{x-3}$

\(\Leftrightarrow \left\{\begin{matrix} x\geq 9\\ (x-9)^2=49(x-3)\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 9\\ x^2-67x+228=0\end{matrix}\right.\Rightarrow x=\frac{67+7\sqrt{73}}{2}\)

a, ĐKXĐ : \(x\ge1\)

Ta có ; \(PT\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}.\sqrt{9}\sqrt{x-1}+24.\sqrt{\dfrac{1}{64}}\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}\left(\dfrac{1}{2}-\dfrac{3}{2}\sqrt{9}+24\sqrt{\dfrac{1}{64}}\right)=-17\)

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

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

\(\Leftrightarrow x=290\left(TM\right)\)

Vậy ....

b, ĐKXĐ : \(x\ge3\)

Ta có : \(PT\Leftrightarrow x-3-7\sqrt{x-3}+12=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=16\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=19\\x=12\end{matrix}\right.\) ( TM )

Vậy ..

a) Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

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

\(\Leftrightarrow x-1=17^2=289\)

hay x=290

Vậy: S={290}

b) Ta có: \(x-7\sqrt{x-3}+9=0\)

\(\Leftrightarrow x-7\sqrt{x-3}=-9\)

\(\Leftrightarrow x-3-2\cdot\sqrt{x-3}\cdot\dfrac{7}{2}+\dfrac{49}{4}=\dfrac{1}{4}\)

\(\Leftrightarrow\left(\sqrt{x-3}-\dfrac{7}{2}\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=4\\\sqrt{x-3}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=16\\x-3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=19\\x=12\end{matrix}\right.\)

Vậy: S={19;12}

a \(ĐKXĐ:x\ge1\) 

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{3^2}{2}\sqrt{x-1}+\dfrac{24}{8}\sqrt{x-1}=-17\Leftrightarrow\dfrac{1}{2}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\) \(\Leftrightarrow-4\sqrt{x-1}+3\sqrt{x-1}=-17\Leftrightarrow-\sqrt{x-1}=-17\Leftrightarrow\sqrt{x-1}=17\Rightarrow x-1=289\Leftrightarrow x=290\left(TM\right)\) b \(ĐKXĐ:x\ge3\) 

\(\Leftrightarrow x-3-7\sqrt{x-3}+12=0\Leftrightarrow\left(\sqrt{x-3}-3\right)\left(\sqrt{x-3}-4\right)=0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=3\\\sqrt{x-3}=4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x-3=9\\x-3=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\left(TM\right)\\x=19\left(TM\right)\end{matrix}\right.\)

\(B=\dfrac{x-4\sqrt{x}-3\sqrt{x}+12}{x-3\sqrt{x}-\sqrt{x}+3}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)

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

`B=sqrtx/(sqrtx+3)+(2sqrtx)/(\sqrtx-3)-(3x+9)/(x-9)(x>0,x ne 9)`
`=(x-3sqrtx+2x+6sqrtx-3x-9)/(x-9)`
`=(3sqrtx-9)/(x-9)`
`=(3(sqrtx-3))/((sqrtx-3)(sqrtx+3))`
`=3/(sqrtx+3)`
`P=A.B=3/x`
`Px+3\sqrt{x-5}=x-2sqrtx+7(x>=5)`
`<=>3+3\sqrt{x-5}=x-2sqrtx+7`
`<=>x-2sqrtx+4-3\sqrt{x-5}=0`
`<=>2x-4sqrtx+8-6sqrt{x-5}=0`
`<=>x-4sqrtx+4+x-5-6sqrt{x-5}+9=0`
`<=>(sqrtx-2)^2+(\sqrt{x-5}-3)^2=0`
Dấu "=" xảy ra khi $\begin{cases}x=4\\x=14\\\end{cases}(l)$
Vậy khong có giá trị của x thể pt có nghiệm