Sửa lại câu c) đặt \(\sqrt{x}+1=\)t \(\Rightarrow\left[2\left(t+\dfrac{1}{2}\right)\right]\left(t-3\right)\)=7⇒\(\left\{{}\begin{matrix}t=3\\t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\\x=\dfrac{9}{4}\end{matrix}\right.\)

a) \(\left(\sqrt{4-3x}\right)^2=8^2\)\(\Leftrightarrow4-3x=64\Rightarrow x=-20\)

b) \(\sqrt{4x-8}+1=12\sqrt{\dfrac{x-2}{9}}\Leftrightarrow2\sqrt{x-2}+1\)\(=\left(12\sqrt{\left(x-2\right).\dfrac{1}{9}}\right)\)

\(\Leftrightarrow2t+1=12.\dfrac{1}{3}t\) (Đặt t = \(\sqrt{x-2}\))

\(\Rightarrow t=\dfrac{1}{2}\) \(\Rightarrow\sqrt{x-2}=\dfrac{1}{2}\)\(\Rightarrow x=\dfrac{9}{4}\)

c) pt\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}+1=7\\\sqrt{x}-2=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\x=4\end{matrix}\right.\)

a) Ta có: \(\sqrt{4-3x}=8\)

\(\Leftrightarrow4-3x=64\)

\(\Leftrightarrow3x=4-64=-60\)

hay x=-20

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

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

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

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

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

hay \(x=\dfrac{9}{4}\)

b: Ta có: \(\sqrt{9x^2-9}+\sqrt{4x^2-4}=\sqrt{16x^2-16}+2\)

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

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

hay \(x\in\left\{\sqrt{5};-\sqrt{5}\right\}\)

a. \(x+\sqrt{x^2-4x+4}=\dfrac{1}{2}\)

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

<=> \(x+\left|x-2\right|=\dfrac{1}{2}\)

<=> \(\left[{}\begin{matrix}x+x-2=\dfrac{1}{2}\\x+\left[-\left(x-2\right)\right]=\dfrac{1}{2}\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}2x=\dfrac{5}{2}\\x-x+2=\dfrac{1}{2}\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{5}{4}\\0=\dfrac{-3}{2}\left(VLí\right)\end{matrix}\right.\)

Vậy nghiệm của PT là \(S=\left\{\dfrac{5}{4}\right\}\)

b. \(\sqrt{9x^2-9}+\sqrt{4x^2-4}=\sqrt{16x^2-16}+2\)

<=> \(\sqrt{9\left(x^2-1\right)}+\sqrt{4\left(x^2-1\right)}=\sqrt{16\left(x^2-1\right)}+2\)

<=> \(3\sqrt{x^2-1}+2\sqrt{x^2-1}-4\sqrt{x^2-1}=2\)

<=> \(\left(3+2-4\right)\sqrt{x^2-1}=2\)

<=> \(\sqrt{x^2-1}=2\)

<=> x² - 1 = 4

<=> x² = 5

<=> x = \(\sqrt{5}\)

f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)

\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)

\(\Leftrightarrow\left|x+1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)

\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

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

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

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

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

d) \(\sqrt{x^2-6x+9}=2\Leftrightarrow\sqrt{\left(x-3\right)^2}=2\Leftrightarrow x-3=2\Leftrightarrow x=5\)

e) đk: \(x\ge2\)\(\sqrt{x^2-3x+2}=\sqrt{x-1}\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)f) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\Leftrightarrow2x-1=x-3\Leftrightarrow x=-2\)

c: Ta có: \(\sqrt{x+4\sqrt{x-4}}=2\)

\(\Leftrightarrow\left|\sqrt{x-4}+2\right|=2\)

\(\Leftrightarrow x-4=0\)

hay x=4

a) \(\sqrt{x-1+2\sqrt{x-1}.1+1^2}=2;đk:x\)≥1
⇔\(\sqrt{\left(\sqrt{x-1}\right)^2+2\sqrt{x-1}.1+1^2}=2\left(hđt-1\right)\)
⇔\(\sqrt{\left(\sqrt{x-1}+1\right)^2=2}\)
⇔|\(\sqrt{x-1}+1\)|=2
⇔\(\left[{}\begin{matrix}\sqrt{x+1}-1=2\\\sqrt{x+1-1}=-2\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}\sqrt{x+1}=3\\\sqrt{x+1}=-1\left(L\right)\end{matrix}\right.\)⇔x+1=9⇔x=10(TM)
→S={10}

a. ĐKXĐ: $x\geq 2$ hoặc $x=1$

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

$\Leftrightarrow \sqrt{x-1}(\sqrt{x-2}-1)=0$

\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=3\end{matrix}\right.\) (đều thỏa mãn)

b.

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

$\Leftrightarrow |x-2|=|2x-3|$

\(\Leftrightarrow \left[\begin{matrix} x-2=2x-3\\ x-2=3-2x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\)

c. ĐKXĐ: $x=2$ hoặc $x\geq 3$

PT $\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}$

$\Leftrightarrow \sqrt{x-2}(\sqrt{x-3}-1)=0$

\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x-3}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=4\end{matrix}\right.\) (đều tm)

d.

PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$

$\Leftrightarrow |2x-1|=|x-3|$

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

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

\(\Leftrightarrow x^2-3x+2=x-1\)

\(\Leftrightarrow x^2-4x+3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

b: Ta có: \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x-2\\2x-3=-x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)

c: Ta có: \(\sqrt{x^2-5x+6}=\sqrt{x-2}\)

\(\Leftrightarrow x^2-5x+6=x-2\)

\(\Leftrightarrow x^2-6x+8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)

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

`a)sqrt{x^2-6x+9}=2`

`<=>sqrt{(x-3)^2}=2`

`<=>|x-3|=2`

`**x-3=2`

`<=>x=5`

`**x-3=-2`

`<=>x=1`

Vậy `S={1,5}`

`b)sqrt{4x-20}+sqrt{x-5}-1/3sqrt{9x-45}=4`

đk:`x>=5`

`pt<=>2sqrt{x-5}+sqrt{x-5}-1/3*3*sqrt{x-5}=4`

`<=>2sqrt{x-5}=4`

`<=>sqrt{x-5}=2`

`<=>x-5=4<=>x=9`

Vậy `S={9}`

Lời giải:

a.

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

$\Leftrightarrow |x-3|=2$

$\Leftrightarrow x-3=\pm 2$

$\Leftrightarrow x=1$ hoặc $x=5$

b. ĐKXĐ: $x\geq 5$

PT $\Leftrightarrow \sqrt{4(x-5)}+\sqrt{x-5}-\frac{1}{3}\sqrt{9(x-5)}=4$

$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$

$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$

$\Leftrightarrow x=2^2+5=9$ (thỏa mãn)

a)Pt\(\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=x+\sqrt{3}\)

\(\Leftrightarrow\left|x+\sqrt{3}\right|=x+\sqrt{3}\)

\(\Leftrightarrow x+\sqrt{3}\ge0\)\(\Leftrightarrow x\ge-\sqrt{3}\)

Vậy...

b)Đk:\(x\ge4\)

Pt\(\Leftrightarrow\sqrt{\left(x-4\right)+2\sqrt{x-4}+1}=2\sqrt{x-4}+1\)

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

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

\(\Leftrightarrow\sqrt{x-4}=0\)

\(\Leftrightarrow x=4\) (tm)

Vậy...

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

\(\Leftrightarrow\left|x+\sqrt{3}\right|=x+\sqrt{3}\)

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

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