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)...ghi lại đề...

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

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

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

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

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

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

\(\Leftrightarrow x-2=1\)(Vì \(x-2\ge0\Leftrightarrow x\ge2\))

\(\Leftrightarrow x=3\)

\(\)

\(a,\sqrt{x^2-3x+2}=\sqrt{x-1}\)

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

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

\(\Rightarrow x^2-x-3x+3=0\)

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

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

Vậy..........

b)...ghi lại đề...

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

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

\(\Leftrightarrow x-2\)(vì \(x-2\ge0\Leftrightarrow x\ge2\))\(=2x-3\)(vì \(2x-3\ge0\Leftrightarrow x\ge\frac{3}{2}\))

\(\Leftrightarrow x-2x=2-3\)

\(\Leftrightarrow-x=-1\)

\(\Leftrightarrow x=1\)

a)

ĐK: $x\geq 2$

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

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

\(\Rightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1(\text{loại vì x}\geq 2)\\ \sqrt{x-2}=1\end{matrix}\right.\)

\(\Rightarrow x=1^2+2=3\) là nghiệm duy nhất thỏa mãn

b)

ĐK: $x\in\mathbb{R}$

Bình phương 2 vế:

\(\Rightarrow x^2-4x+4=4x^2-12x+9\)

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

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

\(\Leftrightarrow (-x+1)(3x-5)=0\Rightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\) (đều thỏa mãn)

Vậy..........

c)

ĐKXĐ: $x\geq 3$

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

\(\Leftrightarrow (x-2)(x-3)=x-2\) (bình phương 2 vế không âm)

\(\Leftrightarrow (x-2)(x-3-1)=0\)

\(\Rightarrow \left[\begin{matrix} x-2=0\\ x-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2(\text{loại vì x}\geq 3)\\ x=4\end{matrix}\right.\)

Vậy $x=4$

d)

ĐK: $x\in\mathbb{R}$

PT \(\Leftrightarrow 4x^2-4x+1=x^2-6x+9\) (bình phương 2 vế không âm)

\(\Leftrightarrow (2x-1)^2=(x-3)^2\Leftrightarrow (2x-1)^2-(x-3)^2=0\)

\(\Leftrightarrow (2x-1-x+3)(2x-1+x-3)=0\)

\(\Leftrightarrow (x+2)(3x-4)=0\Rightarrow \left[\begin{matrix} x+2=0\\ 3x-4=0\end{matrix}\right.\)

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

Vậy.........

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) - 3 =0

<=> (√x-3)(√x+3)-3√x-3=0

<=> (√x-3)(√x+3-3)=0

<=> (√x-3)√x=0

<=> √x-3=0

<=>x=9

b)=x - 3

<=> √(2x -3)² =x-3

<=> 2x-3=x-3

<=>2x-x=-3+3

<=>x=0

c)=3x-1

<=> √(x+3)² =3x-1

<=> x+3=3x-1

<=> -2x=-4

<=>  x=2

Nhớ cho mình 1 tim nha bạn

Lời giải:

a. ĐKXĐ: $x\geq 3$

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

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

$\Leftrightarrow \sqrt{x-3}=0$ hoặc $\sqrt{x+3}-3=0$

$\Leftrightarrow \sqrt{x-3}=0$ hoặc $\sqrt{x+3}=3$

$\Leftrightarrow x=3$ hoặc $x=6$ (tm)

b.

PT \(\Rightarrow \left\{\begin{matrix} x-3\geq 0\\ 4x^2-12x+9=(x-3)^2=x^2-6x+9\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ 3x^2-6x=0\end{matrix}\right.\)

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

$\Rightarrow$ không có giá trị $x$ nào thỏa mãn 

Vậy pt vô nghiệm.

c.

PT \(\Rightarrow \left\{\begin{matrix} 3x-1\geq 0\\ x^2+6x+9=(3x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ x^2+6x+9=9x^2-6x+1\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ 8x^2-12x-8=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ 4(x-2)(2x+1)=0\end{matrix}\right.\Leftrightarrow x=2\)

1

ĐK: \(x\in R\)

\(\sqrt{x^2-4x+4}=\sqrt{4x^2-12+9}\\ \Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(2x-3\right)^2}\\ \Leftrightarrow\left|x-2\right|=\left|2x-3\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2x-3\\2-x=2x-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)

2

ĐK: \(\left\{{}\begin{matrix}x+2\sqrt{x-1}\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow x\ge1\)

Đặt \(t=\sqrt{x-1}\left(t\ge0\right)\Rightarrow t^2=x-1\Rightarrow x=t^2+1\)

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

Do có \(t\ge0\) nên \(\left(1\right)\Leftrightarrow t+1=2\Leftrightarrow t=2-1=1\)

\(\Rightarrow x=t^2+1=1^2+1=2\) (thỏa mãn)

1: =>|2x-3|=|x-2|

=>2x-3=x-2 hoặc 2x-3=-x+2

=>x=1 hoặc 3x=5

=>x=5/3 hoặc x=1

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

=>căn x-1+1=2

=>căn x-1=1

=>x-1=1

=>x=2

1: =>x^2-x=3-x

=>x^2=3

=>x=căn 3 hoặc x=-căn 3

2: =>x^2-4x+3=x^2-4x+4 và x>=2

=>3=4(vô lý)

3: =>2|x-1|=6

=>|x-1|=3

=>x-1=3 hoặc x-1=-3

=>x=-2 hoặc x=4

4: =>|2x-3|=|x-2|

=>2x-3=x-2 hoặc 2x-3=-x+2

=>x=1 hoặc x=5/3

5: =>\(\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)=0\)

=>x+2=0

=>x=-2