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: \(\sqrt{x^2-5x+6}=\sqrt{x-2}\)

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

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

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

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

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

3: Ta có: \(\sqrt{4x+1}=x+1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)

\(\Leftrightarrow3\sqrt{x-1}=15\)

\(\Leftrightarrow x-1=25\)

hay x=26

5: Ta có: \(\sqrt{4x^2-12x+9}=7\)

\(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

a

ĐK: \(x\ge1\left(\sqrt{x-1}\ge0\right)\)

\(PT\Leftrightarrow\sqrt{x^2-x-2x+2}=\sqrt{x-1}\\ \Leftrightarrow\sqrt{x\left(x-1\right)-2\left(x-1\right)}=\sqrt{x-1}\\ \Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\\ \Leftrightarrow\left(\sqrt{x-1}\right)\left(\sqrt{x-2}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x-2}=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

b

ĐK: \(\left\{{}\begin{matrix}x^2-4x+4>0\\4x^2-4x+9>0\end{matrix}\right.\)

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

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

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

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

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

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

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

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}