a) `sqrt(x^2-6x _9) = 4-x`

`<=> sqrt[(x-3)^2] =4-x`

`<=> |x-3| =4-x ( đk :x<=4)`

`<=> |x-3| = |4-x|`

`<=> [(x-3 =4-x),(x-3 = x-4):}`

`<=>[(x = 7/2(t//m)),(0=-1(vl)):}`

Vậy `S = {7/2}`

b) `sqrt(x^2 -9) + sqrt(x^2 -6x +9) =0(đk : x>=3(hoặc) x<=-3)`

`<=>sqrt(x^2 -9) =- sqrt(x^2 -6x +9) `

`<=>(sqrt(x^2 -9))^2 =(- sqrt(x^2 -6x +9))^2`

`<=> x^2 -9 = x^2 -6x +9`

`<=> 6x = 9+9 =18`

`<=> x=3(t//m)`

Vậy `S={3}`

c) `sqrt(x^2 -2x+1) + sqrt(x^2-4x+4) =3`

`<=> sqrt[(x-1)^2] +sqrt[(x-2)^2] =3`

`<=> |x-1| +|x-2| =3`

xét `x<1 =>{(|x-1| =1-x ),(|x-2|=2-x):}`

`=> 1-x +2-x =3`

`=> x = 0(t//m)`

xét `1<=x<2 => {(|x-1|=x-1),(|x-2|= 2-x):}`

`=> x-1 +2-x =3`

`=>1=3 (vl)`

xét `x>=2 => {(|x-1| =x-1),(|x-2|=x-2):}`

`=> x-1+x-2 =3`

`=> x=3(t//m)`

Vậy `S = {0;3}`

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

TH1: x>=3

=>4-x=x-3

=>x=7/2(nhận)

TH2: x<3

=>3-x=4-x(loại)

b: =>căn x-3(căn x+3+căn x-3)=0

=>x-3=0

=>x=3

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

Th1: x<1

=>1-x+2-x=3

=>x=0(nhận)

TH2: 1<=x<2

=>x-1+2-x=3

=>1=3(loại)

TH3: x>=2

=>x-1+x-2=3

=>x=3

\(A=\sqrt{x^2}-\sqrt{x^2-4x+4}\)

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

\(\Leftrightarrow A=x-|x-2|=x-x+2=2\)

A = \(\sqrt{x^2}-\sqrt{x^2-4x+4}=\sqrt{x^2}-\sqrt{\left(x-2\right)^2}=\left|x\right|-\left|x-2\right|=x-x+2=2\)(vì  \(x\ge2\))

B = \(\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}=\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}=\left|x-3\right|-\left|x+3\right|=3-x+x+3=6\)(vì x < 3)

\(B=\sqrt{x^2-6x+3^2}-\sqrt{x^2+6x+3^2}\)

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

\(\Leftrightarrow B=|x-3|-|x+3|=-x+3-x-3=0\)

Lời giải:

a. Đề thiếu

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

$\Leftrightarrow |x-1|+|x-2|=3$
Nếu $x\geq 2$ thì pt trở thành:
$x-1+x-2=3$

$\Leftrightarrow 2x-3=3$

$\Leftrightarrow x=3$ (tm)

Nếu $1\leq x< 2$ thì:

$x-1+2-x=3\Leftrightarrow 1=3$ (vô lý)

Nếu $x< 1$ thì:

$1-x+2-x=3$

$\Leftrightarrow x=0$ (tm)

a)\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

\(\Rightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)

\(\Rightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+x-3=0\)

Đặt \(x-3=t\) pt thành

\(\sqrt{t\left(t-6\right)}-t=0\)

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

\(\Leftrightarrow t=0\)\(\Rightarrow x-3=0\Leftrightarrow x=3\)

b)\(\sqrt{x^2-4}-x^2+4=0\)

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

Đặt \(\sqrt{x^2-4}=t\) pt thành

\(t=t^2\Rightarrow t\left(1-t\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}t=1\\t=0\end{array}\right.\).

Với \(t=0\Rightarrow\sqrt{x^2-4}=0\Rightarrow x=\pm2\) 

Với \(t=1\Rightarrow\sqrt{x^2-4}=1\)\(\Rightarrow x=\pm\sqrt{5}\)

bạn đăg từng bài thui

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

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

a 

ĐK: \(x^2-2x+1>0\)

PT \(\Leftrightarrow\sqrt{\left(x-1\right)^2}+x-6x+9=0\)

\(\Leftrightarrow\left|x-1\right|-5x+9=0\\ \Leftrightarrow\left|x-1\right|=-9+5x\\ \Leftrightarrow\left[{}\begin{matrix}x-1=-9+5x\\1-x=-9+5x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=\dfrac{10}{6}\left(nhận\right)\end{matrix}\right.\)

b

ĐK: \(\left\{{}\begin{matrix}2x^2-3>0\\4x-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>\dfrac{\sqrt{6}}{2}\\x< -\dfrac{\sqrt{6}}{2}\end{matrix}\right.\\x>\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow x>\dfrac{\sqrt{6}}{2}\)

PT \(\Leftrightarrow2x^2-3=4x-3\)

\(\Leftrightarrow2x^2-4x=0\\ \Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

c

ĐK: \(\left\{{}\begin{matrix}1-x^2\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow x=1\)

PT \(\Leftrightarrow1-x^2=x-1\)

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