\(1)\) ĐKXĐ : \(x\ge3\)

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

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

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

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}\sqrt{x-1}=0\\\sqrt{x-3}+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x\in\left\{\varnothing\right\}\end{cases}}}\)

Vậy \(x=1\)

\(2)\)\(\sqrt{x^2-2x+1}-\sqrt{x^2-6x+9}=10\)

\(\Leftrightarrow\)\(\sqrt{\left(x-1\right)^2}-\sqrt{\left(x-3\right)^2}=10\)

\(\Leftrightarrow\)\(\left|x-1\right|-\left|x-3\right|=10\)

+) Với \(\hept{\begin{cases}x-1\ge0\\x-3\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\ge3\end{cases}\Leftrightarrow}x\ge3}\) ta  có : 

\(x-1-x+3=10\)

\(\Leftrightarrow\)\(0=8\) ( loại ) 

+) Với \(\hept{\begin{cases}x-1< 0\\x-3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 1\\x< 3\end{cases}\Leftrightarrow}x< 1}\) ta có : 

\(1-x+x-3=10\)

\(\Leftrightarrow\)\(0=12\) ( loại ) 

Vậy không có x thỏa mãn đề bài 

Chúc bạn học tốt ~ 

PS : mới lp 8 sai đừng chửi nhé :v 

a: Thay m=1 vào pt, ta được:

\(x^2-1=0\)

=>(x-1)(x+1)=0

=>x=1 hoặc x=-1

b: \(\text{Δ}=\left(2m-2\right)^2-4\cdot\left(-m\right)\)

\(=4m^2-8m+4+4m\)

\(=4m^2-4m+4\)

\(=4\left(m^2-m+1\right)\)

\(=4m^2-4m+1+3=\left(2m-1\right)^2+3>0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt

Ta có: \(2\left(x_1+x_2\right)-3x_1x_2+9=0\)

\(\Leftrightarrow2\cdot\left[-2\left(m-1\right)\right]-3\cdot\left(-m\right)+9=0\)

\(\Leftrightarrow-4\left(m-1\right)+3m+9=0\)

=>-4m+4+3m+9=0

=>13-m=0

hay m=13

a, Thay m = 1 ta được 

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

b, 

Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=-2\left(m-1\right)\\x_1x_2=-m\end{matrix}\right.\)

\(-4\left(m-1\right)+3m+9=0\Leftrightarrow-m+13=0\Leftrightarrow m=13\)

Ta có : \(\left(x-1\right)^4-8\left(x-1\right)^2-9=0\)

- Đặt \(\left(x-1\right)^2=a\) ta được phương trình : \(a^2-8a-9=0\)

Ta có : \(a-b+c=1-\left(-8\right)+9=0\)

Nên phương trình có 2 nghiệm \(a_1=-1,a_2=-\frac{c}{a}=9\)

=> \(\left[{}\begin{matrix}\left(x-1\right)^2=-1\left(VL\right)\\\left(x-1\right)^2=9\end{matrix}\right.\)

=> \(\left(x-1\right)^2=9\)

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

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

Vậy .....

`1/9(x-3)^2-1/25(x+5)^2=0`

`<=>(1/3x-1)^2-(1/5x+1)^2=0`

`<=>(1/3x-1-1/5x-1)(1/3x-1+1/5x+1)=0`

`<=>(2/15x-2). 8/15x=0`

`<=>2/15x-2=0` hoặc `8/15x=0`

`<=>x=15`         hoặc `x=0`

Vậy `S=`{`15;0`}

\(ĐK:-1\le x\le1\\ PT\Leftrightarrow13\left(1-2x^2\right)\sqrt{\left(1-x^2\right)\left(1+x^2\right)}+9\left(1+2x^2\right)\sqrt{\left(1+x^2\right)\left(1-x^2\right)}=0\\ \Leftrightarrow\sqrt{1-x^4}\left(13-26x^2+9+18x^2\right)=0\\ \Leftrightarrow\sqrt{1-x^4}\left(22-8x^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1-x^4=0\\22-8x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(1+x^2\right)\left(1-x\right)\left(1+x\right)=0\\x^2=\dfrac{22}{8}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\\left[{}\begin{matrix}x=\dfrac{\sqrt{11}}{2}\left(ktm\right)\\x=-\dfrac{\sqrt{11}}{2}\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

\(\left(3x-1\right)^2-3\left(3x-2\right)=9\left(x+1\right)\left(x-3\right)\)

\(\Leftrightarrow9x^2-6x+1-9x+6=9\left(x^2-2x-3\right)\)

\(\Leftrightarrow9x^2-15x+7=9x^2-18x-27\)

\(\Leftrightarrow-15x+18x+7+27=0\)

\(\Leftrightarrow3x+34=0\)

\(\Leftrightarrow x=\frac{-34}{3}\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-\frac{34}{3}\right\}\)

Nhiều qua trời 

\(4.\left(x+1\right)^2-9.\left(x-1\right)^2=0\)

\(\Leftrightarrow4.\left(x^2+2x+1\right)-9.\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow4x^2+8x+4-9x^2+18x-9=0\)

\(\Leftrightarrow\left(4x^2+8x+4\right)-\left(9x^2-18x+9\right)=0\)

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

\(\Leftrightarrow\left[2x+2-\left(3x-3\right)\right].\left[2x+2+\left(3x-3\right)\right]=0\)

\(\Leftrightarrow\left(2x+2-3x+3\right).\left(2x+2+3x-3\right)=0\)

\(\Leftrightarrow\left(5-x\right).\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5-x=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{1}{5}\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{5;\frac{1}{5}\right\}.\)

Chúc bạn học tốt!

\(4\left(x+1\right)^2-9\left(x-1\right)^2=0\)

\(\Leftrightarrow4\left(x^2+2x+1\right)-9\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow4x^2+8x+4-9x^2-18x-9=0\)

\(\Leftrightarrow-5x^2-10x-5=0\)

\(\Leftrightarrow-5\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow-5\left(x+1\right)^2=0\)

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

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy S = {1}