\(1)\) 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) Ta có: \(\sqrt{49\left(x^2-2x+1\right)}-35=0\)

\(\Leftrightarrow7\left|x-1\right|=35\)

\(\Leftrightarrow\left|x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

b)

ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)

Ta có: \(\sqrt{x^2-9}-5\sqrt{x+3}=0\)

\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x-3}-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=0\\\sqrt{x-3}=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-3=25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=28\left(nhận\right)\end{matrix}\right.\)

c) ĐKXĐ: \(x\ge0\)

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

\(\Leftrightarrow x-1=x+\sqrt{x}-6\)

\(\Leftrightarrow\sqrt{x}-6=-1\)

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

hay x=25(nhận)

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

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

=> (a - 3).(a - 1) = 0

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

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

ĐKXĐ: \(x\ge3\)

\(\Leftrightarrow\sqrt{x-3}=2\sqrt{x^2-9}\)

\(\Leftrightarrow x-3=4\left(x-3\right)\left(x+3\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{4}\left(loại\right)\end{matrix}\right.\)

chuyển 1 trong hai cái qua 1 vế để có dạng √A=√B rồi bình phương hai vế lên giải. nhớ đặt ĐK và kết luận nghiệm

đk: x\(\ge3\)

\(\sqrt{x+3}+\sqrt{x^2-9}=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\left(loại\right)\\\sqrt{\left(x-3\right)\left(x+3\right)=0}\end{matrix}\right.\)

vậy phương trình vô nghiệm

Bài 1: 

\(\Leftrightarrow\left(x^2-6x-7\right)^2-\left(3x^2-12x-9\right)^2=0\)

\(\Leftrightarrow\left(3x^2-12x-9-x^2+6x+7\right)\left(3x^2-12x-9+x^2-6x-7\right)=0\)

\(\Leftrightarrow\left(2x^2-6x-2\right)\left(4x^2-18x-16\right)=0\)

\(\Leftrightarrow\left(x^2-3x-1\right)\left(2x^2-9x-8\right)=0\)

hay \(x\in\left\{\dfrac{3+\sqrt{13}}{2};\dfrac{3-\sqrt{13}}{2};\dfrac{9+\sqrt{145}}{4};\dfrac{9-\sqrt{145}}{4}\right\}\)

\(\sqrt{x+6-4\sqrt{x+2}}-\sqrt{9-4\sqrt{5}}=0\left(đk:x\ge-2\right)\)

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

\(\Leftrightarrow\left|\sqrt{x+2}-2\right|=\left|\sqrt{5}-2\right|\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=5\\x+2=21-8\sqrt{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=19-8\sqrt{5}\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{3;19-8\sqrt{5}\right\}\)

a) Đk: \(\hept{\begin{cases}x^2-4x+1\ge0\\x+1\ge0\end{cases}}\)

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)thỏa mãn điều kiện

Vậy x=0 hoặc x=5

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

Đk: x>=3 hoặc x=1

pt  (1)<=> \(\sqrt{x-1}\left(\sqrt{x-3}+1\right)=0\)

<=> \(\sqrt{x-1}=0\)(vì\(\sqrt{x-3}+1>0\)mọi x )

<=> x-1=0

<=> x=1 ( thỏa mãn điều kiện)

\(a,ĐK:1\le x\le3\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\\\sqrt{3-x}=b\end{matrix}\right.\left(a,b\ge0\right)\)

\(PT\Leftrightarrow a+b-ab=1\Leftrightarrow a+b-ab-1=0\\ \Leftrightarrow\left(a-1\right)\left(1-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=1\\3-x=1\end{matrix}\right.\Leftrightarrow x=2\left(tm\right)\)

\(b,ĐK:0\le x\le9\\ PT\Leftrightarrow9+2\sqrt{x\left(9-x\right)}=-x^2+9x+9\\ \Leftrightarrow2\sqrt{-x^2+9x}-\left(-x^2+9x\right)=0\\ \Leftrightarrow\sqrt{-x^2+9x}\left(2-\sqrt{-x^2+9x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x^2+9x=0\\\sqrt{-x^2+9x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\\x^2-9x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=9\left(n\right)\\x=\dfrac{9+\sqrt{65}}{2}\left(n\right)\\x=\dfrac{9-\sqrt{65}}{2}\left(n\right)\end{matrix}\right.\)