6) \(\sqrt{x^2-4x+1}=x\left(x\ge0\right)\) 

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

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

\(\Leftrightarrow4x=1\)

\(\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\) 

8) \(\sqrt{x^2-x-6}=\sqrt{x-3}\left(x\ge3\right)\) 

\(\Leftrightarrow x^2-x-6=x-3\)

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

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

9) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\left(x\ge1\right)\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

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

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

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

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=1+1\)

\(\Leftrightarrow x=2\left(tm\right)\)

a. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$

$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$

$\Leftrightarrow -\sqrt{x-1}=-17$

$\Leftrightarrow \sqrt{x-1}=17$

$\Leftrightarrow x-1=289$

$\Leftrightarrow x=290$

b. ĐKXĐ: $x\geq \frac{1}{2}$

PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$

$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$

$\Leftrihgtarrow \sqrt{2x-1}=2$

$\Leftrightarrow x=2,5$ (tm)

c. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$

$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$

$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)

Vậy pt vô nghiệm

d. ĐKXĐ: $x>\frac{-2}{3}$

PT $\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{1}{2}\sqrt{9}.\sqrt{\frac{1}{3x+2}}+\sqrt{16}.\sqrt{\frac{1}{3x+2}}-5\sqrt{\frac{1}{4}}\sqrt{\frac{1}{3x+2}}=1$

$\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{3}{2}\sqrt{\frac{1}{3x+2}}+4\sqrt{\frac{1}{3x+2}}-\frac{5}{2}\sqrt{\frac{1}{3x+2}}=1$

$\Leftrightarrow \sqrt{\frac{1}{3x+2}}=1$

$\Leftrightarrow \frac{1}{3x+2}=1$

$\Leftrightarrow 3x+2=1$

$\Leftrightarrow x=-\frac{1}{3}$

\(25\sqrt{\dfrac{x-3}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\left(x\ge3\right)\)

\(=25\sqrt{\dfrac{1}{25}.\left(x-3\right)}-7\sqrt{\dfrac{4}{9}.\left(x-3\right)}-7\sqrt{x^2-9}+18\sqrt{\dfrac{1}{9}.\left(x^2-9\right)}=0\)

\(=5\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{26}{9}\left(l\right)\end{matrix}\right.\)

a: \(\Leftrightarrow2\cdot5\sqrt{x-3}-\dfrac{1}{2}\cdot2\sqrt{x-3}+\dfrac{1}{7}\cdot7\sqrt{x-3}=20\)

=>\(10\cdot\sqrt{x-3}=20\)

=>\(\sqrt{x-3}=2\)

=>x-3=4

=>x=7

b: =>|x-3|=2

=>x-3=2 hoặc x-3=-2

=>x=5 hoặcx=1

Điều kiện: mọi \(x\in R\)

Ta có \(\sqrt{x^2+x+25}=\sqrt{x^2+x+9}+2\)

\(\Leftrightarrow x^2+x+25=x^2+x+9+4.\sqrt{x^2+x+9}+4\)

\(\Leftrightarrow\sqrt{x^2+x+9}=3\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

=>\(\sqrt{\left(x+3\right)^2}\)+ \(\sqrt{\left(x+4\right)^2}\)+\(\sqrt{\left(x+5\right)^2}\)=9x

=> x + 3 + x + 4 + x + 5 = 9x

=> - 6x = - 12

=> x=2

Ủa sao phá đc trị tuyệt đối hay v bạn? (căn a^2 = trị tuyệt đối của a ) 

Vì \(\sqrt{x^2+6x+9}>0\\ \)

\(\sqrt{x^2+8x+16}>0\\ \)

\(\sqrt{x^2+10x+25}>0\\ \)

Suy ra 9x>0. Suy ra x>0 .Nha bạn!

Đặt \(\hept{\begin{cases}\sqrt{x^2+x+25}=a\ge0\\\sqrt{x^2+x+16}=b\ge0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a+b=9\\a^2-b^2=9\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=9\\\left(a+b\right)\left(a-b\right)=9\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=9\\a-b=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=5\\b=4\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\sqrt{x^2+x+25}=5\\\sqrt{x^2+x+16}=4\end{cases}}\)

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

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

Đặt  \(t=x^2+x+16>0\)

pt trên đc viết lại thành

\(\sqrt{t+9}+\sqrt{t}=9\)

\(\Leftrightarrow t+9+t+2\sqrt{t\left(t+9\right)}=81\)

\(\Leftrightarrow2\sqrt{t\left(t+9\right)}=72-t\)

\(\Leftrightarrow\hept{\begin{cases}72-t>0\\4t\left(t+9\right)=\left(72-t\right)^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}t< 72\\3t^2+180t-5184=0\end{cases}}\)

\(\Leftrightarrow t=-30+6\sqrt{73}\) (vì t > 0)

Thử lại thấy ko thỏa mãn

Vậy pt vô nghiệm.

\(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\left(x\ge1\right)\)

\(< =>5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)

\(< =>30\sqrt{x-1}-15\sqrt{x-1}=36+6\sqrt{x-1}\)

\(< =>9\sqrt{x-1}=36\\ < =>\sqrt{x-1}=4\\ < =>x-1=16\\ < =>x=17\left(tm\right)\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{1}{3}\sqrt{x-1}-\sqrt{x-1}=6\)

=>\(1.5\cdot\sqrt{x-1}=6\)

=>\(\sqrt{x-1}=4\)

=>x-1=16

=>x=17

\(x=0\)

\(\sqrt{x2+x+25}\) + \(\sqrt{x2+x+16}\)=9

=\(\sqrt{ }\)(x+5)² +\(\sqrt{ }\)(x+4)²=9

= /x+5/ +/x+4/ =9

= x+5+x+4 =9

= 2x+9=9

= 2x=9-9

=2x=0

x= 0:2

x=0

vậy x = 0