a.

\(\Leftrightarrow4x^2-6x+1+\dfrac{1}{\sqrt{3}}\sqrt{\left(4x^2-2x+1\right)\left(4x^2+2x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{4x^2-2x+1}=a>0\\\sqrt{4x^2+2x+1}=b>0\end{matrix}\right.\) ta được:

\(2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)

\(\Leftrightarrow\left(a-\dfrac{b}{\sqrt{3}}\right)\left(2a+\sqrt{3}b\right)=0\)

\(\Leftrightarrow a=\dfrac{b}{\sqrt{3}}\)

\(\Leftrightarrow3a^2=b^2\)

\(\Leftrightarrow3\left(4x^2-2x+1\right)=4x^2+2x+1\)

\(\Leftrightarrow...\)

b.

\(x^2-3x+1+\dfrac{1}{\sqrt{3}}\sqrt{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)

Lặp lại cách làm câu a

1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)

\(\Leftrightarrow5-2x=36\)

\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)

2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)

\(\Leftrightarrow2-x=x+1\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

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

4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)

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

\(\Leftrightarrow\left|x-5\right|=x-2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)

\(1)\) 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 

Giải PT

a) \(3\sqrt{9x}+\sqrt{25x}-\sqrt{4x} = 3\)

\(\Leftrightarrow\) \(3.3\sqrt{x} +5\sqrt{x} - 2\sqrt{x} = 3 \)

\(\Leftrightarrow\) \(9\sqrt{x}+5\sqrt{x}-2\sqrt{x} = 3 \)

\(\Leftrightarrow\) \(12\sqrt{x} = 3\)

\(\Leftrightarrow\) \(\sqrt{x} = 4 \)

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

\(\Leftrightarrow\) \(x=16\)

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

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

\(\Leftrightarrow\) \(|x-1|=3\)

* \(x-1=3\)

\(\Leftrightarrow\) \(x=4\)

* \(-x-1=3\)

\(\Leftrightarrow\) \(-x=4\)

\(\Leftrightarrow\) \(x=-4\)

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

<=> \(\sqrt{(2x+1)^2} = 3+x\)

<=> \(|2x+1|=3+x\)

* \(2x+1=3+x\)

<=> \(2x-x=3-1\)

<=> \(x=2\)

* \(-2x+1=3+x\)

<=> \(-2x-x = 3-1\)

<=> \(-3x=2\)

<=> \(x=\dfrac{-2}{3}\)

d) \(\sqrt{x-1} = x-3\)

<=> \(\sqrt{(x-1)^2} = (x-3)^2\)

<=> \(|x-1| = x^2-2.x.3+3^2\)

<=> \(|x-1| = x-6x+9\)

<=> \(|x-1| = -5x+9\)

* \(x-1= -5x+9\)

<=> \(x+5x = 9+1\)

<=> \(6x=10\)

<=> \(x= \dfrac{10}{6} =\dfrac{5}{3}\)

* \(-x-1 = -5x+9\)

<=> \(-x+5x = 9+1\)

<=> \(4x = 10\)

<=> \(x= \dfrac{10}{4} = \dfrac{5}{2}\)

Vì \(\sqrt{x^2+1}\)\(\ge\) 0

nên x²+1 \(\ge\)0

mà x²+1 > 0

nên \(\sqrt{4x^2-4x+5}=0\)

\(\Rightarrow\)4x²-4x+5 =0 

mà 4x²-4x+1+4

=(2x-1)²+4>0 

\(\Rightarrow\)Phương trình vô nghiệm

Bạn ơi, bài này vô nghiệm nhé . Có cần Milky Way giải rõ không ? Rất sẵn lòng  ^^

từ cái đề bài \(\Rightarrow\sqrt{x^2+1}=\sqrt{4x^2-4x+5}=0\)

=>x²+1=0

=>không có x

a/ ĐKXĐ: \(x\ge\frac{3}{4}\)

\(\Leftrightarrow6x+1+2\sqrt{5x^2+5x}=6x+1+2\sqrt{8x^2+10x-12}\)

\(\Leftrightarrow\sqrt{5x^2+5x}=\sqrt{8x^2+10x-12}\)

\(\Leftrightarrow5x^2+5x=8x^2+10x-12\)

\(\Leftrightarrow3x^2+5x-12=0\Rightarrow\left[{}\begin{matrix}x=-3< \frac{3}{4}\left(l\right)\\x=\frac{4}{3}\end{matrix}\right.\)

b/ \(\Leftrightarrow x^2+x+1+2\sqrt{x^2+x+1}-3=0\)

Đặt \(\sqrt{x^2+x+1}=t>0\)

\(\Rightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)

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

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

\(5x^2+4x+7-4x\sqrt{x^2+x+2}-4\sqrt{3x+1}=0\)

ĐK: \(x\ge-\frac{1}{3}\)

\(\Leftrightarrow5x^2+4x-9-\left(4x\sqrt{x^2+x+2}-8\right)-\left(4\sqrt{3x+1}-8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x+9\right)-4\frac{x^2\left(x^2+x+2\right)-4}{x\sqrt{x^2+x+2}+2}-4\frac{3x+1-4}{\sqrt{3x+1}+2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x+9\right)-4\frac{\left(x-1\right)\left(x^3+2x^2+4x+4\right)}{x\sqrt{x^2+x+2}+2}-4\frac{3\left(x-1\right)}{\sqrt{3x+1}+2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x+9-4\frac{\left(x^3+2x^2+4x+4\right)}{x\sqrt{x^2+x+2}+2}-4\frac{3}{\sqrt{3x+1}+2}\right)=0\)

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(ĐKXĐ:x\ge\frac{-1}{3}\)

\(5x^2+4x+7-4x\sqrt{x^2+x+2}-4\sqrt{3x+1}=0\)

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

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

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x^2+x+2}=2x\\\sqrt{3x+1}=2\end{cases}}\Leftrightarrow\hept{\begin{cases}x>0\\x^2+x+2=4x\\3x+1=4\end{cases}}\Leftrightarrow x=1\)

Vậy nghiệm duy nhất của phương trình là x = 1

ĐKXĐ: \(\left[{}\begin{matrix}0\le x\le2-\sqrt{3}\\x\ge2+\sqrt{3}\end{matrix}\right.\)

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

\(\Leftrightarrow\left(2x+2-5\sqrt{x}\right)+\left(\sqrt{4x^2-16x+4}-\sqrt{x}\right)=0\)

\(\Leftrightarrow\dfrac{4x^2-17x+4}{2x+2+5\sqrt{x}}+\dfrac{4x^2-17x+4}{\sqrt{4x^2-16x+4}+\sqrt{x}}=0\)

\(\Leftrightarrow\left(4x^2-17x+4\right)\left(\dfrac{1}{2x+2+5\sqrt{x}}+\dfrac{1}{\sqrt{4x^2-16x+4}+\sqrt{x}}\right)=0\)

\(\Leftrightarrow4x^2-17x+4=0\)