\(a,PT\Leftrightarrow\left|3x-1\right|=\left|x-3\right|\Leftrightarrow\left[{}\begin{matrix}3x-1=x-3\\3x-1=3-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\\ b,PT\Leftrightarrow\left|x-4\right|=4-x\Leftrightarrow\left[{}\begin{matrix}x-4=4-x\left(x\ge4\right)\\x-4=x-4\left(x< 4\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

\(A=\frac{1}{x-1}\sqrt{\frac{3x^2-6x+3}{x}}\)\(=\frac{1}{x-1}\sqrt{\frac{3\left(x^2-2x+1\right)}{x}}=\frac{1}{x-1}\sqrt{\frac{3\left(x-1\right)^2}{x}}=\frac{1}{x-1}\cdot\frac{\sqrt{3}\left(x-1\right)}{\sqrt{x}}=\frac{\sqrt{3}}{\sqrt{x}}\)

a, ĐK: \(6x^2-12x+7\ge0\) (*)

\(PT\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\6x^2-12x+7=x^4-4x^3+4x^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\x^4-4x^3-2x^2+12x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\\left(x-1\right)^2\left(x^2-2x-7\right)=0\end{matrix}\right.\) \(\Rightarrow x=1\pm2\sqrt{2}\) (thỏa mãn ĐK)

Vậy...

a, <=> (x-1)^3 + x^2(x-1)=0

<=> (x-1)(x^2-2x+1+x^2)=0

<=> (x-1)(2x^2-2x+1)=0

=> x=1

2x^2-2x+1=0 (*)

giải (*):

2x^2-2x+1=0

<=> (x-1)^2 + x^2 > 0

=> * vô nghiệm

=> Pt có nghiệm là 1.

b, x^2+x-12=0

<=> (x-3)(x+4)=0

=> x=3 hoặc x = -4

vậy....

c, 6x^2-11x-10=0

<=> (x-5/2)(6x+4)=0

=> x=5/2 hoặc x= -2/3.

vậy...

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

\(\Leftrightarrow\left|2x-1\right|=\dfrac{5}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{5}{6}\\2x-1=-\dfrac{5}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=\dfrac{1}{12}\end{matrix}\right.\)

Vậy...

b)Đk:\(x\ge3\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\x-4=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=4\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)

Vậy...

c)Đk:\(x\ge1\)

\(x+\sqrt{x-1}=13\)

\(\Leftrightarrow\sqrt{x-1}=13-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}13-x\ge0\\x-1=x^2-26x+169\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\x^2-27x+170=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\x^2-17x-10x+170=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\\left(x-17\right)\left(x-10\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\\left[{}\begin{matrix}x=17\\x=10\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=10\) (tm)

Vậy...

a. ĐKXĐ: \(-1\le x\le1\)

Đặt \(\sqrt{1+x}+\sqrt{1-x}=t>0\)

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

Pt trở thành:

\(t.t^2=8\Leftrightarrow t^3=8\Leftrightarrow t=2\)

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

\(\Leftrightarrow2+2\sqrt{1-x^2}=2\)

\(\Leftrightarrow1-x^2=0\Rightarrow x=\pm1\)

b.

ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)

\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\)

Pt trở thành:

\(t=t^2-4-16\Leftrightarrow...\)

a) \(\sqrt{x^2-x-4}=\sqrt{x-1}\)

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

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

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

\(\left(x^2-2.x.1+1^2\right)-6=0\)

\(\left(x-1\right)^2=6\)

⇒\(\left\{{}\begin{matrix}x-1=6\\x-1=-6\end{matrix}\right.\)         ⇒\(\left\{{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)

a:Ta có: \(\sqrt{2x+9}=\sqrt{5-4x}\)

\(\Leftrightarrow2x+9=5-4x\)

\(\Leftrightarrow6x=-4\)

hay \(x=-\dfrac{2}{3}\left(nhận\right)\)

b: Ta có: \(\sqrt{2x-1}=\sqrt{x-1}\)

\(\Leftrightarrow2x-1=x-1\)

hay x=0(loại)

c: Ta có: \(\sqrt{x^2+3x+1}=\sqrt{x+1}\)

\(\Leftrightarrow x^2+3x=x\)

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

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

a. \(\sqrt{2x+9}=\sqrt{5-4x}\)

<=> 2x + 9 = 5 - 4x 

<=> 2x + 4x = 5 - 9

<=> 6x = -4

<=> x = \(\dfrac{-4}{6}=\dfrac{-2}{3}\)