1) \(X^2-7x+6=0\)

\(\Leftrightarrow x^2-x-6x+6=0\)

\(\Leftrightarrow x\left(x-1\right)-6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-6\right)=0\)

\(\Leftrightarrow\) x - 1 = 0 hoặc x - 6 = 0

\(\Leftrightarrow\) x = 1 hoặc x = 6

Vậy tập nghiệm của phương trình là S = { 1 ; 6 }

2) \(x^4-4x^3+12x-9=0\)

\(\Leftrightarrow x^4-4x^3+3x^2-3x^2+12x-9=0\)

\(\Leftrightarrow x^2\left(x^2-4x+3\right)-3\left(x^2-4x+3\right)=0\)

\(\Leftrightarrow\left(x^2-4x+3\right)\left(x^2-3\right)=0\)

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

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

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

\(\Leftrightarrow\)x - 3 = 0 hoặc x - 1 = 0 hoặc x² - 3 = 0

\(\Leftrightarrow\)x = 3 hoặc x = 1 hoặc x =+ - \(\sqrt{3}\)

Vậy tập nghiệm của phương trình là S = { 3 ; 1 ; + - \(\sqrt{3}\) }

TUI có cách gọn hơn nek
1) x^2-x-6x+6=0
<=>(x^2-x)-(6x-6)=0
<=>x.(x-1)-6.(x-1)=0
<=>(x-1).(x-6)=0
1) x^4-x^3-3x^3+3x+9x-9=0
<=>(x^4-x^3)-(3x^3-3x)+(9x-9)=0
<=>x^3.(x-1)-3x.(x^2-1)+9.(x-1)=0
<=>x^3.(x-1)-3x.(x-1).(x+1)+9.(x-1)=0
<=>(x-1).[x^3-3x^2-3x+9]=0
<=>(x-1).[x^2.(x-3)-3.(x-3)]=0
<=>(x-1).(x-3).(x^2-3)=0

1,=\(x^2-3x-2x^2+6x=-x^2+3x\)

2,=\(3x^2-x-5+15x=3x^2+14x-5\)

3,=\(5x+15-6x^2-6x=-6x^2-x+15\)

4,=\(4x^2+12x-x-3=4x^2+11x-3\)

5: =>(x+5)^3=0

=>x+5=0

=>x=-5

6: =>(2x-3)^2=0

=>2x-3=0

=>x=3/2

7: =>(x-6)(x-10)=0

=>x=10 hoặc x=6

8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)

=>(x-4)^3=0

=>x-4=0

=>x=4

`a, <=> 5/3 . 3sqrt(x^2+2) + 3/2.2sqrt(x^2+2)-7sqrt6=sqrt(x^2+2)`

`= (5+3-1)sqrt(x^2+2)=7sqrt6`

`<=> 7sqrt(x^2+2)=7sqrt6`.

`<=> x^2+2=36`.

`<=> x^2=34`.

`<=> x=+-sqrt(34)`.

Vậy...

`b, sqrt(4x^2-12x+9)-6=0`

`<=> |2x-3|=6`.

`@ x >=3/2 <=> 2x-3=6.`

`<=> x=9/2 (tm)`.

`@x <3/2 <=> 3-2x=6`

`<=> 2x=-3`

`<=> x=-3/2.`

Vậy...

6) ĐKXĐ: \(x\le-6\)

\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)

\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)

Vậy \(x\le-6\)

7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)

\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)

\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)

Vậy \(x\ge\dfrac{2}{3}\)

8) ĐKXĐ: \(x\ge5\)

\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)

\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)

9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

a: ĐKXĐ: \(x\in R\)

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

=>|2x+3|=5

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

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{9\left(x-2\right)^2}=18\)

=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)

=>\(3\cdot\left|x-2\right|=18\)

=>\(\left|x-2\right|=6\)

=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2

\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)

=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)

=>\(4\sqrt{x-2}=40\)

=>\(\sqrt{x-2}=10\)

=>x-2=100

=>x=102(nhận)

d: ĐKXĐ: \(x\in R\)

\(\sqrt{4\left(x-3\right)^2}=8\)

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

=>|2x-6|=8

=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

e: ĐKXĐ: \(x\in R\)

\(\sqrt{4x^2+12x+9}=5\)

=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)

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

=>|2x+3|=5

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

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

f: ĐKXĐ:x>=6/5

\(\sqrt{5x-6}-3=0\)

=>\(\sqrt{5x-6}=3\)

=>\(5x-6=3^2=9\)

=>5x=6+9=15

=>x=15/5=3(nhận)

Lời giải:

1.

$x^3-x(x^2+2)=6$

$\Leftrightarrow -2x=6\Leftrightarrow x=-3$

2.

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

\(\Leftrightarrow (x-2)(x-5)=0\Rightarrow \left[\begin{matrix} x=2\\ x=5\end{matrix}\right.\)

3.

\(x^2+2x+1=0\Leftrightarrow (x+1)^2=0\Rightarrow x=-1\)

4.

\(4x^2-12x=-9\Leftrightarrow 4x^2-12x+9=0\)

\(\Leftrightarrow (2x)^2-2.2x.3+3^2=0\Leftrightarrow (2x-3)^2=0\)

\(\Rightarrow x=\frac{3}{2}\)

a) \(x^3-6x^2+12x-9=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-1=0\)

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

\(\Leftrightarrow x-2=1\Leftrightarrow x=3\)

b) \(8x^3+12x^2+6x-26=0\)

\(\Leftrightarrow8x^3+12x^2+6x+1-27=0\)

\(\Leftrightarrow\left(2x+1\right)^3=27\)

\(\Leftrightarrow2x+1=3\Leftrightarrow x=1\)

a: \(x^2+12x+36=0\)

\(\Leftrightarrow\left(x+6\right)^2=0\)

\(\Leftrightarrow x+6=0\)

hay x=-6

b: Ta có: \(x^2-1=0\)

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

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

c: Ta có: \(25x^2-9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

Lời giải:
a. $x^2+12x+36=0$

$\Leftrightarrow (x+6)^2=0$

$\Leftrightarrow x+6=0$

$\Leftrightarrow x=-6$

b.

$x^2-1=0$

$\Leftrightarrow (x-1)(x+1)=0$
$\Leftrightarrow x-1=0$ hoặc $x+1=0$

$\Leftrightarrow x=1$ hoặc $x=-1$

c. 

$25x^2-9=0$

$\Leftrightarrow (5x)^2-3^2=0$

$\Leftrightarrow (5x-3)(5x+3)=0$

$\Leftrightarrow 5x-3=0$ hoặc $5x+3=0$

$\Leftrightarrow x=\frac{3}{5}$ hoặc $x=-\frac{3}{5}$