Lời giải:

a. $\sqrt{x^2}=1$

$\Leftrightarrow |x|=1$

$\Leftrightarrow x=\pm 1$

b. $\sqrt{4x^2-4x+1}=3$

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

$\Leftrightarrow 2x-1=\pm 3$

$\Leftrightarrow x=-1$ hoặc $x=2$

3. ĐKXĐ: $x^2\geq 4$

$\sqrt{x^2-4}+\sqrt{x^2+4x+4}=0$

Do $\sqrt{x^2-4}\geq 0; \sqrt{x^2+4x+4}\geq 0$ với mọi $x\in$ ĐKXĐ nên để tổng của chúng bằng $0$ thì:

$\sqrt{x^2-4}=\sqrt{x^2+4x+4}=0$

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

$\Leftrightarrow x=-2$

4. 

PT \(\Leftrightarrow \left\{\begin{matrix} x-3\geq 0\\ x^2-4x+3=(x-3)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ x^2-4x+3=x^2-6x+9\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ 2x=6\end{matrix}\right.\Leftrightarrow x=3\)

Ý 1:

\(\sqrt{x^2}=1\\ \Leftrightarrow\left|x\right|=1\\ Vậy:x=1.hoặc.x=-1\\ S=\left\{\pm1\right\}\)

Ý 2:

\(\sqrt{4x^2-4x+1}=3\\ \Leftrightarrow\sqrt{\left(2x-1\right)^2}=3\\ \Leftrightarrow\left|2x-1\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ Vậy:S=\left\{-1;2\right\}\)

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

=>căn x+2=0

=>x+2=0

=>x=-2

4: =>\(\left\{{}\begin{matrix}x>=3\\x^2-4x+3=x^2-6x+9\end{matrix}\right.\Leftrightarrow x=3\)

c: Ta có: \(x^3+3x^2+3x-7=0\)

\(\Leftrightarrow x+1=2\)

hay x=1

b: Ta có: \(x\left(x-3\right)-4x+12=0\)

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

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

a: \(x^3-4x^2-x+4=0\)

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

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

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

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

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

=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

\(A\left(x\right)+B\left(x\right)=x^3+3x^2-4x-12+2x^3-3x^2+4x+1\)

                     \(=3x^3-11\)

a: =(x^2+x-6)(x^2+x-8)

=(x+3)(x-2)(x^2+x-8)

b: =(x^2+x)^2+4(x^2+x)-12

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

=(x^2+x+6)(x+2)(x-1)

c: =x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12

=(x-1)(x^3+3x^2+8x+12)

=(x-1)(x^3+2x^2+x^2+2x+6x+12)

=(x-1)(x+2)(x^2+x+6)

a)  \(x^3\)\(-\)\(\frac{1}{4}x\)\(=\)\(0\)

\(x\left(x^2-\frac{1}{4}\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2-\frac{1}{4}=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2=0,5^2\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=+-0,5\end{cases}}\)

Vậy .............................

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+2=0\\x-4=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-2\\x=4\end{cases}}\)\(\orbr{\begin{cases}x=\frac{-2}{3}\\x=4\end{cases}}\)

Vậy ................................

c)  \(x^2\)\(\left(x-3\right)\)\(+\)\(12\)\(-\)\(4x\)\(=\)\(0\)

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

\(\left(x^2-4\right)\left(x-3\right)\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x-3=0\end{cases}-4=0}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x=3\end{cases}=2^2}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=+-2\\x=3\end{cases}}\)

a)\(x^3-\frac{1}{4}x=0\)

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

\(\Leftrightarrow x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-\frac{2}{3}\end{cases}}}\)

\(4x\left(x^2-5x+3\right)=4x^3-20x^2+12x\)

=> Chọn A

a) x^2+4x+3=x^2+x+3x+3=x(x+1)+3(x+1)=(x+1)(x+3)

b) 4x^2+4x-3=4x^2+4x+1-4=(2x+1)^2-4=(2x+1-2)(2x+1+2)=(2x-1)(2x+3)

c) x^2-x-12=x^2-4x+3x-12=x(x-4)+3(x-4)=(x-4)(x+3)

d) 4x^4+4x^2y^2-8y^4=4(x^4+x^2y^2-2y^4)=4(x^4-x^2y^2+2x^2y^2-2y^4)=4(x^2-y^2)(x^2+2y^2)=4(x-y)(x+y)(x^2+2y^2)

a) \(x^2+4x+3\)

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

\(=\left(x^2+x\right)+\left(3x+3\right)\)

\(=x\left(x+1\right)+3\left(x+1\right)\)

\(=\left(x+1\right)\left(x+3\right)\)

c) \(x^2-x-12\)

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

\(=\left(x^2-4x\right)+\left(3x-12\right)\)

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

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

\(x^2+4x+3\)

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

\(=x\left(x+1\right)+3\left(x+1\right)\)

\(=\left(x+1\right)\left(x+3\right)\)

Làmmmm

1)\(\left(x^2+x\right)^2+4x^2+4x-12\)

\(=\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)

\(=a^2+4a-12\)

\(=\left(a-2\right)\left(a+6\right)\)

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

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

2) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

\(=\left(a+1\right)\left(a+2\right)-12\)

\(=a^2+3a+2-12=a^2+3a-10\)

\(=\left(a-2\right)\left(a+5\right)\)

\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)

=\(\left(x-1\right)\left(x+2\right)\left(x^2+x+5\right)\)

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

\(=\left(x^2+5x+5\right)^2-1-24\)

\(=\left(x^2+5x+5\right)-5^2\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

\(=x\left(x+5\right)\left(x^2+5x+10\right)\)