3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)

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

4x.(x+1)-8(x+1)=0

(4x-8)(x+1)=0

suy ra x=2 hoặc x=-1

1) \(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow4x^2+4x=8x+8\Leftrightarrow4x^2-4x-8=0\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

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

a) \(\sqrt{x}\left(\sqrt{x}-1\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

b) \(\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\\sqrt{x}+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\\sqrt{x}=-3\left(vôlí\right)\end{cases}}\)

c) \(\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+1=0\\\sqrt{x}+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=-1\left(vôlí\right)\\\sqrt{x}=-3\left(vôlí\right)\end{cases}}\)

b

\(\left|6+x\right|\ge0;\left(3+y\right)^2\ge0\Rightarrow\left|6+x\right|+\left(3+y\right)^2\ge0\)

Suy ra \(\left|6+x\right|+\left(3+y\right)^2=0\)\(\Leftrightarrow\hept{\begin{cases}6+x=0\\3+y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-6\\y=-3\end{cases}}\)

a

Ta có:\(\left|3x-12\right|=3x-12\Leftrightarrow3x-12\ge0\Leftrightarrow3x\ge12\Leftrightarrow x\ge4\)

\(\left|3x-12\right|=12-3x\Leftrightarrow3x-12< 0\Leftrightarrow3x< 12\Leftrightarrow x< 4\)

Với \(x\ge4\) ta có:

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

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\left(KTMĐK\right)\)

Với  \(x< 4\) ta có:

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

\(\Rightarrow10=x\left(KTMĐK\right)\)

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

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

\(\text{1 , ( x - 3 ) . ( 4 - x ) = 0}\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\4-x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\in Z\\x=4\in Z\end{cases}}\)

vậy______

\(2,\left(x-5\right)\left(x^2+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x^2+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\in Z\\x\in\varnothing\end{cases}}\)

vậy x = 5

3, ( x + 1 ) + ( x + 2 ) + (x + 3 ) + ... +( x + 99 ) = 0

(x+x+x+....+x)+(1+2+3+.....+99) = 0

(x.99) + 5050 = 0

x.99 = 0-5050

x.99 = -5050

x = -5050 : 99

x = \(\frac{-5050}{99}\notin Z\Rightarrow x\in\varnothing\)

vậy_____

1, suy ra x-3=0 suy ra x=3

  suy ra  4-x=0 suy ra x=4