a: \(x^4-2x^3-25x^2+50x=0\)

=>\(x^3\left(x-2\right)-25x\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x^3-25x\right)=0\)

=>x(x-2)(x^2-25)=0

=>x(x-2)(x+5)(x-5)=0

=>x∈{0;2;-5;5}

b: \(x^2\left(x-1\right)-4x^2+8x-4=0\)

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

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

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

=>x∈{1;2}

c: \(9x^2-4-2\left(3x-2\right)^2=0\)

=>(3x-2)(3x+2)-(3x-2)(6x-4)=0

=>(3x-2)(3x+2-6x+4)=0

=>(3x-2)(-3x+6)=0

=>(x-2)(3x-2)=0

=>x∈{2;2/3}

d: \(9x^2+90x+225-\left(x-7\right)^2=0\)

=>\(\left(3x+15\right)^2-\left(x-7\right)^2=0\)

=>(3x+15-x+7)(3x+15+x-7)=0

=>(2x+22)(4x+8)=0

=>2(x+11)*4*(x+2)=0

=>(x+11)(x+2)=0

=>x∈{-11;-2}

e: \(x^3-8+\left(x-2\right)\left(x+1\right)=0\)

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

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

=>\(\left(x-2\right)\left(x^2+3x+5\right)=0\)

mà \(x^2+3x+5=x^2+3x+\frac94+\frac{11}{4}=\left(x+\frac32\right)^2+\frac{11}{4}>0\forall x\)

nên x-2=0

=>x=2

g: (x+1)(x+2)(x+3)(x+4)-24=0

=>\(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

=>\(\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-24=0\)

=>\(\left(x^2+5x\right)\left(x^2+5x+10\right)=0\)

mà \(x^2+5x+10=x^2+5x+\frac{25}{4}+\frac{75}{4}=\left(x+\frac52\right)^2+\frac{75}{4}\ge\frac{75}{4}>0\forall x\)

nên \(x^2+5x=0\)

=>x(x+5)=0

=>x∈{0;-5}

a) \(2x^2-2x-x^2+6=0\) 

\(\Leftrightarrow x^2-2x+1+5=0\)

\(\Leftrightarrow\left(x-1\right)^2=-5\) ( vô lý)

Vậy không có x thoả mãn \(2x.\left(x-1\right)-x^2+6=0\)

b) \(x^4-2x^2.\left(3+2x^2\right)+3x^2.\left(x^2+1\right)=-3\) 

\(\Leftrightarrow x^4-6x^2-4x^4+3x^4+3x^2+3=0\)

\(\Leftrightarrow3-3x^2=0\)

\(\Leftrightarrow3x^2=3\Leftrightarrow x^2=1\) \(\Leftrightarrow x\in\left\{-1;1\right\}\)

Vậy \(x\in\left\{-1;1\right\}\)

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

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

\(\Leftrightarrow x^3+1-2x-x^3+4x=0\)

\(\Leftrightarrow1+2x=0\Leftrightarrow x=\dfrac{-1}{2}\)

Vậy x=\(\dfrac{-1}{2}\)

d) \(\left(x+3\right).\left(x^2-3x+9\right)-x.\left(x-2\right).\left(x+2\right)=15\)

\(\Leftrightarrow x^3+27-x.\left(x^2-4\right)-15=0\)

\(\Leftrightarrow x^3-27-x^3+4x-15=0\)

\(\Leftrightarrow4x-42=0\)

\(\Leftrightarrow x=10,5\)

Vậy x=10,5

\(a,=x\left(x^2+9\right)\\ b,=\left(3x+4\right)\left(x^2+9\right)\\ c,=a\left(a+b\right)-M\left(a+b\right)=\left(a-M\right)\left(a+b\right)\\ d,=\left(x+1\right)^2-y^2=\left(x+y+1\right)\left(x-y+1\right)\)

a.x^3-1^3

b.x^3-5^3

c)(2x)^3+3^3

d)x^3+1/2^3

1. (x - 1)^3 + 3.(x - 3)^2 - (x + 2).(x^2 - 2x + 4) = (x + 2)^3 - (x - 3).(x^2 + 9) - 6x^2 + 5 
<=> x^3 - 3x^2 + 3x - 1 + 3(x^2 - 6x + 9) - (x^3 + 2^3) 
= x^3 + 6x^2 + 12x + 8 - (x^3 - 3x^2 + 9x -27) - 6x^2 + 5 
<=> x^3 - 3x^2 + 3x - 1 + 3x^2 - 18x + 27 - x^3 - 8 
= x^3 + 6x^2 + 12x + 8 - x^3 + 3x^2 - 9x + 27 - 6x^2 + 5 
<=> 3x - 18x -12x - 3x^2 + 9x = 27 + 5 + 8 + 8 + 1 - 27 
<=> - 3x^2 - 18x - 22 = 0 
<=> 3x^2 + 18x + 22 = 0 

Nửa chu vi mảnh đất là: 

                                               120 : 2 = 60 (m)

Chiều dài hơn chiều rộng là:

                                               5 + 5 = 10 (m)

Chiều rộng là:

                                          ( 60 - 10 ) : 2 = 25 (m)

Chiều dài là:

                                                25 + 10 = 35 (m)

Diện tích là:

                                               25  35 = 875 ( )

1,2,3,4 không tính được.

`5)(2x-1/2)^2`

`=(2x)^2-2+(1/2)^2`

`=4x^2-2+1/4`

`6)(x+1/4)^2`

`=x^2+1/2x+1/16`

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

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

Bài 1: 

b: \(3x-6=x^2-16\)

\(\Leftrightarrow x^2-3x-10=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)

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