a) x^2+ 2x + 1

= x^2+2.x.1+1^2

=(x+1)^2
b) x^2 + 6x + 9

= x^2+2.x.3+3^2

=(x+3)^2
c) x^2 - 6x +9

=x^2-2.x.3+3^2

=(x-3)^2
d) x^2 - 2x + 1

=x^2-2.x.1+1^2

=(x-1)^2
e) 4x^2 + 4xy + y^2

=(2x)^2+2.2x.y+y^2

=(2x+y)^2

a) x^2 - x = x - 1

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

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

\(\left(x-1\right)^2=0=>x-1=0=>x=1\)
b) (x^2 - 36) - (x+6) = 0

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

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

=> (x-6)(x+6-1) =0

=> (x-6)(x+5)=0

=> x=6 hoặc x= (-5)
c) (2x-1)^2 - (4x^2 - 1)= 0

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

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

=> (2x-1)(2x-1-2x-1)=0

=> (2x-1)(-2)=0

=> 2x-1=0 => 2x=1 => x= \(\dfrac{1}{2}\)
d) x^2(x^2 - 4) - (x^2 - 4 ) = 0

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

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

=> x=2;-2;1 hoặc (-1)

a) x^2 + 2x + 1

=\(x^2+2.x.1+1^2\)

\(=\left(x+1\right)^2\)
b) x^2 + 6x + 9

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

\(=\left(x+3\right)^2\)
c) x^2 - 6x + 9

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

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

\(=x^2-2.x.1+1^2\)

\(=\left(x-1\right)^2\)
e ) 4x^2 + 4xy +y^2

\(=\left(2x\right)^2+2.2x.y+y^2\)

\(=\left(2x+y\right)^2\)
f) x^2 + 4xy + 4y^2

\(=x^2+2.x.2y+\left(2y\right)^2\)

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

b,\(x^2-10x=-25\)

=> \(x^2-2.x.5+5^2=0\)

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

a, \(x^3-0,25x=0\)

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

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

=> \(\left[{}\begin{matrix}x=0\\x-0,5=0\\x+0,5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=0,5\\x=\left(-0,5\right)\end{matrix}\right.\)

b,\(x^2-10x=-25\)

x(x-10)=(-25)

\(\left[{}\begin{matrix}x=0\\x-10=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)

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

\(9^2-\left(3x+2\right)^2=\left(9-3x-2\right)\left(9+3x+2\right)=\left(7-3x\right)\left(11+3x\right)\)

Ta có :

\(\left(x-y\right)^3=2^3=8\)

=> \(x^3-2x^2y+2x^2y-y^3=8\)

=>\(x^3-y^3-2xy\left(x-y\right)=8\)

=>\(x^3-y^3-2.48.2=8\)

=>\(x^3-y^3-192=8=>x^3-y^3=200\)

Vậy ...

A=9 rồi kìa

Xem lại đề bài bạn ơi

b, \(\left(x-2\right)^2-x^2+4=0\)

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

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

=> -4x+4+4=0

=> 8-4x=0

=> 4(2-x)=0

=> 2-x=0=> x=2

Vậy x=2