a) (x+y)² + (x-y)²

= x² + 2xy + y² + x²  - 2xy+ y²

= 2x² + 2y²

b) 2(x-y)(x+y) + (x-y)² + (x+y)²

= (x+y + x-y)²

= 4x²

c) (x-y+z)² +  (z-y)² +2 (x-y+z) (y-z)

= (x-y+z)² + 2(x-y+z)(y-z) + (y-z)²

= (x-y+z+ y-z)²

=x²

x^2+6x+9

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

=(x+3)^2

x^2+x+1/4

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

=(x+1/2)^2

a) x^2+7x+12

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

=x(x+3)+4(x+3)

=(x+4)(x+3)

b)x^2-10x+16

=x^2-8x-2x=16

=(x^2-8x)-(2x-16)

=x(x-8)(x-2)

c)x^4+4

=x^4+4+4x^2-4x^2

=(x^2+2)^2-4x^2

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

\(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}=3\sqrt[3]{9}\)

Dấu \(=\)khi \(a=b=c=\sqrt[3]{3}\).

\(1.\left(x+4\right)^2-\left(x-1\right)\left(x+1\right)=16\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x=-1\Leftrightarrow x=-\frac{1}{8}\)

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

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

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

\(4.\left(3x-1\right)^2+\left(5x-2\right)^2-2\left(3x-1\right)\left(5x-2\right)=9\Leftrightarrow\left(3x-1-5x+2\right)^2=9\)

\(\Leftrightarrow\left(2x-1\right)^2=9\Leftrightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

5.\(\left(x-1\right)\left(x^2+x+1\right)-x\left(x-2\right)\left(x+2\right)=5\Leftrightarrow x^3-1-\left(x^3-4x\right)=5\)

\(\Leftrightarrow4x=6\Leftrightarrow x=\frac{3}{2}\)

6.\(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(x-2\right)\left(x+2\right)=2\)

\(\Leftrightarrow x^3-3x^2+3x-1-\left(x^3+27\right)+x^2-4=2\)

\(\Leftrightarrow-2x^2+3x-34=0\text{ vô nghiệm}\)

a) \(x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\)

\(\left(x^2+4x+4\right)\div\left(x+2\right)=x+2\)

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

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

c) \(x^3+6x^2+12x+8=x^3+3.x^2.2+3.x.2^2+2^3=\left(x+2\right)^3\)

\(\left(x^3+6x^2+12x+8\right)\div\left(x+2\right)=\left(x+2\right)^2\)