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

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

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

\(\left(x+5\right)\left(x^2-5x+25\right)\)

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

\(=x^3+5^3\)

\(=x^3+125\)

3) \(27-y^3\)

\(=3^3-y^3\)

\(=\left(3-y\right)\left(9-3y+y^2\right)\)

mk ko bt lm

\(B=\left(x-y\right)^2+2\left(x-y\right)+1+2\left(y^2-4y+4\right)+8\) 

\(=\left(x-y+1\right)^2+2\left(y-2\right)^2+8\ge8\forall x;y\)

" = " <=> x - y + 1 = 0 ; y - 2 = 0 

<=> y = 2 ; x = 1 

\(\left(x+\frac{1}{2}\right)^2-\left(x+\frac{1}{2}\right)\left(x+6\right)=8\)

\(\Leftrightarrow\frac{\left(2x+1\right)^2}{4}-\frac{\left(2x+1\right)\left(x+6\right)}{2}=8\)

\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)\left(x+6\right)=32\)

\(\Leftrightarrow-22x-43=0\Leftrightarrow x=\frac{43}{-22}\)

\(a,a^2-6a+9\)

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

\(=\left(a-3\right)^2\)

\(b,\frac{1}{4}x^2+2xy^2+4y^4\)

\(=\left(\frac{1}{2}x\right)^2+2.\frac{1}{4}x.4y^2+\left(4y^2\right)^2\)

\(=\left(\frac{1}{2}x+4y^2\right)\)

câu c thì sao bạn