a) x² + 2x +1 

= (x + 1)²

b) 9x² + y² + 6xy

= (3x + y)²

c) 25a² + 4b² - 20ab

= (5a - 2b)²

d) x² - x + 1/4

= (x - 1/2)²

ta có \(a^2-b^2=\left(a+b\right)\left(a-b\right)\) => \(\frac{a^2-b^2}{a-b}=a+b\)

        \(a^4-b^4=\left(a^2-b^2\right)\left(a^2+b^2\right)\)=> \(\frac{a^4-b^4}{a^2-b^2}=a^2+b^2\) 

        \(a^8-b^8=\left(a^4-b^4\right)\left(a^4+b^4\right)\) => \(\frac{a^8-b^8}{a^4-b^4}=a^4+b^4\)

        ...............................................................................................

        \(a^{64}-b^{64}=\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)\) => \(\frac{a^{64}-b^{64}}{a^{32}-b^{32}}=a^{32}+b^{32}\)

thay vào ta được 

\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)......\left(a^{32}+b^{32}\right)\)

\(=\frac{a^2-b^2}{a-b}.\frac{a^4-b^4}{a^2-b^2}.\frac{a^8-b^8}{a^4-b^4}.............\frac{a ^{64}-b^{64}}{a^{32}-b^{32}}\)

\(=\frac{a^{64}-b^{64}}{a-b}\)

mà a-b= 1 nên \(\frac{a^{64}-b^{64}}{a-b}=a^{64}-b^{64}\)

a) \(a\left(a-6\right)+10=a^2-6a+10\)

                                      \(=a^2-6a+9+1\)

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

vì \(\left(a-3\right)^2\ge0\) với mọi a nên \(\left(a-3\right)^2+1>0\) hay \(a\left(a-6\right)+10>0\)

b) \(\left(x-3\right)\left(x-5\right)+4\)

\(=x^2-8x+15+4\)

\(=x^2-8x+16+3\)

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

vì \(\left(x-4\right)^2\ge0\) với mọi x nên \(\left(x-4\right)^2+3>0\) hay \(\left(x-3\right)\left(x-5\right)+4>0\)

Ta có: \(4x^2-x-\frac{3}{16}=4x^2-2.2x.\frac{1}{4}+\left(\frac{1}{4}\right)^2-\left(\frac{1}{4}\right)^2-\frac{3}{16}\)

\(=\left(2x-\frac{1}{4}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Vậy Min = -1/4 khi 2x - 1/4 = 0 => 2x = 1/4 => x = 1/8