a)  \(4a^2+16xa+16x^2=\left(2a+4x\right)^2\)

b)  \(32x^2+32x+8=8\left(4x^2+4x+1\right)=8\left(2x+1\right)^2=\)

\(=\left[\sqrt{8}\left(2x+1\right)\right]^2=\left(4\sqrt{2}+2\sqrt{2}\right)^2\)

p/s: chúc bạn học tốt

Ta có : 

\(m^2-4m+4=m^2-2.m.2+2^2=\left(m-2\right)^2\)

\(m^2+6mn+9n^2=m^2+2.m.3n+\left(3n\right)^2=\left(m+3n\right)^2\)

m² - 4m + 4 = m² - 2.2.m + 2²

= ( m - 2 )²    đpcm

m² + 6mn + 9n² = m² + 2.m.3n + (3n)²

= ( m + 3n )²   đpcm

Vậy ................( đề bài  )

Giả sử : \(x^2+y^2\ge2xy\forall x;y\)

\(\Leftrightarrow x^2+y^2-2xy\ge0\forall x;y\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\forall x;y\)

( điều này luôn đúng ) 

\(\Rightarrow x^2+y^2\ge2xy\forall x;y\)

\(\Rightarrowđpcm\)

ta có (x+y)² _{> hoặc = với mọi x,y} 

=> x²+ 2xy+y² > hoặc = 0 nên x²+y²> hoặc bằng 2xy với mọi xy 

\(\left(1-x\right)\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\left(1+x^{32}\right)\left(1+x^{64}\right)\)

\(=\left(1-x^2\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\left(1+x^{32}\right)\left(1+x^{64}\right)\)

\(=\left(1-x^4\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\left(1+x^{32}\right)\left(1+x^{64}\right)\)

\(=\left(1-x^8\right)\left(1+x^8\right)\left(1+x^{16}\right)\left(1+x^{32}\right)\left(1+x^{64}\right)\)

\(=\left(1-x^{16}\right)\left(1+x^{16}\right)\left(1+x^{32}\right)\left(1+x^{64}\right)\)

\(=\left(1-x^{32}\right)\left(1+x^{32}\right)\left(1+x^{64}\right)\)

\(=\left(1-x^{64}\right)\left(1+x^{64}\right)\)

\(=1-x^{128}\)

\(\frac{x^2+\left(x-z\right)^2}{y^2}hay x^2+\frac{\left(x-z\right)^2}{y^2}\)

ban ghi ro de ra

\(C=\frac{30}{4x-4x^2-6}=\frac{-30}{4x^2-4x+6}=\frac{-30}{\left(2x-1\right)^2+5}\)

Vì \(\left(2x-1\right)^2\ge0\Rightarrow\left(2x-1\right)^2+5\ge5\Rightarrow\frac{1}{\left(2x-1\right)^2+5}\le\frac{1}{5}\Rightarrow C=\frac{-30}{\left(2x-1\right)^2+5}\ge\frac{-30}{5}=-6\)

Dấu "=" xảy ra khi x=1/2

Vậy Cmin=-6 khi x=1/2

\(E=\frac{1000}{x^2+y^2-20x-20y+2210}=\frac{1000}{\left(x-10\right)^2+\left(y-10\right)^2+2010}\)

Vì \(\left(x-10\right)^2\ge0;\left(y-10\right)^2\ge0\Rightarrow\left(x-10\right)^2+\left(y-10\right)^2\ge0\)

\(\Rightarrow\left(x-10\right)^2+\left(y-10\right)^2+2010\ge2010\)

\(\Rightarrow\frac{1}{\left(x-10\right)^2+\left(y-10\right)^2+2010}\le\frac{1}{2010}\)

\(\Rightarrow E=\frac{1000}{\left(x-10\right)^2+\left(y-10\right)^2+2010}\le\frac{1000}{2010}=\frac{100}{201}\)

Dấu "=" xảy ra khi x=y=10

Vậy Emax = 100/201 khi x=y=10