Bài 1 : 

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

b, \(x^3+2x^2y+xy^2-9x=x\left(x^2+2xy+y^2-9\right)=x\left[\left(x+y\right)^2-9\right]=x\left(x+y-3\right)\left(x+y+3\right)\)

Bài 2 : 

a, \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)=8x^3-4x^2y+2xy^2-4x^2y+2xy^2-y^3\)

\(=8x^3-8x^2y+4xy^2-y^3\)

b, \(6x^5y^2-9x^4y^3:3x^3y^2=\left(6x^5y^2:3x^3y^2\right)+\left(-9x^4y^3:3x^3y^2\right)=2x^2-3xy\)

a, \(\frac{14x^5y^3z^2}{21x^2y^4z}=\frac{2x^3z}{3y}\)

b, \(\frac{25x^2y\left(x+1\right)^3}{30xy\left(x+1\right)}=\frac{5x\left(x+1\right)^2}{6}\)

c, \(\frac{30x\left(5-x\right)}{12\left(x-5\right)^3}=\frac{-30x\left(x-5\right)}{12\left(x-5\right)^3}=\frac{-5x}{2\left(x-5\right)^2}\)

d, \(\frac{60xy\left(3x-2\right)^3}{45xy^2\left(2-3x\right)}=\frac{60xy\left(3x-2\right)^3}{-45xy^2\left(3x-2\right)}=-\frac{4\left(3x-2\right)^2}{3y}\)

Bài làm

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2=\frac{\left(a+\frac{1}{b}\right)^2}{1}+\frac{\left(b+\frac{1}{a}\right)^2}{1}\ge\frac{\left(a+\frac{1}{b}+b+\frac{1}{a}\right)^2}{1+1}=\frac{\left(1+\frac{1}{a}+\frac{1}{b}\right)^2}{2}\)

Tiếp tục áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\frac{1}{a}+\frac{1}{b}\ge\frac{\left(1+1\right)^2}{a+b}=\frac{4}{1}=4\)

=> \(\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\ge\frac{\left(1+\frac{1}{a}+\frac{1}{b}\right)^2}{2}\ge\frac{\left(1+4\right)^2}{2}=\frac{25}{2}\)

=> \(\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\ge\frac{25}{2}\)( đpcm )

Đẳng thức xảy ra <=> a = b = 1/2

Bài làm

a) a² + 7a + 12 = a² + 3a + 4a + 12 = a( a + 3 ) + 4( a + 3 ) = ( a + 3 )( a + 4 )

b) 5x² + 20x + 15 = 5( x² + 4x + 3 ) = 5( x² + 3x + x + 3 ) = 5[ x( x + 3 ) + ( x + 3 ) ] = 5( x + 3 )( x + 1 )

a, \(a^2+7a+12=a^2+4a+3a+12=a\left(a+3\right)+4\left(a+3\right)\)

\(=\left(a+4\right)\left(a+3\right)\)

b, \(5x^2+20x+15=5\left(x^2+4x+3\right)=5\left(x^2+x+3x+3\right)\)

\(=5\left[x\left(x+1\right)+3\left(x+1\right)\right]=5\left(x+3\right)\left(x+1\right)\)