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

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