TA CÓ: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Do đó: \(\frac{a^3+b^3+c^3}{4abc}=\frac{3}{4}+\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{4abc}\)

\(=\frac{3}{4}+\frac{1}{4}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(\ge\frac{3}{4}+\frac{1}{4}.\frac{9}{ab+bc+ca}\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(=\frac{3}{4}+\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{9}{4}=\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{3}{2}\)

\(\Rightarrow P\ge\frac{1}{30}+\frac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}+\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}-\frac{3}{2}\)

\(=\frac{-22}{15}+\frac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}+\frac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}\)

\(\ge\frac{-22}{15}+2\sqrt{\left[\frac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}\right]\left[\frac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}\right]}=\frac{-22}{15}+\frac{2}{15}=\frac{-4}{3}\)

Dấu '=' xảy ra <=> a=b=c

Vậy GTNN của P là -4/3 khi a=b=c

trả lời hộ cái

2xy-x²-y²+16

=16-(x^2-2xy+y^2)

=16-(x-y)^2

=(4+x-y)(4(-x+y)

\(2x\left(x-1\right)-\left(x-1\right)\left(x-2\right)\)

\(=\left(x-1\right)\left(2x-x-2\right)\)

\(=\left(x-1\right)\left(x-2\right)\)

x^5+x^2+1

=(x^5-x)+(x^2+x+1)

=x(x^6-1)+(x^2+x+1)

=x(x^3+1)(x^3-1)+(x^2+x+1)

=x(x^3+1)(x-1)(x^2+x+1)+(x^2+x+1)

=(x^2+x+1)[x(x^3+1)(x-1)+1]

=(x^2+x+1)(x^5-x^4+x^2-x+1)

Ta có : ( 3x - 1 )( 2x + 7 ) - ( 12x³ + 8x² - 14x ) : 2x

= 6x² + 21x - 2x - 7 - [ ( 12x³ : 2x ) + ( 8x² : 2x ) - ( 14x : 2x ) ]

= 6x² + 19x - 7 - ( 6x² + 4x - 7 )

= 6x² + 19x - 7 - 6x² - 4x + 7

= 15x