|3x-2y| ≥ 0

|2z-5y| ≥ 0

|xy+yz+zx-174| ≥ 0

=> |3x-2y|+|2z-5y|+|xy+yz+zx-174| ≥ 0

=> p ≥ 2017

vậy GTNN của p là 2017 

Bạn ghi lại đề đi bạn

\(-3ab.\left(a^2-3b\right)\)

\(=-3a^3b+9ab^2\)

\(\left(x^2-2xy+y^2\right)\left(x-2y\right)\)

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

\(=x^3-4x^2y+5xy^2-2y^3\)

a) \(-3ab.\left(a^2-3b\right)=-3ab.a^2+3ab.3b=-3a^3b+9ab^2\)

b) \(\left(x^2-2xy+y^2\right).\left(x-2y\right)=\left(x-2y\right).x^2-\left(x-2y\right).2xy+\left(x-2y\right).y^2\)

\(=xx^2-2yx^2-2xyx+2xy2y+xy^2-2yy^2\)

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

\(=x^3-4yx^2+5xy^2-2y^3\)

mk chỉ có thể thu gọn đc thôi, mk ko tính đc đâu!

\(\left(xy+yz+zx\right)^2\ge3xyz\left(x+y+z\right)=9\Rightarrow xy+yz+zx\ge3\)

\(2\left(x^2+y^2\right)-xy\ge\left(x+y\right)^2-\dfrac{1}{4}\left(x+y\right)^2=\dfrac{3}{4}\left(x+y\right)^2\)

Tương tự và nhân vế với vế:

\(VT\ge\dfrac{27}{64}\left[\left(x+y\right)\left(y+z\right)\left(z+x\right)\right]^2\)

Mặt khác ta có:

\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)

\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)

\(\ge\left(x+y+z\right)\left(xy+yz+xz\right)-\dfrac{1}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)

\(=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\ge\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)}.\left(xy+yz+zx\right)\)

\(\Rightarrow VT\ge\dfrac{27}{64}.\dfrac{64}{81}.3\left(xy+yz+zx\right)^3\ge3^3=27\) (đpcm)

\(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{zx}{z+x}\Rightarrow\frac{xyz}{z\left(x+y\right)}=\frac{xyz}{x\left(y+z\right)}=\frac{xyz}{y\left(z+x\right)}\)

 \(\frac{xyz}{z\left(x+y\right)}=\frac{xyz}{x\left(y+z\right)}\Rightarrow z\left(x+y\right)=x\left(y+z\right)\Rightarrow xz+yz=xy+xz\Rightarrow yz=xy\Rightarrow z=x\)

CM tương tự ta cũng có : \(x=y;y=z\)

\(\Rightarrow x=y=z\) Thay vào B ta được :

\(B=\frac{x^3+y^3+z^3}{x^2y+y^2z+z^2x}=\frac{x^3+x^3+x^3}{x^2x+x^2x+x^2x}=\frac{3x^3}{3x^3}=1\)