x³ - 19x - 30 

= x³ - 4x - 15x - 30

= x(x² - 4) - 15(x + 2)

= x(x - 2)(x + 2) - 15(x + 2)

= (x² - 2x) (x + 2) - 15(x + 2)

= (x + 2)(x² - 2x - 15)

= (x + 2)(x² - 5x + 3x - 15)

= (x + 2)(x - 5)(x + 3)

2bc(b + 2c) + 2ac(c - 2a) - 2ab(a + 2b) - 7abc

= 2b²c + 4bc² + 2ac² - 4a²c - 2ab(a + 2b) - 7abc

= 2b²c + abc + 4bc² + 2ac² - 4a²c - 8abc - 2ab(a + 2b)

= bc(2b + a) + 2c²(2b + a) - 4ac(a + 2b) - 2ab(a + 2b)

= (a + 2b)(bc + 2c² - 4ac - 2ab)

= (a + 2b)[c(b + 2c) - 2a(2c + b)]

= (a + 2b)(b + 2c)(c - 2a) 

\(pkkikkkkkk\min\limits_{kkkkk\max\limits_{ }kkkk\lim\limits_{\rightarrow}kkkk\sqrt{ }kkk\sqrt{ }\sqrt{ }\sqrt{ }\sqrt{ }\sqrt{ }\sqrt{ }\sqrt{ }\sqrt{ }\sqrt{ }k\sqrt{ }k\sqrt{ }\sqrt{ }\sqrt{ }k\sqrt{ }\sqrt{ }k\sqrt{ }k\sqrt{ }k\sqrt{ }\sqrt{ }\sqrt{ }\sqrt{ }k\sqrt{ }\sqrt{ }\sqrt{ }\sqrt{ }}\)

\(x^2+y^2+z^2=xy+yz+zx\)

\(2.\left(x^2+y^2+z^2\right)=2.\left(xy+yz+zx\right)\)

\(\Rightarrow2.\left(x^2+y^2+z^2\right)-2xy-2yz-2zx=0\)

\(\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

Ta có: \(VT\ge0\forall x;y;z\)( tự c/m. nếu b ko c/m được thì bảo mình )

Mà \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Leftrightarrow}}\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Leftrightarrow x=y=z}\)

Có \(x^{2014}+y^{2014}+z^{2014}=3\)

\(\Rightarrow3.x^{2014}=3\)

\(\Rightarrow x^{2014}=1\)

\(\Rightarrow x=1\)

\(\Rightarrow x=y=z=1\)

Có: \(P=x^{25}+y^4+z^{2015}\)

\(\Rightarrow P=1^{25}+1^4+1^{2015}\)

\(P=1+1+1\)

\(P=3\)

Vậy \(P=3\)

Tham khảo nhé~

Ta có: x²+y²+z²=xy+yz+zx

<=>2x²+2y²+2z²=2xy+2yz+2zx

<=>2x²+2y²+2z²-2xy-2yz-2zx=0

<=>(x²-2xy+y²)+(y²-2yz+z²)+(z²-2zx+x²)=0

<=>(x-y)²+(y-z)²+(z-x)²=0

Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{cases}\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0}\)

=>\(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Rightarrow x=y=z}\)

=>x²⁰¹⁴=y²⁰¹⁴=z²⁰¹⁴

Lại có: x²⁰¹⁴+y²⁰¹⁴+z²⁰¹⁴ = 3

=>3x²⁰¹⁴ = 3 => x²⁰¹⁴ = 1 => \(x=\pm1\)

=>\(x=y=z=\pm1\)

Thay x,y,z vào P rồi tính

Cái thứ 2 là b. (a^2+c^2) đúng ko bạn

đúng rồi nha

\(=\frac{15\left(x-y\right)^5}{5\left(x-y\right)^3}-\frac{10\left(x-y\right)^4}{5\left(x-y\right)^3}+\frac{20\left(x-y\right)^3}{5\left(x-y\right)^3}\)

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

sửa dòng cuối

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