Ta có: a² + b² + c² + 3 = 2(a + b + c)

=> a² + b² + c² + 3 - 2(a + b + c) = 0

=> a² + b² + c² + 3 - 2a - 2b - 2c = 0

=> (a² - 2a + 1) + (b² - 2b + 1) + (c² - 2c + 1) = 0

=> (a - 1)² + (b - 1)² + (c - 1)² = 0

       <=> \(\hept{\begin{cases}a-1=0\\b-1=0\\c-1=0\end{cases}}\) <=> a = b = c = 1

=> đpcm

\(a^2+b^2+c^2+3=2\left(a+b+c\right)\)

\(\Leftrightarrow a^2+b^2+c^2+3-2\left(a+b+c\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2+1+1+1-2a-2b-2c=0\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a-1=0\\b-1=0\\c-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=1\\c=1\end{cases}\Leftrightarrow}a=b=c=1}\)

Ta có: \(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right).\)

Thật vậy: \(VP=\left(x-y\right)^3+3xy\left(x-y\right)=x^3-3x^2y+3xy^2-y^3+3x^2y-3xy^2=x^3-y^3=VT\)

Áp dụng định lý trên và thay \(x-y=7;\) \(xy=15\)vào \(x^3-y^3\)ta có: \(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right).\)

                                                                                                                                \(=7^3+3.15.7=343+315=658\)

\(P=x^2-5x\)

\(=x^2-5x+\frac{25}{4}-\frac{25}{4}\)

\(=\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\ge\frac{-25}{4}\forall x\)

Dấu"=" xảy ra<=> \(\left(x-\frac{5}{2}\right)^2=0\)

                   \(\Leftrightarrow x=\frac{5}{2}\)

\(P=x^2-5x\)

\(P=x^2-2x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}\)

\(P=\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\ge\frac{25}{4}\)

Dấu '' = '' xảy ra 

\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=0\)

\(\Leftrightarrow x=\frac{5}{2}\)

Vậy Min P = 25/4 <=> x = 5/2.

a) \(A=\left(x^2-\frac{1}{2}x\right)^2+\frac{3}{4}\left(x+\frac{2}{3}\right)^2+\frac{2}{3}>0\)

Ko biết xét khoảng:v