Question

1. Cho x+y+z = 1, chứng minh x² + y² + z² \(\ge\) \(\dfrac{1}{3}\)

2. Giải bất phương trình sau:

\(\left|x^2-x+2\right|-3x-7>0\)

3. Tìm x biết : \(\left(9-x^2\right)^2-12x=1\)

Answer 1

1. Đặt : x = a + \(\dfrac{1}{3}\) ; y = b + \(\dfrac{1}{3}\) ; z = \(c+\dfrac{1}{3}\)

Ta có : x + y + z = 1

⇒ a + b + c = 0

Ta có : x² + y² + z² = ( a + \(\dfrac{1}{3}\))² + ( b + \(\dfrac{1}{3}\))² + ( c + \(\dfrac{1}{3}\))²

= a² + \(\dfrac{2}{3}a+\dfrac{1}{9}+b^2+\dfrac{2}{3}b+\dfrac{1}{9}+c^2+\dfrac{2}{3}c+\dfrac{1}{9}\)

= \(\dfrac{1}{3}+\dfrac{2}{3}\left(a+b+c\right)+a^2+b^2+c^2\)

= \(\dfrac{1}{3}+a^2+b^2+c^2\) ≥ \(\dfrac{1}{3}\)

Dâu "=" xảy ra khi và chỉ khi : a = b = c = 0 ⇔ x = y = z = \(\dfrac{1}{3}\)