Question

cho 3 số a b c thỏa mãn 3a-3b+c=0 và 6ab+2bc-3ac=0 tính P =(a-1)²⁰¹⁹+(b-1)²⁰²⁰+(c-1)²⁰²¹

 

Answer 1

Ta có: 

\(\left(3a-2b+c\right)^2=9a^2+4b^2+c^2+2\left(3ac-6ab-2bc\right)\)

\(\Rightarrow b^2=9a^2+4b^2+c^2\)

(vì \(3a-3b+c=0\Leftrightarrow3a-2b+c=-b\), \(6ab+2bc-3ac=0\))

\(\Leftrightarrow9a^2+3b^2+c^2=0\)

\(\Leftrightarrow a=b=c=0\). 

Khi đó: \(P=\left(-1\right)^{2019}+\left(-1\right)^{2020}+\left(-1\right)^{2021}=-1\)

Answer 2

Ta có: 

(3a−2b+c)2=9a2+4b2+c2+2(3ac−6ab−2bc)

⇒b2=9a2+4b2+c2

(vì 3a−3b+c=0⇔3a−2b+c=−b, 6ab+2bc−3ac=0)

⇔9a2+3b2+c2=0

⇔a=b=c=0. 

Khi đó: P=(−1)2019+(−1)2020+(−1)2021=−1