Question

[Nguyễn Trung Kiên]

CHo 2 đa thức P(x)=x\(^2\)-2ax+a\(^2\)và Q(x)=x\(^2\)+(3a+1).x+a\(^2\)

Tìm a sao cho P(x)=Q(x)

Answer 1

[Huỳnh Quang Sang]

Để P(x) = Q(x)

Thì x² - 2ax + a² = x² + (3a + 1)x + a²

=> x² - 2ax + a² = x² + 3ax + x + a²

=> (x² - 2ax + a²) - (x² + 3ax + x + a²) = 0

=> x² - 2ax + a² - x² + 3ax - x - a² = 0

=> (x² - x²) + (-2ax + 3ax) + (a² - a²) - x = 0

=> ax - x = 0

=> x(a - 1) = 0

Vậy a = 1

Answer 2

[Nobi Nobita]

Để \(P\left(x\right)=Q\left(x\right)\)thì \(x^2-2ax+a^2=x^2+\left(3a+1\right).x+a^2\)

\(\Leftrightarrow-2ax=\left(3a+1\right).x\)\(\Leftrightarrow\left(3a+1\right).x+2ax=0\)

\(\Leftrightarrow\left(3a+1+2a\right).x=0\)\(\Leftrightarrow\left(5a+1\right).x=0\)

\(\Leftrightarrow5a+1=0\)\(\Leftrightarrow5a=-1\)\(\Leftrightarrow a=\frac{-1}{5}\)

Vậy \(a=\frac{-1}{5}\)