Question

Tìm nghiệm của đa thức: \(Q\left(x\right)=\dfrac{1}{3}x^2+\dfrac{2}{5}x-\dfrac{11}{15}\)

Answer 1

Ta có: \(Q\left(x\right)=0\Leftrightarrow\dfrac{1}{3}x^2+\dfrac{2}{5}x-\dfrac{11}{15}=0\)

\(\Leftrightarrow\dfrac{1}{3}x^2-\dfrac{1}{3}x+\dfrac{11}{15}x-\dfrac{11}{15}=0\)

\(\Leftrightarrow\dfrac{1}{3}x\left(x-1\right)+\dfrac{11}{15}\left(x-1\right)=0\)

\(\Leftrightarrow\left(\dfrac{1}{3}x+\dfrac{11}{15}\right)\left(x-1\right)=0\)

\(\Leftrightarrow\dfrac{1}{3}x+\dfrac{11}{15}=0\) hoặc x - 1 = 0

\(\Leftrightarrow x=-\dfrac{11}{5}\) hoặc x = 1

Vậy: ...

Answer 2

x=1 ; x = -11/5

Answer 3

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