Question

Cho a,b t/m: \(\left\{{}\begin{matrix}a^3-a^2+a-5=0\\b^3-2b^2+2b+4=0\end{matrix}\right.\)

Tính \(a+b?\)

Answer 1

Đặt \(a-1=c\Rightarrow a=c+1\)

\(\Rightarrow\left(c+1\right)^3-\left(c+1\right)^2+c+1-5=0\)

\(\Leftrightarrow c^3+2c^2+2c-4=0\)

Cộng vế với vế pt chứa b:

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

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

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

Do: \(b^2-bc+c^2-2b+2c+2=\left(b-\dfrac{c}{2}-1\right)^2+\dfrac{3}{4}\left(c+\dfrac{2}{3}\right)^2+\dfrac{2}{3}>0\)

\(\Rightarrow b+c=0\)

\(\Rightarrow a+b-1=0\)

\(\Rightarrow a+b=1\)