Question

cho a+b=1 va a^2+b^2=13 . Tính giá trị biểu thức C=a^3+b^3. mk cảm ơn trc ha

Answer 1

\(C=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=a^2-ab+b^2=13-ab\)

Có : \(a^2+2ab+b^2=\left(a+b\right)^2\)

\(\Leftrightarrow13+2ab=1^2\)

\(\Leftrightarrow2ab=-12\)

\(\Leftrightarrow ab=-6\)

\(\Leftrightarrow C=16-\left(-6\right)=13+6=19\)

Answer 2

\(a+b=1\Rightarrow\left(a+b\right)^2=1\)\(\Rightarrow a^2+b^2+2ab=1\)

Mà \(a^2+b^2=13\Rightarrow ab=-6\)

\(\Rightarrow a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)=1.\left[13-\left(-6\right)\right]=19\)