Ta có: \(a^3+b^3+3\left(a^2+b^2\right)+4\left(a+b\right)+4=0\)
\(\Leftrightarrow a^3+b^3+3a^2+3b^2+4a+4b+4=0\)
\(\Leftrightarrow\left(a+1\right)^3+\left(b+1\right)^3+a+b+2=0\)
\(\Leftrightarrow\left(a+b+2\right)\left[\left(a+1\right)^2-\left(a+1\right)\left(b+1\right)+\left(b+1\right)^2\right]+\left(a+b+2\right)=0\)
\(\Leftrightarrow\left(a+b+2\right)\left[\left(a+1\right)^2-\left(a+1\right)\left(b+1\right)+\left(b+1\right)^2+1\right]=0\left(1\right)\)
Đặt \(a+1=x;b+1=y\)
Xét \(x^2-xy+y^2+1=x^2-2.x.\frac{y}{2}+\frac{y^2}{4}-\frac{y^2}{4}+y^2+1\)
\(=\left(x-\frac{y}{2}\right)^2+\frac{3}{4}y^2+1\)
Mà \(\hept{\begin{cases}\left(x-\frac{y}{2}\right)^2\ge0;\forall x,y\\\frac{3}{4}y^2\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(x-\frac{y}{2}\right)^2+\frac{3}{4}y^2\ge0;\forall x,y\)
\(\Rightarrow\left(x-\frac{y}{2}\right)^2+\frac{3}{4}y^2+1\ge1>0;\forall x,y\)
Hay \(\left(a+1\right)^2-\left(a+1\right)\left(b+1\right)+\left(b+1\right)^2+1>0\)
Từ đó\(\left(1\right)\)xảy ra \(\Leftrightarrow a+b+2=0\)
\(\Leftrightarrow a+b=-2\)Thay vào biểu thức M ta đuợc:
\(M=2018.\left(-2\right)^2=8072\)
Vậy ...
ơ bài nào v ...................
Cho 2 số a,b thỏa mãn \(a^3+b^3+3\left(a^2+b^2\right)+4\left(a+b\right)+4=0\)
Tính giá trị của biểu thức \(M=2018\left(a+b\right)^2\)
