\(a,x^3-3x^2+3x-9=0\\ \Leftrightarrow x^2\left(x-3\right)+3\left(x-3\right)=0\\\Leftrightarrow \left(x-3\right)\left(x^2+3\right)=0\\ \Leftrightarrow x-3=0\left(dox^2+3\ge3>0\right)\\ \Leftrightarrow x=3\)
Vậy...
\(b,x^2+3y^2+2xy+4y+2x+3=0\\ \Leftrightarrow\left(x^2+2xy+y^2\right)+2\left(x+y\right)+1+\left(2y^2+2y+2\right)=0\\ \Leftrightarrow\left(x+y+1\right)^2+2\left[\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0\)
Với mọi x;y thì \(\left(x+y+1\right)^2\ge0\\ 2\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{2}\ge\dfrac{3}{2}\)
\(\Rightarrow\left(x+y+1\right)^2+2\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
Do đó ko tìm đc gtri nào củax;y thoa mãn
Đúng 0
Bình luận (0)
