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\(x^3+y^3+1=3xy\)
\(\Leftrightarrow\left(x^3+3x^2y+3xy^2+y^3\right)+1=3xy+3x^2y+3xy^2\)
\(\Leftrightarrow\left(x+y\right)^3+1=3xy\left(1+x+y\right)\)
\(\Leftrightarrow\left(x+y+1\right)\left[\left(x+y\right)^2-\left(x+y\right)+1\right]=3xy\left(1+x+y\right)\)
\(\left(x+y+1\right)\left(x^2+y^2+2xy-x-y+1\right)-3xy\left(1+x+y\right)=0\)
\(\Leftrightarrow\left(x+y+1\right)\left(x^2+y^2-xy-x-y+1\right)=0\)
Với \(x+y+1\ne0\) thì \(x^2+y^2-xy-x-y+1=0\)
\(\Leftrightarrow x^2+y^2-xy-x-y+1=0\)
\(\Leftrightarrow2x^2+2y^2-2xy-2x-2y+2=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2=0\Rightarrow x=y=1\)(thỏa mãn \(x+y+1\ne0\))
\(\Rightarrow P=\left(1+\frac{x_0}{y_0}\right)\left(1+y_0\right)\left(1+\frac{1}{x_0}\right)=\left(1+\frac{1}{1}\right)\left(1+1\right)\left(1+\frac{1}{1}\right)=8\)
Trần Hoàng Việt thế này có đúng ko ạ?
\(\hept{\begin{cases}x=3\\y=3\end{cases}\Rightarrow}3=a.1\Rightarrow a=3\)
\(Px_o,y_o\in y=3x\Rightarrow y_o=3.x_o\)
\(P=\frac{x_o+1}{3x_o+1}=\frac{x_o+1}{3"x_o+1"}\)
\(\hept{\begin{cases}x_o=-1\Rightarrow P=kXD\\x_o\ne-1\Rightarrow P=\frac{1}{3}\end{cases}}\)
P/s: Ko chắc :D
Theo vi-et ta có: \(\hept{\begin{cases}x_1+x_2=-2\left(m+1\right)\\x_1x_2=2m^2+9m+7\end{cases}}\)
Theo đề bài ta có:
\(\left|\frac{7\left(x_1+x_2\right)}{2}-x_1x_2\right|\le18\)
\(\Leftrightarrow\left|\frac{7\left(-2\left(m+1\right)\right)}{2}-\left(2m^2+9m+7\right)\right|\le18\)
\(\Leftrightarrow\left|-2m-16m-14\right|\le18\)
Xét VT ta có:
| - 2m2 - 16m - 14| = | ( - 2m2 - 16m - 32) + 18|
= |- 2(m + 4)2 + 18| \(\le\)|18| = 18
Vì pt đã cho là pt bậc 2 \(\Rightarrow a\ne0\)
Do x0 là nghiệm \(\Rightarrow-ax_0^2=bx_0+c\)
\(\Rightarrow-x_0^2=\frac{b}{a}x_0+\frac{c}{a}\)
\(\Rightarrow\left|-x_0\right|^2=\left|\frac{b}{a}x_0+\frac{c}{a}\right|\le\left|\frac{b}{a}\right|\left|x_0\right|+\left|\frac{c}{a}\right|\le M\left|x_0\right|+M\)
\(\Rightarrow\left|x_0\right|^2-1< M\left(\left|x_0\right|+1\right)\)
\(\Rightarrow\left(\left|x_0\right|-1\right)\left(\left|x_0\right|+1\right)< M\left(\left|x_0\right|+1\right)\)
\(\Rightarrowđpcm\)