Câu hỏi của Ánh Dương

Question

1.Chứng minh rằng $\frac{1}{x}+\frac{1}{y}\le-2$ biết $x^3+y^3+3\left(x^2+y^2\right)+4\left(x+y\right)+4=0$ và xy>0

2. cho x, y, z thỏa mãn \(\lef...

Nguyễn Việt Lâm · Accepted Answer

$x^3+3x^2+3x+1+y^3+3y^3+3y+1+x+y+2=0$

$\Leftrightarrow\left(x+1\right)^3+\left(y+1\right)^3+x+y+2=0$

$\Leftrightarrow\left(x+y+2\right)\left(\left(x+1\right)^2+\left(y+1\right)^2-\left(x+1\right)\left(y+1\right)\right)+\left(x+y+2\right)=0$

$\Leftrightarrow\left(x+y+2\right)\left(\left(x+1\right)^2+\left(y+1\right)^2-\left(x+1\right)\left(y+1\right)+1\right)=0$

$\Leftrightarrow x+y+2=0$

(phần trong ngoặc $\left(x+1\right)^2-\left(x+1\right)\left(y+1\right)+\frac{\left(y+1\right)^2}{4}+\frac{3\left(y+1\right)^2}{4}+1$

$=\left(x+1-\frac{y+1}{4}\right)^2+\frac{3\left(y+1\right)^2}{4}+1$ luôn dương)

$\Rightarrow x+y=-2$

Mà $xy>0\Rightarrow\left\{{}\begin{matrix}x< 0\y< 0\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}-x>0\-y>0\end{matrix}\right.$

Ta có: $\frac{1}{-x}+\frac{1}{-y}\ge\frac{4}{-\left(x+y\right)}=2$ $\Leftrightarrow\frac{1}{x}+\frac{1}{y}\le-2$ (đpcm)

Dấu "=" xảy ra khi và chỉ khi $x=y=-1$

Câu trả lời:

$x^3+3x^2+3x+1+y^3+3y^3+3y+1+x+y+2=0$

$\Leftrightarrow\left(x+1\right)^3+\left(y+1\right)^3+x+y+2=0$

$\Leftrightarrow\left(x+y+2\right)\left(\left(x+1\right)^2+\left(y+1\right)^2-\left(x+1\right)\left(y+1\right)\right)+\left(x+y+2\right)=0$

$\Leftrightarrow\left(x+y+2\right)\left(\left(x+1\right)^2+\left(y+1\right)^2-\left(x+1\right)\left(y+1\right)+1\right)=0$

$\Leftrightarrow x+y+2=0$

(phần trong ngoặc $\left(x+1\right)^2-\left(x+1\right)\left(y+1\right)+\frac{\left(y+1\right)^2}{4}+\frac{3\left(y+1\right)^2}{4}+1$

$=\left(x+1-\frac{y+1}{4}\right)^2+\frac{3\left(y+1\right)^2}{4}+1$ luôn dương)

$\Rightarrow x+y=-2$

Mà $xy>0\Rightarrow\left\{{}\begin{matrix}x< 0\y< 0\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}-x>0\-y>0\end{matrix}\right.$

Ta có: $\frac{1}{-x}+\frac{1}{-y}\ge\frac{4}{-\left(x+y\right)}=2$ $\Leftrightarrow\frac{1}{x}+\frac{1}{y}\le-2$ (đpcm)

Dấu "=" xảy ra khi và chỉ khi $x=y=-1$

Nguyễn Việt Lâm · Answer

2/ $x;y;z\ne0$

$\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$

$\Leftrightarrow\frac{x+y}{xy}+\frac{1}{z}-\frac{1}{x+y+z}=0$

$\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{xz+yz+z^2}=0$

$\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{xz+yz+z^2}\right)=0$

$\Leftrightarrow\left(x+y\right)\left(\frac{xy+yz+xz+z^2}{xyz\left(x+y+z\right)}\right)=0$

$\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0$

$\Leftrightarrow\left[{}\begin{matrix}x=-y\y=-z\z=-x\end{matrix}\right.$ dù trường hợp nào thì thay vào ta đều có $B=0$

3/ $\Leftrightarrow mx-2x+my-y-1=0$

$\Leftrightarrow m\left(x+y\right)-\left(2x+y+1\right)=0$

Gọi $A\left(x_0;y_0\right)$ là điểm cố định mà d đi qua

$\Leftrightarrow\left\{{}\begin{matrix}x_0+y_0=0\2x_0+y_0+1=0\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}x_0=-1\y_0=1\end{matrix}\right.$

Vậy d luôn đi qua $A\left(-1;1\right)$ với mọi m