\(x^3+3x^2+3x+1+y^3+3y^3+3y+1+x+y+2=0\)

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

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

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

\(\Leftrightarrow x+y+2=0\)

(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 \(x=y=-1\)

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

ĐKXĐ: ...

\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{x}+y+\frac{1}{y}=-4\\x\left(y+\frac{1}{y}\right)+\frac{1}{x}\left(y+\frac{1}{y}\right)=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)=-4\\\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)=4\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+\frac{1}{x}=a\\y+\frac{1}{y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=-4\\ab=4\end{matrix}\right.\)

Theo Viet đảo, a và b là nghiệm:

\(t^2+4t+4=0\Rightarrow t=-2\)

\(\Rightarrow\left\{{}\begin{matrix}x+\frac{1}{x}=-2\\y+\frac{1}{y}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\)

a) \(\text{ }x^4+y^4\ge x^3y+xy^3\)

\(\Leftrightarrow x^4+y^4-x^3y-xy^3\ge0\)

\(\Leftrightarrow x^3\left(x-y\right)-y^3\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)(ĐPCM) 

*NOTE: chứng minh đc vì (x-y)^2  >= 0 ;  x^2  +xy +y^2 > 0

mình cũng làm đến nơi rồi nhưng sợ x^2+xy+y^2 chưa chắc lớn hơn 0 thanks bạn nhé

ta có \(\left(x-y\right)^2\ge0\)

<=> \(x^2+y^2\ge2xy\)

<=>\(x^2+y^2+2xy\ge4xy\)

<=>\(\left(x+y\right)^2\ge4xy\)

<=>\(\frac{x+y}{xy}\ge\frac{4}{x+y}\)

<=>\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

ĐKXĐ: ...

\(\left\{{}\begin{matrix}x+\frac{1}{x}+y+\frac{1}{y}=-4\\x\left(y+\frac{1}{y}\right)+\frac{1}{x}\left(y+\frac{1}{y}\right)=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{x}+y+\frac{1}{y}=-4\\\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)=4\end{matrix}\right.\)

Theo Viet đảo, \(x+\frac{1}{x}\) và \(y+\frac{1}{y}\) là nghiệm của:

\(t^2+4t+4=0\Rightarrow t=-2\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=-2\\y+\frac{1}{y}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x+1=0\\y^2+2y+1=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

Tham khảo tại đây:

Câu hỏi của nguyen tan 12 - Toán lớp 8 - Học toán với OnlineMath

Chỉ cần đặt \(x=a^2;y=b^2\)