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\(x+y=1\)\(\Leftrightarrow\hept{\begin{cases}x-1=-y\\y-1=-x\end{cases}}\)
Ta có: \(\frac{x}{y^3-1}-\frac{y}{x^3-1}=\frac{x}{\left(y-1\right)^3+3y\left(y-1\right)}-\frac{y}{\left(x-1\right)^3+3x\left(x-1\right)}\)
\(=\frac{x}{-x^3-3xy}-\frac{y}{-y^3-3xy}=\frac{x}{-x\left(x^2+3y\right)}-\frac{y}{-y\left(y^2+3x\right)}\)
\(=\frac{-1}{x^2+3y}+\frac{1}{y^2+3x}=\frac{-\left(y^2+3x\right)+\left(x^2+3y\right)}{\left(x^2+3y\right)\left(y^2+3x\right)}=\frac{-y^2-3x+x^2+3y}{x^2y^2+3x^3+3y^3+9xy}\)
\(=\frac{\left(x^2-y^2\right)-3\left(x-y\right)}{x^2y^2+3\left(x^3+y^3\right)+9xy}=\frac{\left(x-y\right)\left(x+y\right)-3\left(x-y\right)}{x^2y^2+3\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]+9xy}\)
\(=\frac{\left(x-y\right)-3\left(x-y\right)}{x^2y^2+3\left(1-3xy\right)+9xy}=\frac{-2\left(x-y\right)}{x^2y^2+3-9xy+9xy}=\frac{-2\left(x-y\right)}{x^2y^2+3}\)
\(\Rightarrow\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=\frac{-2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)( đpcm )
By Titu's Lemma we easy have:
\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)
\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)
\(=\frac{17}{4}\)
Mk xin b2 nha!
\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)
\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)
\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)
\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)
Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)
Ta có:
\(\left(y^2+y+1\right)\left(x^2+x+1\right)\)
\(=x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+x+y+1\)
\(=x^2y^2+x^2+y^2+2xy+2=x^2y^2+3\)
Ta lại có:
\(\left(y^2+y+1\right)-\left(x^2+x+1\right)=\left(y^2-x^2\right)+\left(y-x\right)\)
\(=\left(y-x\right)\left(x+y+1\right)=-2\left(x-y\right)\)
Theo đề bài ta có: (sửa đề luôn)
\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{x}{\left(y-1\right)\left(y^2+y+1\right)}-\frac{y}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-1}{y^2+y+1}+\frac{1}{x^2+x+1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{\left(y^2+y+1\right)-\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(y^2+y+1\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=-\frac{2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)
Nếu\(a^3+b^3+c^3=3abc\Rightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
Thật vậy:\(a+b+c=0\Rightarrow a+b=-c\\ \Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\Rightarrow a^3+b^3+c^3=3abc\)
Tương tự \(a=b=c\Rightarrow\orbr{\begin{cases}3abc=3a^3\\a^3+b^3+c^3=3a^3\end{cases}\Rightarrow a^3+b^3+c^3=3abc}\)
Áp dụng ta có:\(\orbr{\begin{cases}xy+yz+zx=0\\xy=yz=zx\Rightarrow x=y=z\end{cases}}\)
Khi x=y=z,ta có P=(1+1)(1+1)(1+1)=8
Khi xy+yz+zx=0,ta có:\(xy+yz=-zx\)
Tương tự:\(yz+zx=-xy\)
\(xy+zx=-yz\)
Ta có \(P=2+\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}=2+\frac{xz+yz}{z^2}+\frac{xy+xz}{x^2}+\frac{zy+xy}{y^2}\)\(=2-\left(\frac{z}{x}+\frac{x}{y}+\frac{y}{z}\right)\)\(=2-\frac{xy+yz+zx}{xyz}=2-\frac{0}{xyz}=2\)
Vậy P=8 khi x=y=z
P=2 khi xy+yz+zx=0
\(\frac{x}{y^3-1}-\frac{y}{x^3-1}\)
\(=\frac{1-y}{\left(y-1\right)\left(y^2+y+1\right)}-\frac{1-x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{-1}{y^2+y+1}+\frac{1}{x^2+x+1}\)
\(=\frac{-x^2-x-1+y^2+y+1}{\left(x^2+x+1\right)\left(y^2+y+1\right)}\)
\(=\frac{\left(y^2-x^2\right)+y-x}{x^2y^2+x^2y+x^2+xy^2+xy+x+y^2+y+1}\)
\(=\frac{\left(y-x\right)\left(y+x\right)+y-x}{x^2y^2+x^2y+xy^2+x^2+xy+y^2+x+y+1}\)
\(=\frac{y-x+y-x}{x^2y^2+xy\left(x+y\right)+x\left(x+y\right)+y^2+x+y+1}\)
\(=\frac{2\left(y-x\right)}{x^2y^2+xy+x+y^2+x+y+1}\)
\(=\frac{2\left(y-x\right)}{x^2y^2+x\left(y+1\right)+y^2+x+y+1}\)
\(=\frac{2\left(y-x\right)}{x^2y^2+\left(1-y\right)\left(y+1\right)+y^2+\left(x+y\right)+1}\)
\(=\frac{2\left(y-x\right)}{x^2y^2+1-y^2+y^2+1+1}\)
\(=\frac{2\left(y-x\right)}{x^2y^2+3}\)