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1: \(C=\left(x-\dfrac{4xy}{x+y}+y\right):\left(\dfrac{x}{x+y}+\dfrac{y}{y-x}+\dfrac{2xy}{x^2-y^2}\right)\)
\(=\dfrac{\left(x+y\right)^2-4xy}{x+y}:\left(\dfrac{x}{x+y}-\dfrac{y}{x-y}+\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}\right)\)
\(=\dfrac{x^2+2xy+y^2-4xy}{x+y}:\dfrac{x\left(x-y\right)-y\left(x+y\right)+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{x^2-2xy+y^2}{x+y}:\dfrac{x^2-xy-xy-y^2+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{\left(x-y\right)^2}{x+y}\cdot\dfrac{x^2-y^2}{x^2-y^2}=\dfrac{\left(x-y\right)^2}{x+y}\)
2: \(\left(x^2-y^2\right)\cdot C=-8\)
=>\(\left(x-y\right)\left(x+y\right)\cdot\dfrac{\left(x-y\right)^2}{x+y}=-8\)
=>\(\left(x-y\right)^3=-8\)
=>x-y=-2
=>x=y-2
\(M=x^2\left(x+1\right)-y^2\left(y-1\right)-3xy\left(x-y+1\right)+xy\)
\(=\left(y-2\right)^2\left(y-2+1\right)-y^2\left(y-1\right)-3xy\left(-2+1\right)+xy\)
\(=\left(y-1\right)\left[\left(y-2\right)^2-y^2\right]+3xy+xy\)
\(=\left(y-1\right)\left(-4y+4\right)+4xy\)
\(=-4\left(y-1\right)^2+4y\left(y-2\right)\)
\(=-4y^2+8y-4+4y^2-8y\)
=-4
Ta có: \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}=\dfrac{y-x+x-z}{\left(x-y\right)\left(x-z\right)}\)\(=\dfrac{y-x}{\left(x-y\right)\left(x-z\right)}+\dfrac{x-z}{\left(x-y\right)\left(x-z\right)}\) \(=\dfrac{1}{z-x}+\dfrac{1}{x-y}\)
Tương tự:
\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)
\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)
\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)
\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)
\(\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{x+y}.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\right]:\dfrac{x^3+y^3}{x^2y^2}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{x+y}.\dfrac{x+y}{xy}\right].\dfrac{x^2y^2}{x^3+y^3}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{xy}\right].\dfrac{x^2y^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\dfrac{y^2+x^2+2xy}{x^2y^2}.\dfrac{x^2y^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\dfrac{\left(x+y\right)^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)
=\(=\dfrac{x+y}{x^2-xy+y^2}-\dfrac{x+y}{x^2-xy+y^2}=0\)
\(A=\dfrac{x^2+y^2}{xy}+\dfrac{2xy}{x^2+y^2}=\dfrac{x^2+y^2}{2xy}+\dfrac{x^2+y^2}{2xy}+\dfrac{2xy}{x^2+y^2}\)
\(A\ge\dfrac{2xy}{2xy}+2\sqrt{\left(\dfrac{x^2+y^2}{2xy}\right)\left(\dfrac{2xy}{x^2+y^2}\right)}=3\)
Dấu "=" xảy ra khi \(x=y\)
\(B=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{4xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{4xy}{\left(x+y\right)^2}-4\)
\(B=\dfrac{\left(x+y\right)^2}{4xy}+\dfrac{4xy}{\left(x+y\right)^2}+\dfrac{3}{4}.\dfrac{\left(x+y\right)^2}{xy}-4\)
\(B\ge2\sqrt{\dfrac{\left(x+y\right)^2.4xy}{4xy.\left(x+y\right)^2}}+\dfrac{3}{4}.\dfrac{4xy}{xy}-4=1\)
\(B_{min}=1\) khi \(x=y\)
Áp dụng BĐT cauchy ta có:\(\left\{{}\begin{matrix}x^2+y^2\ge2xy\\y^2+z^2\ge2yz\\x^2+z^2\ge2xz\end{matrix}\right.\)
\(P\le\dfrac{1}{4xy+4x+4}+\dfrac{1}{4yz+4y+4}+\dfrac{1}{4xz+4z+4}=\dfrac{1}{4}\left(\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+x+1}\right)\)
xét biểu thức \(\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{zx+z+1}=\dfrac{1}{xy+x+1}+\dfrac{x}{1+yx+x}+\dfrac{xy}{x+1+xy}=\dfrac{xy+x+1}{xy+x+1}=1\)do đó \(P\le\dfrac{1}{4}\)
dấu = xảy ra khi x=y=z=1
Trước tiên ta tính:
\(\dfrac{1}{x+xy+1}+\dfrac{1}{y+yz+1}+\dfrac{1}{z+zx+1}\)
Đặt: \(\left\{{}\begin{matrix}x=\dfrac{a}{b}\\y=\dfrac{b}{c}\\z=\dfrac{c}{a}\end{matrix}\right.\left(a,b,c\ne0\right)\)
Thì ta có: \(\dfrac{1}{\dfrac{a}{b}+\dfrac{a}{b}.\dfrac{b}{c}+1}+\dfrac{1}{\dfrac{b}{c}+\dfrac{b}{c}.\dfrac{c}{a}+1}+\dfrac{1}{\dfrac{c}{a}+\dfrac{c}{a}.\dfrac{a}{b}+1}\)
\(=\dfrac{bc}{ab+ac+bc}+\dfrac{ca}{ab+bc+ca}+\dfrac{ab}{ab+bc+ca}=1\)
Quay về bài toán ban đầu. Ta có:
\(P=\dfrac{1}{\left(x+2\right)^2+y^2+2xy}+\dfrac{1}{\left(y+2\right)^2+z^2+2yz}+\dfrac{1}{\left(z+2\right)^2+x^2+2xz}\)
\(=\dfrac{1}{x^2+4x+4+y^2+2xy}+\dfrac{1}{y^2+4y+4+z^2+2yz}+\dfrac{1}{z^2+4z+4+z^2+2xz}\)
\(=\dfrac{1}{\left(x-y\right)^2+4x+4xy+4}+\dfrac{1}{\left(y-z\right)^2+4y+4yz+4}+\dfrac{1}{\left(z-x\right)^2+4z+4zx+4}\)
\(\le\dfrac{1}{4x+4xy+4}+\dfrac{1}{4y+4yz+4}+\dfrac{1}{4z+4zx+4}\)
\(=\dfrac{1}{4}.\left(\dfrac{1}{x+xy+1}+\dfrac{1}{y+yz+1}+\dfrac{1}{z+zx+1}\right)=\dfrac{1}{4}\)
