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\(1,\\ a,\dfrac{x^2}{x+1}+\dfrac{x}{x+1}=\dfrac{x^2+x}{x+1}=\dfrac{x\left(x+1\right)}{x+1}=x\)
\(b,\left(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}\right):\dfrac{x+y}{2x}=\left(\dfrac{4xy}{2\left(x-y\right)\left(x+y\right)}+\dfrac{\left(x-y\right)^2}{2\left(x-y\right)\left(x+y\right)}\right).\dfrac{2x}{x+y}=\dfrac{4xy+x^2-2xy+y^2}{2\left(x-y\right)\left(x+y\right)}.\dfrac{2x}{x+y}=\dfrac{2x\left(x^2+2xy+y^2\right)}{2\left(x-y\right)\left(x+y\right)^2}=\dfrac{2x\left(x+y\right)^2}{2\left(x-y\right)\left(x+y\right)^2}=\dfrac{x}{x-y}\)
Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)với a,b>0
Ta có: \(\frac{4xy}{z+1}=\frac{4xy}{2z+x+y}\le\frac{xy}{x+z}+\frac{xy}{y+z}\)
Tương tự: \(\frac{4yz}{x+1}\le\frac{yz}{x+y}+\frac{yz}{x+z}\)
\(\frac{4zx}{y+1}\le\frac{zx}{y+x}+\frac{zx}{y+z}\)
\(\Rightarrow4\left(\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}\right)\le\frac{xy}{x+z}+\frac{xy}{y+z}+\frac{yz}{x+y}+\frac{yz}{x+z}+\frac{zx}{y+x}+\frac{zx}{y+z}=x+y+z=1\)
\(\Rightarrow\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}\le\frac{1}{4}\)
Dấu "=" xảy ra khi: x=y=z>0
Bài 2:
+) Với y=0 <=> x=0
Ta có: 1-xy= 12 (đúng)
+) Với \(y\ne0\)
Ta có: \(x^6+xy^5=2x^3y^2\)
\(\Leftrightarrow x^6-2x^3y^2+y^4=y^4-xy^5\)
\(\Leftrightarrow\left(x^3-y^2\right)^2=y^4\left(1-xy\right)\)
\(\Rightarrow1-xy=\left(\frac{x^3-y^2}{y^2}\right)^2\)
a )\(2x\left(xy-3\right)+3xy\left(x+1-y\right)+3x\left(y^2-1\right)=2x^2y-6x+3x^2y+3xy-3xy^2+3xy^2-3x=5x^2y-9x+3xy\)
=> Phụ thuộc vào giá trị của biến
b) \(\left(x+2y\right)\left(x-2y\right)-x\left(x+4y^2\right)+5=x^2-4y^2-x^2-4xy^2+5=-4y^2-4xy^2+5\)
=> Phụ thuộc vào giá trị của biến
c) \(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(3x+2\right)=27x^3+8-9x^2+4=27x^3-9x^2+12\)
=> Phụ thuộc vào giá trị của biến
a: Ta có: \(2x\left(xy-3\right)+3xy\left(x-y+1\right)+3x\left(y^2-1\right)\)
\(=2x^2y-6x+3x^2y-3xy^2+3xy+3xy^2-3x\)
\(=5x^2y+3xy-9x\)
c: Ta có: \(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(3x+2\right)\)
\(=27x^3+8-9x^2+4\)
\(=27x^3-9x^2+12\)
Theo bđt Cauchy schwarz dạng Engel
\(P\ge\frac{\left(2x+2y+\frac{1}{x}+\frac{1}{y}\right)^2}{1+1}=\frac{\left[2\left(x+y\right)+\frac{1}{x}+\frac{1}{y}\right]^2}{2}\)
Ta có \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)(bđt phụ)
\(\Rightarrow P\ge\frac{\left[2.1+4\right]^2}{2}=\frac{36}{2}=18\)
Dấu ''='' xảy ra khi \(x=y=\frac{1}{2}\)
\(P=\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(2x+\dfrac{1}{x}+2y+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(2x+2y+\dfrac{4}{x+y}\right)^2=18\)
\(P_{min}=18\) khi \(x=y=\dfrac{1}{2}\)
\(\frac{1}{x}-\frac{1}{2y}=\frac{1}{2x+y}\)
=> \(\frac{2y-x}{2xy}=\frac{1}{2x+y}\)
=> (2y - x)(2x + y) = 2xy
=> 4xy + 2y2 - 2x2 - xy = 2xy
=> 2(y2- x2) = -xy
=> [2(y2 - x2)]2 = (-xy)2
=> 4(y2 - x2)2 = (xy)2
=> 4(y4 - 2(xy)2 + x4) = (xy)2
=> 4y4 - 8(xy)2 + 4x4 = (xy)2
=> 4(y4 + x4) = 9(xy)2
=> y4 + x4 = \(\frac{9}{4}\left(xy\right)^2\)
Khi đó \(\frac{x^2}{y^2}+\frac{y^2}{x^2}=\frac{x^4+y^4}{\left(xy\right)^2}=\frac{\frac{9}{4}\left(xy\right)^2}{\left(xy\right)^2}=\frac{9}{4}\)