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vì x2+y2+z2=1 mà x2+y2+z2>=xy+yz+xz suy ra 1>= xy+yz+xz
x2+y2+z2=1 suy ra (x-y)2=1-2xy-z2 ,(y-z)2=1-2yz-x2,(x-z)2=(x-z)2=1-2xz-y2
\(\sqrt{3}+\frac{1}{2\sqrt{3}}[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2]=\)
\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)(do (x-y)2=1-2xy-z2(y-z)2=1-2yz-x2,(x-z)2=(x-z)2=1-2xz-y2)
theo bdt cosi ta có:
\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)
\(\le\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2z\sqrt{2xy}+2y\sqrt{2xz}+2x\sqrt{2yz}\right)]\)
\(\le\sqrt{3}+\frac{1}{2\sqrt{3}}[3-3\sqrt[3]{\left(2z\sqrt{2xy}.2y\sqrt{2xz}.2x\sqrt{2yz}\right)}\)
\(=\sqrt{3}+\frac{\sqrt{3}}{2}[1-2\sqrt{2}.\sqrt[3]{xyz^2}]\)\(=\sqrt{3}\left(1+\frac{1}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)=\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\)
suy ra
\(\frac{x+y+z}{xy+yz+xz}\ge3.\sqrt[3]{xyz}\left(doxy+yz+xz\le1\right)\)
ta giả sử:
\(3\sqrt[3]{xyz}\ge\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\Leftrightarrow\sqrt{3}\ge\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\) mà \(\sqrt{3}>\frac{3}{2}\)
suy ra \(\frac{3}{2}\ge\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\)(luôn đúng) suy ra điều giả sử trên là đúng
hay \(3\sqrt[3]{xyz}\ge\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\)
mà \(\frac{x+y+z}{xy+yz+xz}\ge3.\sqrt[3]{xyz}\),\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)\(\le\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\)
suy ra \(\frac{x+y+z}{xy+yz+xz}\ge\)\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)
suy ra \(\frac{x+y+z}{xy+yz+xz}\ge\)\(\sqrt{3}+\frac{1}{2\sqrt{3}}[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2]\)(đpcm)
em mới có lớp 8, nếu em làm sai cho em xin lỗi nha anh
Xét \(pt(2):\) \(\left(2x+4y-1\right)\sqrt{2x-y-1}=\left(4x-2y-3\right)\sqrt{x+2y}\)
\(\Leftrightarrow\left(2x+4y-1\right)^2\left(2x-y-1\right)-\left(4x-2y-3\right)^2\left(x+2y\right)=0\)
\(\Leftrightarrow-8x^3+12x^2y+12x^2+44xy^2+8xy-3x-24y^3-32y^2-11y-1=0\)
\(\Leftrightarrow-\left(x-3y-1\right)\left(8x^2+12xy-4x-8y^2-8y-1\right)=0\)
\(\Rightarrow x=3y+1\) thay vào \(pt(1)\) ta có
\(pt\left(1\right)\Leftrightarrow\left(3y+1\right)^2-5y^2-8y=3\)
\(\Leftrightarrow\left(y-1\right)\left(2y+1\right)=0\Leftrightarrow\left[{}\begin{matrix}y=1\Leftrightarrow x=4\\y=-\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
Lời giải:
\((x+\sqrt{x^2+2})(y-1+\sqrt{y^2-2y+3})=2(*)\)
Nhân 2 vế của $(*)$ với $x-\sqrt{x^2+2}$ thu được:
\([x^2-(x^2+2)](y-1+\sqrt{y^2-2y+3})=2(x-\sqrt{x^2+2})\)
\(\Leftrightarrow y-1+\sqrt{y^2-2y+3}=\sqrt{x^2+2}-x\)
\(\Leftrightarrow x+y-1=\sqrt{x^2+2}-\sqrt{y^2-2y+3}(1)\)
Nhân 2 vế của $(*)$ với $y-1-\sqrt{y^2-2y+3}$ thu được:
\((x+\sqrt{x^2+2})[(y-1)^2-(y^2-2y+3)]=2(y-1-\sqrt{y^2-2y+3})\)
\(\Leftrightarrow x+\sqrt{x^2+2}=\sqrt{y^2-2y+3}-(y-1)\)
\(\Leftrightarrow x+y-1=\sqrt{y^2-2y+3}-\sqrt{x^2+2}(2)\)
Lấy \((1)+(2)\Rightarrow 2(x+y-1)=0\Rightarrow x+y-1=0\)
\(\Rightarrow x+y=1\)
Khi đó:
\(x^3+y^3+3xy=(x+y)^3-3xy(x+y)+3xy\)
\(=1^3-3xy.1+3xy=1\) (đpcm)