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Aki Tsuki Mysterious Person Phùng Khánh Linh Nhã DoanhQuoc Tran Anh Le Nguyễn Thị Ngọc Thơ lê thị hương giang giúp mình vs
a) \(x\ne yvàx;y>0\)
ta có : \(P=\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}-\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}-y\)
\(\Leftrightarrow P=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}-y\)
\(\Leftrightarrow P=\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}-\sqrt{y}\right)-y\)
\(\Leftrightarrow P=2\sqrt{y}-y\)
b) ta có : \(P-1=2\sqrt{y}-y-1=-\left(\sqrt{y}-1\right)^2\le0\)
\(\Rightarrow P\le1\)
bài này không thể chứng minh \(P< 1\) đc .
ĐKXĐ: x≠y,x>0,y>0
a) \(P=\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}-\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}-y=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}-\dfrac{x+2\sqrt{xy}+y-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}-y=\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}-y=\sqrt{x}+\sqrt{y}-\sqrt{x}+\sqrt{y}-y=2\sqrt{y}-y\)b) Ta có \(\left(\sqrt{y}-1\right)^2>0\Leftrightarrow y-2\sqrt{y}+1>0\Leftrightarrow1>2\sqrt{y}-y\Leftrightarrow P< 1\)
Có : \(x-2y-\sqrt{xy}+\sqrt{x}-2\sqrt{y}=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{x}-2\sqrt{y}=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+1\right)=0\)
\(\Leftrightarrow\sqrt{x}=2\sqrt{y}\) (Do \(\sqrt{x}+\sqrt{y}+1>0,\forall x;y>0\))
\(\Leftrightarrow x=4y\)
Khi đó \(P=\dfrac{7y}{\left(2\sqrt{y}+3\sqrt{y}\right).\left(\sqrt{x}+2\sqrt{y}\right)}\)
\(=\dfrac{7y}{5\sqrt{y}.4\sqrt{y}}=\dfrac{7}{20}\)
\(P=\left(\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)}{1-xy}\right):\left(\frac{x+y+2xy+1-xy}{1-xy}\right)\)
\(=\left(\frac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\right):\left(\frac{\left(x+1\right)\left(y+1\right)}{1-xy}\right)\)
\(=\frac{2\sqrt{x}\left(y+1\right)}{\left(1-xy\right)}.\frac{\left(1-xy\right)}{\left(x+1\right)\left(y+1\right)}=\frac{2\sqrt{x}}{x+1}\)
\(x=\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}-1\)
\(\Rightarrow P=\frac{2\left(\sqrt{3}-1\right)}{5-2\sqrt{3}}=\frac{2+6\sqrt{3}}{13}\)
Ta có \(1-P=1-\frac{2\sqrt{x}}{x+1}=\frac{x-2\sqrt{x}+1}{x+1}=\frac{\left(\sqrt{x}-1\right)^2}{x+1}\ge0\) \(\forall x\ge0\)
\(\Rightarrow1-P\ge0\Rightarrow P\le1\)
a) \(\sqrt{\dfrac{x-2\sqrt{x+1}}{x+2\sqrt{x+1}}}\) = \(\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}\) = \(\dfrac{\sqrt{x-1}}{\sqrt{x+1}}\)
b) \(\dfrac{x-1}{\sqrt{y}-1}\)\(\sqrt{\dfrac{y-2\sqrt{y+1}}{\left(x-1\right)^4}}\)
= \(\dfrac{x-1}{\sqrt{y}-1}\) \(\sqrt{\dfrac{\left(y-1\right)^4}{\left(x-1\right)^4}}\)
= \(\dfrac{x-1}{\sqrt{y}-1}\)\(\dfrac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^2}\)
= \(\dfrac{\sqrt{y-1}}{x-1}\)
Chúc bạn học tốt :3
C.hóa \(x+y=1\) và dùng C-S:
\(VT^2\le\frac{2x}{\left(y+1\right)^2}+\frac{2y}{\left(x+1\right)^2}\le\frac{8}{9}=VP^2\)
\(BDT\Leftrightarrow\frac{x}{\left(2-x\right)^2}+\frac{y}{\left(2-y\right)^2}\le\frac{4}{9}\left(1\right)\)
Ta có BĐT phụ \(\frac{x}{\left(2-x\right)^2}\le\frac{20}{27}x-\frac{4}{27}\)
\(\Leftrightarrow-\frac{\left(2x-1\right)^2\left(5x-16\right)}{27\left(x-2\right)^2}\le0\) *Đúng*
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT_{\left(1\right)}\le\frac{20}{27}\left(x+y\right)-\frac{4}{27}\cdot2=\frac{4}{9}=VP_{\left(1\right)}\)
"=" khi \(x=y=\frac{1}{2}\)
\(a.R=\left(\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}+\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)}{1-\sqrt{xy}}+1\right):\left(1-\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}-\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)}{\sqrt{xy}-1}\right)\)
\(R=\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)+xy-1}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right]:\left[\dfrac{xy-1-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right]\)
\(R=\dfrac{x\sqrt{y}-\sqrt{x}+\sqrt{xy}-1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}+xy-1}{xy-1}:\dfrac{xy-1-x\sqrt{y}+\sqrt{x}+\sqrt{xy}+1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}}{xy-1}\)
\(R=\dfrac{-2\sqrt{x}-2}{xy-1}:\dfrac{-2x\sqrt{y}-2\sqrt{xy}}{xy-1}\)
\(R=\dfrac{-2\left(\sqrt{x}+1\right)}{xy-1}.\dfrac{xy-1}{-2\left(x\sqrt{y}+\sqrt{xy}\right)}\)
\(R=\dfrac{\sqrt{x}+1}{x\sqrt{y}+\sqrt{xy}}\)
\(b.C=\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{7\sqrt{x}+4}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
\(C=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\dfrac{7\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(C=\dfrac{2x-6\sqrt{x}+7\sqrt{x}+4-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(C=\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(C=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
\(c.M=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+x}=\dfrac{\sqrt{x}+1+x}{x+\sqrt{x}}.\dfrac{\sqrt{x}+x}{\sqrt{x}}=\dfrac{\sqrt{x}+1+x}{\sqrt{x}}\)