HOC24
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\(\sqrt{\dfrac{a+b}{c+ab}}+\sqrt{\dfrac{b+c}{a+bc}}+\sqrt{\dfrac{c+a}{b+ac}}\)
\(2y^3+y+2x\sqrt{1-x}=3\sqrt{1-x}\)
\(\Leftrightarrow2y\left(x+y^2-1\right)+2x\left(\sqrt{1-x}-y\right)=3\left(\sqrt{1-x}-y\right)\)
\(\Leftrightarrow2y\left(x+y^2-1\right)+2x\cdot\dfrac{1-x-y^2}{\sqrt{1-x}+y}=3\cdot\dfrac{1-x-y^2}{\sqrt{1-x}+y}\)
\(\Leftrightarrow2y\left(x+y^2-1\right)-2x\cdot\dfrac{x+y^2-1}{\sqrt{1-x}+y}+3\cdot\dfrac{x+y^2-1}{\sqrt{1-x}+y}=0\)
\(\Leftrightarrow\left(x+y^2-1\right)\left(2y-\dfrac{2x}{\sqrt{1-x}+y}+\dfrac{3}{\sqrt{1-x}+y}\right)=0\)
\(\Leftrightarrow x+y^2-1=0\)\(\Leftrightarrow x=1-y^2\)
Thay vào pt(2) ta có: \(\sqrt{2y^2+1}+y=4+\sqrt{5-y^2}\)
\(\Leftrightarrow y=2\Leftrightarrow x=-3\)
hint:pt(1) cho ta nhan tu x+y^2-1
\(P'\left(x;y\right)=\dfrac{\left(x+2y+3\right)'\cdot\left(x+y+6\right)-\left(x+2y+3\right)\cdot\left(x+y+6\right)'}{\left(x+y+6\right)^2}\)
\(=\dfrac{\left(1+2y'\right)\cdot\left(x+y+6\right)-\left(x+2y+3\right)\cdot\left(1+y'\right)}{\left(x+y+6\right)^2}\)
\(=\dfrac{\left(x+9\right)y'-y+3}{\left(x+y+6\right)^2}=0\)
\(\left(x+9\right)y'-y+3=0\)\(\Leftrightarrow y'=-\dfrac{3}{x+9}+\dfrac{y}{x+9}\) la` pt vi phan tuyen tinh cap 1
\(\Leftrightarrow y=c_1x+9c_1+3\) khi do ta co:
\(P=\dfrac{x+2\left(c_1x+9c_1+3\right)+3}{x+c_1x+9c_1+3+6}=\dfrac{2c_1+1}{c_1+1}\)
Voi \(x=0\) khi do \(c_1=\dfrac{y\left(0\right)-3}{9}\)
Khi do tu dieu kien \(log_{\sqrt{3}}\left(\dfrac{x+y}{x^2+y^2+xy+2}\right)=x\left(x-3\right)+y\left(y-3\right)+xy\) cho \(2\) nghiem la \(y=1;y=2\)
*)Voi \(y=1\rightarrow c_1=-\dfrac{2}{9}\rightarrow P=\dfrac{5}{7}\)
*)Voi \(y=2\rightarrow c_1=-\dfrac{1}{9}\rightarrow P=\dfrac{7}{8}\)
De thay: \(\dfrac{5}{7}>\dfrac{7}{8}\rightarrow P_{min}=\dfrac{5}{7}\)
Is that true ?