Lightning Farron

chuc mung cac em

\(\sqrt{\dfrac{a+b}{c+ab}}+\sqrt{\dfrac{b+c}{a+bc}}+\sqrt{\dfrac{c+a}{b+ac}}\)

\(2y^3+y+2x\sqrt{1-x}=3\sqrt{1-x}\)

\(\Leftrightarrow2y\left(x+y^2-1\right)+2x\left(\sqrt{1-x}-y\right)=3\left(\sqrt{1-x}-y\right)\)

\(\Leftrightarrow2y\left(x+y^2-1\right)+2x\cdot\dfrac{1-x-y^2}{\sqrt{1-x}+y}=3\cdot\dfrac{1-x-y^2}{\sqrt{1-x}+y}\)

\(\Leftrightarrow2y\left(x+y^2-1\right)-2x\cdot\dfrac{x+y^2-1}{\sqrt{1-x}+y}+3\cdot\dfrac{x+y^2-1}{\sqrt{1-x}+y}=0\)

\(\Leftrightarrow\left(x+y^2-1\right)\left(2y-\dfrac{2x}{\sqrt{1-x}+y}+\dfrac{3}{\sqrt{1-x}+y}\right)=0\)

\(\Leftrightarrow x+y^2-1=0\)\(\Leftrightarrow x=1-y^2\)

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 ?

tag ko co thong bao de mai t nghien cuu

\(\sqrt[3]{x+3}-\sqrt[3]{6-x}=1\)

\(\Leftrightarrow\sqrt[3]{x+3}-2-\left(\sqrt[3]{6-x}-1\right)=0\)

\(\Leftrightarrow\dfrac{x+3-8}{\sqrt[3]{x+3}^2+4+2\sqrt[3]{x+3}}-\dfrac{6-x-1}{\sqrt[3]{6-x}^2+1+\sqrt[3]{6-x}}=0\)

\(\Leftrightarrow\dfrac{x-5}{\sqrt[3]{x+3}^2+4+2\sqrt[3]{x+3}}+\dfrac{x-5}{\sqrt[3]{6-x}^2+1+\sqrt[3]{6-x}}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\dfrac{1}{\sqrt[3]{x+3}^2+4+2\sqrt[3]{x+3}}+\dfrac{1}{\sqrt[3]{6-x}^2+1+\sqrt[3]{6-x}}\right)=0\)

Dễ thấy: \(\dfrac{1}{\sqrt[3]{x+3}^2+4+2\sqrt[3]{x+3}}+\dfrac{1}{\sqrt[3]{6-x}^2+1+\sqrt[3]{6-x}}>0\)

\(\Rightarrow x-5=0\Leftrightarrow x=5\)

\(x=\sqrt[3]{5-\sqrt{17}}+\sqrt[3]{5+\sqrt{17}}\)

\(\Leftrightarrow x^3=5-\sqrt{17}+5+\sqrt{17}+3\left(\sqrt[3]{5-\sqrt{17}}+\sqrt[3]{5+\sqrt{17}}\right)\sqrt[3]{5-\sqrt{17}}\sqrt[3]{5+\sqrt{17}}\)

\(\Leftrightarrow x^3=10+3x\sqrt[3]{\left(5-\sqrt{17}\right)\left(5+\sqrt{17}\right)}\)

\(\Leftrightarrow x^3=10+3x\sqrt[3]{8}\)\(\Leftrightarrow x^3=10+6x\)

\(\Leftrightarrow x^3-6x-10=0\)

Hay ta co DPCM

\(K=\sqrt{\dfrac{1}{a^2+b^2}+\dfrac{1}{\left(a+b\right)^2}+\sqrt{\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{\left(a^2+b^2\right)^2}}}\)

\(=\sqrt{\dfrac{1}{a^2+b^2}+\dfrac{1}{\left(a+b\right)^2}+\sqrt{\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)^2-\dfrac{2}{a^2+b^2}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)+\dfrac{1}{\left(a^2+b^2\right)^2}}}\)

\(=\sqrt{\dfrac{1}{a^2+b^2}+\dfrac{1}{\left(a+b\right)^2}+\sqrt{\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}-\dfrac{1}{a^2+b^2}\right)^2}}\)

\(=\sqrt{\dfrac{1}{\left(a+b\right)^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}}\)\(=\sqrt{\dfrac{1}{\left(a+b\right)^2}+\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2-\dfrac{2}{\left(a+b\right)}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}\)

\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a+b}\right)^2}\)\(=\left|\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a+b}\right|\)

WLOG \(x\ge y \ge z\)

Áp dụng BĐT AM-GM và BĐT Rearrangement ta có:

\(VT=\dfrac{x+1}{y+1}+\dfrac{y+1}{z+1}+\dfrac{z+1}{x+1}\)

\(=\dfrac{\left(x+y+z\right)^2+3\left(x+y+z\right)+xy^2+yz^2+xz^2+3}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(\le\dfrac{21+xy^2+yz^2+xz^2}{xy+yz+xz+4}\)\(\le\dfrac{21+x^2y+xyz+yz^2}{3\sqrt[3]{4\left(xy+yz+xz\right)}}\)

\(\le\dfrac{21+y\left(x+z\right)^2}{3\sqrt[3]{4\left(xy+yz+xz\right)}}\)\(\le\dfrac{21+\dfrac{\left(\dfrac{2\left(x+y+z\right)}{3}\right)^3}{2}}{3\sqrt[3]{4\left(xy+yz+xz\right)}}\)

\(=\dfrac{21+4}{3\sqrt[3]{4\left(xy+yz+xz\right)}}=\dfrac{25}{3\sqrt[3]{4\left(xy+yz+xz\right)}}=VP\)

Dấu "=" khi \(\left(x;y;z\right)=\left(2;1;0\right)\) và h.vị

Còn bài nào khó hơn không? |(.-.)|