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? |(.-.)|

bài này dễ thôi bạn, quan trọng là nó hơi dài nên mình không có hứng làm chi tiết

BĐT đã cho viết lại thành

\(\left(a^3+b^3+c^3\right)\left(a+b+c\right)^2+72abc\left(ab+bc+ca\right)-\left(a+b+c\right)^5\le0\)

\(\Leftrightarrow-\dfrac{3}{2}\left(8a^3+7a^2b+7a^2c-7ab^2-7ac^2+9b^2c+9bc^2\right)\left(b-c\right)^2-\dfrac{3}{2}\left(8b^3+7b^2c-7bc^2+9ac^2+7ab^2+9a^2c-7a^2b\right)\left(c-a\right)^2-\dfrac{3}{2}\left(9a^2b+9ab^2+7ac^2-7a^2c-7b^2c+7bc^2+8c^3\right)\left(a-b\right)^2\le0\)

B1: Từ \(\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}=\dfrac{2}{xy+1}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\xy=1\end{matrix}\right.\)

*)Với \(x=y\) khi đó ta có: \(P=\dfrac{1}{x^2+1}+\dfrac{1}{1+x^2}+\dfrac{2}{1+x^2}=\dfrac{4}{x^2+1}\)

*)Với \(xy=1\) khi đó ta có: \(P=\dfrac{1}{x^2+xy}+\dfrac{1}{y^2+xy}+\dfrac{2}{1+1}\)

\(=\dfrac{x+y}{xy\left(x+y\right)}+1=2\)

vẽ cái hình xem nào

\(x;y;z\ge0;p;q;n\in N^{star};\sum_{cyc}x^n=3;2p+2q>2n\text{.Prove}\sum_{cyc}\dfrac{x^p}{y^q}\ge3\)

\(\dfrac{\left(b+c\right)^2}{5a^2+\left(b+c\right)^2}+\dfrac{\left(c+a\right)^2}{5b^2+\left(c+a\right)^2}+\dfrac{\left(a+b\right)^2}{5c^2+\left(a+b\right)}\ge\dfrac{4}{3}\)

\(\Leftrightarrow\dfrac{-20a^2+10bc+5b^2+c^2}{9\left(5a^2+\left(b+c\right)^2\right)}+\dfrac{-20b^2+10ac+5c^2+5a^2}{9\left(5b^2+\left(c+a\right)^2\right)}+\dfrac{-20c^2+10ab+5a^2+5b^2}{9\left(5c^2+\left(a+b\right)\right)}\ge0\)

\(\Leftrightarrow\sum_{cyc}\dfrac{\left(c-a\right)\left(10a+5b+5c\right)-\left(a-b\right)\left(10a+5b+5c\right)}{9\left(5a^2+\left(b+c\right)^2\right)}\ge0\)

\(\Leftrightarrow\sum_{cyc}\left(\dfrac{-\left(a-b\right)\left(10a+5b+5c\right)}{9\left(5a^2+\left(b+c\right)^2\right)}+\dfrac{\left(a-b\right)\left(10b+5a+5c\right)}{9\left(5b^2+\left(a+c\right)^2\right)}\right)\ge0\)

\(\Leftrightarrow\sum_{cyc}\left(\left(a-b\right)\left(\dfrac{10b+5a+5c}{9\left(5b^2+\left(a+c\right)^2\right)}-\dfrac{10a+5b+5c}{9\left(5a^2+\left(b+c\right)^2\right)}\right)\right)\ge0\)

\(\Leftrightarrow\sum_{cyc}\left(\left(a-b\right)^2\dfrac{5\left(a^2+b^2-c^2+4ab\right)}{3\left(a^2+2ac+5b^2+c^2\right)\left(5a^2+b^2+2bc+c^2\right)}\right)\ge0\)

Dau "=" khi \(a=b=c\)

Thì admin cũng cố gắng lắm rồi, còn bận trăm công nghìn việc cố được đến thế nào thì cố còn phải chịu thôi =)) không có thời gian thì làm sao mà duy trì được web tốt được ai vẫn muốn onl thì cứ onl =))

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

Xét \(pt\left(1\right)\Leftrightarrow2x^2+y^2-3xy-4x+3y+2=0\)

\(\Leftrightarrow\left(x-y-1\right)\left(2x-y-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=x-1\\y=2x-2\end{matrix}\right.\)

*)\(y=x-1\) thay vao \(pt(2)\) :

\(pt\Leftrightarrow\sqrt{x^2-x+4}=2\Leftrightarrow x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=-1\\y=0\end{matrix}\right.\)

*)\(y=2x-2\) thay vao \(pt(2)\):

\(pt\Leftrightarrow\sqrt{x^2-2x+5}+\sqrt{x-1}=2\)

\(\Leftrightarrow\dfrac{x^2-2x+1}{\sqrt{x^2-2x+5}+2}+\sqrt{x-1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x-1}{\sqrt{x^2-2x+5}+2}+\dfrac{1}{\sqrt{x-1}}\right)=0\)

\(\Leftrightarrow x=1\)\(\Leftrightarrow y=0\)

Ap dung BDT Cauchy-Schwarz ta co:

\(\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\left(\dfrac{b^3}{a}+\dfrac{c^3}{b}+\dfrac{a^3}{c}\right)\ge\left(a^2+b^2+c^2\right)^2\)

Can chung minh \(\dfrac{b^3}{a}+\dfrac{c^3}{b}+\dfrac{a^3}{c}\ge a^2+b^2+c^2\)

\(VT=\dfrac{a^4}{ac}+\dfrac{b^4}{ab}+\dfrac{c^4}{bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)}{ab+bc+ca}=a^2+b^2+c^2=VP\)

\("="\Leftrightarrow a=b=c\)

DK: \(x\ge5\)

\(BPT\Leftrightarrow x^2-7x+2-2\sqrt{x^2-7x+10}< 0\)

\(\Leftrightarrow t^2-8-2t< 0\)\(\left(t=\sqrt{x^2-7x+10}\ge0\right)\)

\(\Leftrightarrow\left(t+2\right)\left(t-4\right)< 0\)

\(\Leftrightarrow-2< t< 4\)\(\Leftrightarrow-2< \sqrt{x^2-7x+10}< 4\)

\(\Leftrightarrow\sqrt{x^2-7x+10}< 4\)\(\Leftrightarrow x^2-7x-6< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}5\le x< \dfrac{7+\sqrt{73}}{2}\\\dfrac{7-\sqrt{73}}{2}< x\le2\end{matrix}\right.\)