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