a) AM-GM với điểm rơi x=8

b) bunya

553,5        

\(PT\Leftrightarrow2\sqrt{x-a}+2\sqrt{y-b}+2\sqrt{z-c}=x+y+z\)

\(\Leftrightarrow x-a-2\sqrt{x-a}+y-b-2\sqrt{y-b}+z-c-2\sqrt{z-c}+a+b+c=0\)

\(\Leftrightarrow\left(\sqrt{x-a}-1\right)^2+\left(\sqrt{y-b}-1\right)^2+\left(\sqrt{z-c}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-a=1\\y-b=1\\z-c=1\end{matrix}\right.\)

Kết quả giống bên trên x=y=z=1 hoặc x=1;y=2;z=3 và các hoán vị

Bình phương và chuyển vế,rút gọn ta thu được phương trình sau:

\(16a^3-24a^2+23a-8=0\)( \(a=\sqrt{x}\))

\(\Leftrightarrow a^3-\dfrac{3}{2}a^2+\dfrac{23}{16}a-\dfrac{1}{2}=0\)

Đặt \(a=m+\dfrac{1}{2}\),PT trở thành

\(\left(m+\dfrac{1}{2}\right)^3-\dfrac{3}{2}\left(m+\dfrac{1}{2}\right)^2+\dfrac{23}{16}\left(m+\dfrac{1}{2}\right)-\dfrac{1}{2}=0\)

\(\Leftrightarrow m^3+\dfrac{3}{2}m^2+\dfrac{3}{4}m+\dfrac{1}{8}-\dfrac{3}{2}\left(m^2+m+\dfrac{1}{4}\right)+\dfrac{23}{16}\left(m+\dfrac{1}{2}\right)-\dfrac{1}{2}=0\)

\(\Leftrightarrow m^3+\dfrac{11}{16}m-\dfrac{1}{32}=0\) (*)

Xét pt dạng TQ \(x^3+ax+b=0\).Ta sẽ tách x thành m+n ( m khác PT (*) )

\(\Leftrightarrow\left(m+n\right)^3+a\left(m+n\right)+b=0\)

\(\Leftrightarrow\left(m^3+n^3+b\right)+\left(m+n\right)\left(3mn+a\right)=0\)

Chọn m,n sao cho \(\left\{{}\begin{matrix}m^3+n^3=-b\\mn=-\dfrac{a}{3}\end{matrix}\right.\)

Theo định lý viete: \(m^3;n^3\) là nghiệm của Phương trình \(X^2+bX-\dfrac{a^3}{27}=0\)

\(\Delta=b^2+\dfrac{4a^3}{27}\)\(\Rightarrow m^3=\dfrac{-b+\sqrt{b^2+\dfrac{4a^3}{27}}}{2};n^3=\dfrac{-b-\sqrt{b^2+\dfrac{4a^3}{27}}}{2}\)

Áp dụng :với \(a=\dfrac{11}{16};b=-\dfrac{1}{32}\)

\(\Delta=\dfrac{1}{32^2}+\dfrac{\dfrac{4.11}{16}^3}{27}=\dfrac{679}{13824}>0\)

\(m=\sqrt[3]{\dfrac{\dfrac{1}{32}+\sqrt{\dfrac{679}{13824}}}{2}};n=\sqrt{\dfrac{\dfrac{1}{32}-\sqrt{\dfrac{679}{13824}}}{2}}\)

m(*) \(\approx0,0453191..\)

khi đó \(a\approx0,5453191...\)\(\Rightarrow x\approx0,297372986..\)

lập fương

\(a=\sqrt{\sqrt[3]{x^6}+\sqrt[3]{x^4y^2}}+\sqrt{\sqrt[3]{y^6}+\sqrt[3]{y^4x^2}}\)

\(=\sqrt{\sqrt[3]{x^4}\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)}+\sqrt{\sqrt[3]{y^4}\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)}\)

\(=\sqrt{\sqrt[3]{x^2}+\sqrt[3]{y^2}}\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)\)\(\Rightarrow a=\left(\sqrt{\sqrt[3]{x^2}+\sqrt[3]{y^2}}\right)^3\)

\(\Rightarrow\sqrt[3]{a^2}=\sqrt[3]{x^2}+\sqrt[3]{y^2}\)

ÁP dụng BĐT bunyakovsky:

\(\left(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}\right)^2\le3\left(x+y+z-a-b-c\right)=3\left(x+y+z\right)-9\)

Mà \(3\left(x+y+z\right)-9\le\dfrac{\left(x+y+z\right)^2}{4}\)(*)

Vì : (*)\(\Leftrightarrow\left(x+y+z\right)^2-12\left(x+y+z\right)+36\ge0\Leftrightarrow\left(x+y+z-6\right)^2\ge0\)

do đó \(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}\le\dfrac{1}{2}\left(x+y+z\right)\)

Dấu = xảy ra khi \(\left\{{}\begin{matrix}x-a=y-b=z-c\\x+y+z=6\\a+b+c=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y=z=2\\\left(x;y;z\right)~\left(1;2;3\right)\end{matrix}\right.\)_ Và các hoán vị

P/s: khi đó a=b=2 hoặc (a;b;c) ~(0;1;2) và các hoán vị

\(PT\Leftrightarrow x+2+x-2+3\sqrt[3]{\left(x+2\right)\left(x-2\right)}\left(\sqrt[3]{x+2}+\sqrt[3]{x-2}\right)=5x\)

\(\Leftrightarrow\sqrt[3]{\left(x+2\right)\left(x-2\right).5x}=x\)

\(\Leftrightarrow x^3=5x\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow x\left(x^2-5x^2+20\right)=0\)

\(\Leftrightarrow4x\left(5-x^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)

\(PT\Leftrightarrow4x^3+6x^2+12x+8=0\)

\(\Leftrightarrow\left(x+2\right)^3=-3x^3\)

\(\Leftrightarrow x+2=\sqrt[3]{-3}x\)

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

Hình đây :v