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Điều kiện: \(x;y>1\)
\(A=\dfrac{\left(x^3+y^3\right)-\left(x^2+y^2\right)}{\left(x-1\right)\left(y-1\right)}\)
\(=\dfrac{x^2\left(x-1\right)+y^2\left(y-1\right)}{\left(x-1\right)\left(y-1\right)}\)
\(=\dfrac{x^2}{y-1}+\dfrac{y^2}{x-1}\)
\(\ge\dfrac{\left(x+y\right)^2}{x+y-2}\)
Đặt \(x+y=a\left(a>2\right)\)
\(\Rightarrow A=\dfrac{a^2}{a-2}=\dfrac{8\left(a-2\right)+\left(a^2-8a+16\right)}{a-2}=8+\dfrac{\left(a-4\right)^2}{a-2}\ge8\)
Dấu "=" xảy ra khi x = y = 2
Vậy \(Min_A=8\Leftrightarrow x=y=2\)
\(P\ge\dfrac{1}{2}\left(\dfrac{1}{x-2}+\dfrac{1}{3-x}\right)^2+\dfrac{4}{\left(x-2+3-x\right)^2}=\dfrac{1}{2}\left(\dfrac{1}{x-2}+\dfrac{1}{3-x}\right)^2+4\)
\(P\ge\dfrac{1}{2}\left(\dfrac{4}{x-4+3-x}\right)^2+4=12\)
Dấu "=" xảy ra khi \(x-2=3-x\Rightarrow x=\dfrac{5}{2}\)
\(a,ĐK:x>0;x\ne9\\ b,A=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ A=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ c,A>\dfrac{2}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{2}{5}>0\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{5}>0\\ \Leftrightarrow\dfrac{2-\sqrt{x}}{5\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow2-\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)
Ta có \(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{\left(a+b\right)^2}{2}\right)^2}{2}=\dfrac{\left(a+b\right)^4}{8}\). Áp dụng cho biểu thức A, suy ra \(A\ge\dfrac{\left(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\right)^4}{8}\). Ta tìm GTNN của \(P=x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\). Ta có
\(P=x^2+\dfrac{1}{16x^2}+y^2+\dfrac{1}{16y^2}+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2\)
\(P\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}\left(\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}{2}\right)+2\)
\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{15}{16}.\left(\dfrac{4^2}{2}\right)+2\) \(=\dfrac{21}{2}\). Do đó \(P\ge\dfrac{21}{2}\) \(\Leftrightarrow A\ge\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\). Vậy GTNN của A là \(\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\), ĐTXR \(\Leftrightarrow x=y=\dfrac{1}{2}\)
Gợi ý: \(\dfrac{a^4+b^4}{2}\ge\left(\dfrac{a+b}{2}\right)^4\)
Tìm GTNN của biểu thức M
M = \(\left(x-1\right)^4+\left(3-x\right)^4+6\left(x^2-4x+3\right)^2+2013\)
a)...........................
b)\(\Leftrightarrow A=\dfrac{\dfrac{x^2}{4}+x^2y+\dfrac{y}{4}+y^2+x^2y^2+\dfrac{1}{4}+\dfrac{3y}{4}}{x^2y^2+1+y^2-x^2y-y+x^2}\)
\(\Leftrightarrow A=\dfrac{\dfrac{x^2}{4}+\dfrac{1}{4}+y+x^2y+y^2+x^2y^2}{x^2\left(y^2-y+1\right)+\left(y^2-y+1\right)}\)
\(\Leftrightarrow A=\dfrac{\dfrac{\left(x^2+1\right)}{4}+y\left(x^2+1\right)+y^2\left(x^2+1\right)}{\left(y^2-y+1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow A=\dfrac{\left(x^2+1\right)\left(\dfrac{1}{4}+y+y^2\right)}{\left(y^2-y+1\right)\left(x^2+1\right)}=\dfrac{4y^2+4y+1}{4\left(y^2-y+1\right)}\)(không phụ vào x)
\(\Rightarrowđpcm\)
c) Bạn tự làm đi tới đây dễ rồi
Áp dụng bất đẳng thức AM - GM, ta có:
\(S=\dfrac{1}{\left(x-1\right)^2}+\dfrac{1}{\left(2-x\right)^2}+\dfrac{1}{\left(x-1\right)\left(2-x\right)}\)
\(\ge3\sqrt[3]{\dfrac{1}{\left(x-1\right)^2}\times\dfrac{1}{\left(2-x\right)^2}\times\dfrac{1}{\left(x-1\right)\left(2-x\right)}}\)
\(=\dfrac{3}{\left(x-1\right)\left(x-2\right)}=\dfrac{3}{-x^2+3x-2}\)
Vì \(-x^2+3x-2=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
nên \(S\ge\dfrac{3}{\dfrac{1}{4}}=12\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{\left(x-1\right)^2}=\dfrac{1}{\left(2-x\right)^2}=\dfrac{1}{\left(x-1\right)\left(2-x\right)}\\x-\dfrac{3}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{3}{2}\left(\text{ nhận }\right)\)
Vậy \(Min_S=12\Leftrightarrow x=\dfrac{3}{2}\)
Đặt \(x+3=t\ne0\Rightarrow x=t-3\)
\(A=\dfrac{\left(t+2\right)\left(t-4\right)}{t^2}=\dfrac{t^2-2t-8}{t^2}=-\dfrac{8}{t^2}-\dfrac{2}{t}+1=-8\left(\dfrac{1}{t}+\dfrac{1}{8}\right)^2+\dfrac{9}{8}\le\dfrac{9}{8}\)
\(A_{max}=\dfrac{9}{8}\) khi \(t=-8\) hay \(x=-11\)