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\(P=2+\dfrac{2}{b}+a+\dfrac{a}{b}+2+\dfrac{2}{a}+b+\dfrac{b}{a}=\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(a+\dfrac{1}{2a}\right)+\left(b+\dfrac{1}{2b}\right)+\left(\dfrac{3}{2a}+\dfrac{3}{2b}\right)+4\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}+2\sqrt{a.\dfrac{1}{2a}}+2\sqrt{b.\dfrac{1}{2b}}+2\sqrt{\dfrac{3}{2a}.\dfrac{3}{2b}}+4=6+2\sqrt{2}+\dfrac{3}{\sqrt{ab}}\)
Ta lại có: \(a^2+b^2\ge2\sqrt{a^2.b^2}=2ab\left(BĐT.Cauchy\right)\Rightarrow2\left(a^2+b^2\right)\ge4ab\Rightarrow\sqrt{ab}\le\dfrac{\sqrt{2\left(a^2+b^2\right)}}{2}=\dfrac{\sqrt{2}}{2}\)
\(\Rightarrow P\ge6+2\sqrt{2}+\dfrac{3}{\sqrt{ab}}\ge6+2\sqrt{2}+\dfrac{3}{\dfrac{\sqrt{2}}{2}}=6+5\sqrt{2}\)
\(minP=6+5\sqrt{2}\Leftrightarrow a=b=\dfrac{\sqrt{2}}{2}\)
\(A=a^3b^3+\dfrac{1}{a^3b^3}+2=a^3b^3+\dfrac{1}{2^{12}.a^3b^3}+\dfrac{2^{12}-1}{2^{12}a^3b^3}+2\)
\(A\ge2\sqrt{\dfrac{a^3b^3}{2^{12}.a^3b^3}}+\dfrac{2^{12}-1}{2^{12}.\left(\dfrac{a+b}{2}\right)^6}+2=\dfrac{2}{2^6}+\dfrac{2^{12}-1}{2^6}+2=\dfrac{2^{12}+1}{2^6}+2\) (casio)
Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)
a.
\(2x-x^2+7=-\left(x^2-2x+1\right)+8=-\left(x-1\right)^2+8\le8\)
\(\Rightarrow2+\sqrt{2x-x^2+7}\le2+\sqrt{8}=2+2\sqrt{2}\)
\(\Rightarrow\dfrac{3}{2+\sqrt{2x-x^2+7}}\ge\dfrac{3}{2+2\sqrt{2}}=\dfrac{3\sqrt{2}-3}{2}\)
\(A_{min}=\dfrac{3\sqrt{2}-3}{2}\) khi \(x=1\)
b. ĐKXĐ: \(x\le1\)
\(B=-\left(1-x-\sqrt{2\left(1-x\right)}+\dfrac{1}{2}-\dfrac{1}{2}-1\right)\)
\(B=-\left(1-x-\sqrt{2\left(1-x\right)}+\dfrac{1}{2}\right)+\dfrac{3}{2}\)
\(B=-\left(\sqrt{1-x}-\dfrac{\sqrt{2}}{2}\right)^2+\dfrac{3}{2}\le\dfrac{3}{2}\)
\(B_{max}=\dfrac{3}{2}\) khi\(x=\dfrac{1}{2}\)
\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)
\(P=-\dfrac{3}{5}\) sao suy ra đc \(\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\) thế
\(A=\dfrac{1}{a}+\dfrac{1}{b}-\left(\dfrac{a}{b}+\dfrac{b}{a}-2\right)=\dfrac{1-a+b}{b}+\dfrac{1-b+a}{a}\)
Vì \(a^2+b^2=1\) và \(a,b>0\Leftrightarrow0< a< 1;0< b< 1\Leftrightarrow1+a-b>0;1-b+a>0\)
\(\Leftrightarrow A\ge2\sqrt{\dfrac{\left(1-a+b\right)\left(1-b+a\right)}{ab}}=2\sqrt{\dfrac{1-a^2-b^2+2ab}{ab}}=2\sqrt{2}\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\dfrac{1-a+b}{b}=\dfrac{1-b+a}{a}\end{matrix}\right.\Leftrightarrow a=b=\dfrac{1}{\sqrt{2}}\)
Ta có \(A=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)=1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{ab}=1+\dfrac{a+b}{ab}+\dfrac{1}{ab}=1+\dfrac{a+b+1}{ab}=1+\dfrac{1+1}{ab}=1+\dfrac{2}{ab}\)
Áp dụng bđt cosi ta có
\(a+b\ge2\sqrt{ab}\Leftrightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\Leftrightarrow ab\le\dfrac{1}{4}\Leftrightarrow\dfrac{2}{ab}\ge8\Leftrightarrow1+\dfrac{2}{ab}\ge9\Leftrightarrow A\ge9\)
Dấu bằng xảy ra khi \(\left\{{}\begin{matrix}a=b\\a+b=1\end{matrix}\right.\)\(\Leftrightarrow\)\(a=b=0,5\)
Vậy GTNN của A là 9 và xảy ra khi a=b=0,5
\(A=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\)
\(A=1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{ab}\)
\(A=1+\dfrac{a+b}{ab}+\dfrac{1}{ab}\)
Mà a+b=1
nên \(A=1+\dfrac{1}{ab}+\dfrac{1}{ab}=1+\dfrac{2}{ab}\)
Ta có:
a+b=1
Áp dụng bđt Cosi
\(a+b\ge2\sqrt{ab}\Rightarrow1\ge2\sqrt{ab}\)
\(\Rightarrow1\ge4ab\Leftrightarrow ab\le\dfrac{1}{4}\)
Ta có:
\(A=1+\dfrac{2}{ab}\ge1+\dfrac{\dfrac{2}{1}}{4}=1+8=9\)
Dấu bằng xảy ra khi \(\) \(\left\{{}\begin{matrix}a+b=1\\a=b\end{matrix}\right.\)
\(\Rightarrow a=b=\dfrac{1}{2}\)