~ Bài 3:

Hình tự vẽ.

Theo giả thiết, ta có:

\(\widehat{MEB}=\widehat{EBF}=\widehat{BFM}=90^0\)

\(\Rightarrow EBFM\) là hình chữ nhật

\(\Rightarrow ME=FB;MF=EB\)

\(\Delta EAM\) vuông cân tại E \(\left(\widehat{BAC}=45^0\right)\)

\(\Rightarrow AE=ME=BF\)

\(\Delta FMC\) vuông cân tại F \(\left(\widehat{BCA}=45^0\right)\)

\(\Rightarrow FC=MF=BE\)

Ta có:

\(S_{DFE}=S_{ABCD}-S_{ADE}-S_{DCF}-S_{BFE}\)

\(=a^2-\dfrac{1}{2}\times a\times AE-\dfrac{1}{2}\times a\times CF-\dfrac{1}{2}\times BE\times BF\)

\(=a^2-\dfrac{1}{2}\times a\times\left(AE+EB\right)-\dfrac{1}{2}\times AE\times BE\)

Mặt khác, theo bất đẳng thức AM - GM, ta có:

\(S_{DFE}\ge a^2-\dfrac{1}{2}\times a^2-\dfrac{1}{2}\times\dfrac{\left(AE+BE\right)^2}{4}\)

\(=\dfrac{a^2}{2}-\dfrac{a^2}{8}=\dfrac{3}{8}a^2\)

Dấu "=" xảy ra khi AE = BE

<=> E là trung điểm của AB mà ME // BC (do cùng _I_ AB)

=> M là trung điểm của AC

Vậy \(Min_{S_{DFE}}\) \(=\dfrac{3}{8}a^2\) <=> M là trung điểm của AC.

d) \(5\sqrt{x}+\dfrac{5}{2\sqrt{x}}=2x+\dfrac{1}{2x}+4\)

\(\Leftrightarrow2\left(x+\dfrac{1}{4x}\right)+4=5\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)\)

\(\Leftrightarrow2\left[\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-1\right]-5\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)+4=0\)

\(\Leftrightarrow2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-5\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)+2=0\)

Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a\left(a\ge\sqrt{2}\right)\)

\(\Rightarrow2a^2-5a+2=0\)

\(\Leftrightarrow\left(a-2\right)\left(2a-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2\left(\text{nhận}\right)\\a=\dfrac{1}{2}\left(\text{loại}\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\dfrac{1}{2\sqrt{x}}=2\)

\(\Leftrightarrow2x-4\sqrt{x}+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{2+\sqrt{2}}{2}\\\sqrt{x}=\dfrac{2-\sqrt{2}}{2}\end{matrix}\right.\) (nhận)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+2\sqrt{2}}{2}\\x=\dfrac{3-2\sqrt{2}}{2}\end{matrix}\right.\) (nhận)

a) \(\sqrt{x^2-3x+3}+\sqrt{x^2-3x+6}=3\)

Đặt \(\sqrt{x^2-3x+3}=a;\sqrt{x^2-3x+6}=b\left(a;b>0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=3\\b^2-a^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=3\\\left(b+a\right)\left(b-a\right)=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+a=3\\b-a=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=2\\a=1\end{matrix}\right.\) (nhận)

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

\(\Leftrightarrow x^2-3x+3=1\)

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

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

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

Đặt \(\sqrt{3-x+x^2}=a;\sqrt{2+x-x^2}=b\left(a;b>0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\a^2+b^2=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\\left(b^2+2b+1\right)+b^2-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\2\left(b-1\right)\left(b+2\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) (vì \(b+2>0\)) (nhận)

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

\(\Leftrightarrow2+x-x^2=1\)

\(\Leftrightarrow x^2-x-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\) (nhận)

b) Áp dụng bất đẳng thức Cauchy Shwarz dạng Engel, ta có:

\(A=\dfrac{x}{\sqrt{y}-1}+\dfrac{y}{\sqrt{z}-1}+\dfrac{z}{\sqrt{x}-1}\)

\(\ge\dfrac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{\sqrt{x}+\sqrt{y}+\sqrt{z}-3}\)

Đặt \(\sqrt{x}+\sqrt{y}+\sqrt{z}=a\left(a>0\right)\)

\(\Rightarrow A\ge\dfrac{a^2}{a-3}\)

\(=\dfrac{12\left(a-3\right)+\left(a^2-12a+36\right)}{a-3}\)

\(=12+\dfrac{\left(a-6\right)^2}{a-3}\ge12\)

Dấu "=" xảy ra khi x = y = z = 2

a) Áp dụng bất đẳng thức Cauchy Shwarz dạng Engel, ta có:

\(A=\dfrac{x^2}{x-1}+\dfrac{y^2}{y-1}\)

\(\ge\dfrac{\left(x+y\right)^2}{x+y-2}\)

Đặt \(x+y=a\left(a>0\right)\)

\(\Rightarrow A\ge\dfrac{a^2}{a-2}\)

\(=\dfrac{8\left(a-2\right)+\left(a^2-8a+16\right)}{a-2}\)

\(=8+\dfrac{\left(a-4\right)^2}{a-2}\ge8\)

Dấu "=" xảy ra khi \(x=y=2\)