A B C D E F K

d) Kẻ AK vuông góc với BC

Ta có: \(S_{ABC}=S_{ABE}+S_{AEC}=\frac{1}{2}AK.BE+\frac{1}{2}AK.EC=AK.BE\)(vì BE = EC (gt)) (1)

\(S_{AECF}=\frac{1}{2}AK.\left(AF+CE\right)=\frac{1}{2}AK.2.EC=AK.EC=AK.BE\)(vì AECF là hình bình hành => AF = EC) (2)

Từ (1) và (2) => \(S_{ABC}=S_{AECF}\)

\(a^3+b^3+1=3ab\)

Theo bất đẳng thức AM - GM cho 3 số \(a^3,b^3,1\)ta có: 

\(a^3+b^3+1\ge3\sqrt[3]{a^3.b^3.1}=3ab\)

Dấu \(=\)xảy ra khi \(a=b=1\).

\(a^2+2ab+b^2=a+b+2\)

\(\Leftrightarrow\left(a+b\right)^2-\left(a+b\right)-2=0\)

\(\Leftrightarrow\orbr{\begin{cases}a+b=2\\a+b=-1\end{cases}}\)

mà \(a,b>0\)nên \(a+b=2\Leftrightarrow b=2-a\).

Với \(b=2-a\)thế vào biểu thức \(M\)ta được: 

\(M=a^2+3\left(2-a\right)^2+2a-5=4a^2-10a+7=\left(2a-\frac{5}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu \(=\)xảy ra tại \(2a=\frac{5}{2}\Leftrightarrow a=\frac{5}{4}\Rightarrow b=\frac{3}{4}\).

\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-3}+...+\frac{1}{x-4}-\frac{1}{x-5}\)

\(=\frac{1}{x}-\frac{1}{x-5}=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}=\frac{-5}{x\left(x-5\right)}\)

\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+...+\frac{1}{x-4}-\frac{1}{x-5}\)

\(=\frac{1}{x}-\frac{1}{x-5}\)

\(=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}\)

\(=\frac{x-5-x}{x\left(x-5\right)}\)

\(=-\frac{5}{x\left(x-5\right)}\)

 a) Vậy A = 3/4 <=> x = -1/2 A = x 2 + x + 1 A = x 2 + 2. x + + 1 2 1 4 3 4 A = (x + ) 2 + ≥ 1 2 3 4 3 4

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

\(=\left(x^2-2x\frac{1}{2}+\frac{1}{4}\right)+1-\frac{1}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

Dấu"=" xảy ra khi \(x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

Vậy \(Min_A=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)