\(\left(5-7x\right)\left(\dfrac{3}{7}-3x\right)=0\)

\(=>\left\{{}\begin{matrix}5-7x=0\\\dfrac{3}{7}-3x=0\end{matrix}\right.=>\left\{{}\begin{matrix}x=\dfrac{5}{7}\\x=\dfrac{1}{7}\end{matrix}\right.\)

\(tanDEF=\dfrac{DF}{EF}=\dfrac{2BH}{EF}\)

\(tanECH=\dfrac{HE}{HC}=\dfrac{2BH}{AH}=\dfrac{2BH}{EF}\)

\(\)

Kẻ \(DF\perp AE\) , cắt \(EC\) tại `J`

Xét  △`AFD` có:

`AB=DB`

`BH` song song `DF`

`=>BH` là đường trung bình của △`AFD`

\(=>\left\{{}\begin{matrix}2BH=DF\\AH=HF=EF\end{matrix}\right.\)

Ta có: `AH^2=HB*HC` (theo hệ thức giữa cạnh và đường cao trong tam giác )

\(=>HC=\dfrac{AH^2}{HB}\)

`tanDEF=DF/EF`

`=>tanDEF=2BH/EF`          `(1)`

`tanECH=HE/HC`

`=>tanECH=2BH/AH`

`=>tanECH=2BH/EF`            `(2)`

Từ `(1)` và `(2)` 

\(=>\widehat{DEF}=\widehat{ECH}\)

\(=>\widehat{DEF}+\widehat{HEC}=\widehat{ECH}+\widehat{HEC}\)

\(=>DEC=90^o\)  `(đpcm)`

\(a,A=2011.2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1\)

\(=>A=2012^2-1< 2012^2\)

\(=>A< B\)

\(b,A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(=>2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)

\(=>2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(=>2A=\left(3^8-1\right)...\left(3^{64}+1\right)\)

\(=>2A=3^{128}-1\)

\(=>A=\dfrac{3^{128}-1}{2}< 3^{128}-1\)

\(=>A< B\)