a) We see: xAz = AOC = 65^o They are in the same position as Az//Oy

b) Because Ax//Oy should be AOC + OAz = 180^o (two cornerson the same side)

=> OAz = 115^o = CBz

They are in the same position => Ox // Bt

c) Because KBC and CBz two adjacent angles KBC = 180^o - 115^o 

=> OCB = 90^o - 65^o 

=> OCB = 25^o

d) We see: HAK = AHO = 90^o

=> HAK + AKC = 180^o. They are in the same position => AH // CK

P/s: Next time you write in Vietnamese, you go to olm or hoc24

Sorry, edit the section \(\sqrt[8]{25+6}\) in \(\sqrt[8]{256}\) 

\(\sqrt{\sqrt{\sqrt{256}}}=\left(\sqrt{256}\right)^{\dfrac{1}{2}.\dfrac{1}{2}}=\left(256\right)^{\dfrac{1}{4}}=\sqrt[4]{\sqrt[]{256}}=\left(256^{\dfrac{1}{2}}\right)^{\dfrac{1}{4}}=256^{\dfrac{1}{2}.\dfrac{1}{4}}=256^{\dfrac{1}{8}}=\sqrt[8]{25+6}\)

\(=2\Rightarrow\sqrt{\sqrt{\sqrt{256}}=2}\)

Quoc Tran Anh Le You can not rely on the number of reviews to accuse me of how many nick

Quoc Tran Anh Le  I did not do anything wrong that I have to fear =)) You just reported it :))) Thanks in advance

A very perfect lawsuit :) Acclaim

Do you have evidence against me?

You use the word carefully :)

\(\left(x-1\right)\left(2x-1\right)=9-x\)

\(\Leftrightarrow2x^3-3x+1=9-x\)

\(\Leftrightarrow2x^2-3x+1=9-x+x\)

\(\Leftrightarrow2x^2+2x+1=9\)

\(\Leftrightarrow2x^2-2x+1-9=9-9\)

\(\Leftrightarrow2x^2-2x-8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4.2\left(-8\right)}}{2.2}\\x=\dfrac{-\left(2\right)+\sqrt{\left(-2\right)^2-4.\left(-8\right)}}{2.2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{17}}{2}\\\dfrac{1-\sqrt{17}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{17}}{2}\\x=\dfrac{1-\sqrt{17}}{2}\end{matrix}\right.\)

\(D=\left(-2x^3-xy^2\right)+\left(xy^2-1\right)-\left(x^2y-xy^2+3x^2\right)\)

\(D=-2x^3-xy^2+x^2-1-\left(x^2y-xy^2+3x^2\right)\)

\(D=-2x^3-xy^2+xy^2-1-x^2y+xy^2-3x^2\)

\(D=-2x^3-x^2y-3x^2+xy^2-1\)

Answer: Year 2099

Different way:

\(\dfrac{x-91}{37}+\dfrac{x-86}{42}+\dfrac{x-78}{50}+\dfrac{x-49}{79}=4\)

\(=\dfrac{82950}{\left(x-91\right)}+\dfrac{73075}{\left(x-86\right)}+\dfrac{61383}{\left(x-78\right)}+\dfrac{38850}{\left(x-49\right)}=12276600\)

\(=256258x-20524424=1276600\)

\(=256258x=32801024\)

=> x = 128

So: x = 128

\(x^2-3x+2=0\)

I consider two cases

Case 1: \(\dfrac{-\left(-3\right)+\sqrt{\left(-3\right)^2-4.1.2}}{2.1}\)

\(=\dfrac{3+\sqrt{\left(-3\right)^2-4.1.2}}{2.1}\)

\(=\dfrac{4}{2.1}\)

= 2

Case 2: \(\dfrac{-\left(-3\right)-\sqrt{\left(-3\right)^2-4.1.2}}{2.1}\)

\(=\dfrac{3-\sqrt{\left(-3\right)^2-4.1.2}}{2.1}\)

\(=\dfrac{2}{2.1}\)

= 1

=> x = {2; 1}

So: x = {2; 1}