a: Ta có: \(\widehat{xBy}=\widehat{xAz}\)(hai góc đồng vị)

mà hai góc này là hai góc ở vị trí đồng vị

nên By//Az

b: AC là phân giác của góc xAz

=>\(\widehat{xAC}=\widehat{zAC}=\dfrac{\widehat{xAz}}{2}=30^0\)

=>\(\widehat{BAC}=30^0\)

Ta có: \(\widehat{CBA}+\widehat{CBx}=180^0\)(hai góc kề bù)

=>\(\widehat{CBA}+60^0=180^0\)

=>\(\widehat{CBA}=120^0\)

Xét ΔBAC có \(\widehat{BAC}+\widehat{CBA}+\widehat{ACB}=180^0\)

=>\(\widehat{ACB}+30^0+120^0=180^0\)

=>\(\widehat{ACB}=30^0\)

c: BD là phân giác của góc yBA

=>\(\widehat{ABD}=\dfrac{\widehat{yBA}}{2}=60^0\)

Xét ΔBDA có \(\widehat{DBA}+\widehat{DAB}=30^0+60^0=90^0\)

nên ΔBDA vuông tại D

=>AC\(\perp\)BD tại D

Bài 5:

\(A=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\\ 3A=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\\ 3A+A=\left(3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\right)+\left(3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\right)\\ 4A=3^{101}+1\\ A=\dfrac{3^{101}+1}{4}\) 

\(1)\left(\dfrac{1}{5}\right)^5\cdot5^5\\ =\left(\dfrac{1}{5}\cdot5\right)^5\\ =1^5\\ =1\\ 2)\left(\dfrac{2}{5}\right)^9\cdot5^9\\ =\left(\dfrac{2}{5}\cdot5\right)^9\\ =2^9\\ 3)\left(\dfrac{4}{9}\right)^3\cdot3^3\\ =\left(\dfrac{4}{9}\cdot3\right)^3\\ =\left(\dfrac{4}{3}\right)^3\\ 4)\left(\dfrac{3}{7}\right)^2\cdot\left(-7\right)^4\\ =\left(\dfrac{3}{7}\right)^2\cdot\left[\left(-7\right)^2\right]^2\\ =\left(\dfrac{3}{7}\right)^2\cdot49^2\\ =\left(\dfrac{3}{7}\cdot49\right)^2\\ =\left(3\cdot7\right)^2\\ =21^2\\ 5)\left(-11\right)^{12}\cdot\left(\dfrac{4}{11}\right)^6\\ =\left[\left(-11\right)^2\right]^6\cdot\left(\dfrac{4}{11}\right)^6\\ =121^6\cdot\left(\dfrac{4}{11}\right)^6\\ =\left(121\cdot\dfrac{4}{11}\right)^6\\ =\left(4\cdot11\right)^6\\ =44^6\\ 6)\left(-6\right)^8\cdot\left(\dfrac{5}{6}\right)^7\\ =\left(-6\right)\cdot\left(-6\right)^7\cdot\left(\dfrac{5}{6}\right)^7\\ =\left(-6\right)\cdot\left(-6\cdot\dfrac{5}{6}\right)^7\\ =\left(-6\right)\cdot\left(-5\right)^7\)

  Kẻ H\(x\) // FG 

 Ta có : \(\widehat{xHI}\) = \(\widehat{JIH}\) = 45⁰ (Hai góc so le trong)

  \(\widehat{xHG}\) + \(\widehat{FGH}\) = 180⁰ (hai góc trong cùng phía)

⇒ \(\widehat{xHG}\) = 180⁰ - 135⁰ = 45⁰

\(\widehat{IGH}\) = \(\widehat{xHG}\) + \(\widehat{xHI}\) = 45⁰ + 45⁰ = 90⁰

Vậy HG vuông góc với HI

  Kẻ H\(x\) // FG 

 Ta có : \(\widehat{xHI}\) = \(\widehat{JIH}\) = 45⁰ (Hai góc so le trong)

  \(\widehat{xHG}\) + \(\widehat{FGH}\) = 180⁰ (hai góc trong cùng phía)

⇒ \(\widehat{xHG}\) = 180⁰ - 135⁰ = 45⁰

\(\widehat{IGH}\) = \(\widehat{xHG}\) + \(\widehat{xHI}\) = 45⁰ + 45⁰ = 90⁰

Vậy HG vuông góc với HI