Bài 3: Gọi H là giao điểm của CD với AB

\(\widehat{HCB}+\widehat{DCB}=180^0\)(hai góc kề bù)

=>\(\widehat{HCB}+143^0=180^0\)

=>\(\widehat{HCB}=180^0-143^0=37^0\)

Xét ΔHCB có \(\widehat{HCB}+\widehat{HBC}=37^0+53^0=90^0\)

nên ΔHCB vuông tại H

=>CD\(\perp\)AB tại H

Bài 2:

a: Ta có: \(\widehat{DAB}=\widehat{xAM}\)(hai góc đối đỉnh)

mà \(\widehat{xAm}=124^0\)

nên \(\widehat{DAB}=124^0\)

Ta có: \(\widehat{DAB}+\widehat{ABC}=124^0+56^0=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên AD//BC

=>xy//zt

b: xy//zt

=>\(\widehat{BCD}+\widehat{ADC}=180^0\)(hai góc trong cùng phía)

=>\(\widehat{BCD}+90^0=180^0\)

=>\(\widehat{BCD}=90^0\)

Ak là phân giác của góc DAB

=>\(\widehat{DAC}=\dfrac{124^0}{2}=62^0\)

ΔDAC vuông tại D

=>\(\widehat{DAC}+\widehat{DCA}=90^0\)

=>\(\widehat{DCA}+62^0=90^0\)

=>\(\widehat{DCA}=28^0\)

c;     C = \(\dfrac{28^{28}+28^{24}+...+28^4+1}{28^{30}+28^{28}+...+28^2+1}\)

        A =         1 + 28⁴ + 28⁸ + 28¹² + ...28²⁸

  28⁴A = 28⁴ + 28⁸ + 28¹² + ... + 28²⁸ + 28³²

28⁴A - A = 28⁴+ 28⁸+...+28²⁸+ 28³²- (1 + 28⁴ + 28⁸+...+28²⁸)

(28⁴ - 1)A = 28⁴ + 28⁸+ ...+ 28²⁸ + 28³² - 1 - 28⁴- ...- 28²⁸

(28⁴ - 1)A = (28³² - 1) + (28⁴ - 28⁴) + (28⁸ - 28⁸) + ... + (28²⁸ - 28²⁸)

(28⁴ - 1)A = 28³² - 1 + 0 + 0... + 0

            A = (28³² - 1): (28⁴ - 1)

  Đặt B = 28³⁰ + 28²⁸ + ... + 28² + 1

  28²B = 28³² + 28³⁰ + ... + 28⁴ + 28²

28²B - B = 28³² + 28³⁰ + ... + 28⁴ + 28² - (28³⁰ + 28²⁸ +...+1)

(28² - 1)B = 28³² + 28³⁰+...+28⁴ + 28² - 28³⁰ - 28²⁸ - ... 28²- 1

(28² - 1)B = (28³² - 1) + (28³⁰ - 28³⁰) +...+(28² - 28²)

(28² - 1)B = (28³² - 1) + 0 + 0 +...+ 0

(28² - 1)B = 28³² - 1 

             B = (28³² - 1): (28² - 1)

C = \(\dfrac{A}{B}\) = \(\dfrac{28^{32}-1}{28^4-1}\) : \(\dfrac{28^{32}-1}{28^2-1}\)

C = \(\dfrac{28^{32}-1}{28^4-1}\) \(\times\) \(\dfrac{28^2-1}{28^{32}-1}\)

C = \(\dfrac{28^2-1}{28^4-1}\)

C = \(\dfrac{1}{785}\) 

                Câu d:

 \(\dfrac{x-1}{99}\) + \(\dfrac{x-2}{98}\) + \(\dfrac{x-3}{97}\) = \(\dfrac{x-4}{96}\) + \(\dfrac{x-5}{95}\) + \(\dfrac{x-6}{94}\)

(\(\dfrac{x-1}{99}\)-1)+(\(\dfrac{x-2}{98}\)-1)+(\(\dfrac{x-3}{97}\)-1) = (\(\dfrac{x-4}{96}\)-1) + (\(\dfrac{x-5}{95}\)-1)+(\(\dfrac{x-6}{94}\)-1)

\(\dfrac{x-100}{99}\)+\(\dfrac{x-100}{98}\)+\(\dfrac{x-100}{97}\) = \(\dfrac{x-100}{96}\)+\(\dfrac{x-100}{95}\)+\(\dfrac{x-100}{94}\)

\(\dfrac{x-100}{99}\)+\(\dfrac{x-100}{98}\)+\(\dfrac{x-100}{97}\)- \(\dfrac{x-100}{96}\)-\(\dfrac{x-100}{95}\)-\(\dfrac{x-100}{94}\) = 0

(\(x-100\)).(\(\dfrac{1}{99}\)+\(\dfrac{1}{98}\)+\(\dfrac{1}{97}\) - \(\dfrac{1}{96}\)-\(\dfrac{1}{95}\)-\(\dfrac{1}{94}\)) = 0

Vì\(\dfrac{1}{98}< \dfrac{1}{98}< \dfrac{1}{97}< \dfrac{1}{96}< \dfrac{1}{95}< \dfrac{1}{94}\)

Nên (\(\dfrac{1}{99}\) + \(\dfrac{1}{98}\) + \(\dfrac{1}{97}\) )- (\(\dfrac{1}{96}\) + \(\dfrac{1}{95}\) +\(\dfrac{1}{94}\) )< 0 

⇒\(x-100\) = 0

Vậy \(x\) = 100

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