Bài 1:

a: Hai cạnh đáy là AB,CD

Hai cạnh bên là AD,BC

b: Các cặp góc kề cạnh đáy là:

\(\widehat{BAD};\widehat{ABC}\)

\(\widehat{ADC};\widehat{BCD}\)

Các cặp góc kề cạnh bên là:

\(\widehat{BAD};\widehat{ADC}\)

\(\widehat{ABC};\widehat{BCD}\)

c: Hai đường chéo là AC,BD

Bài 2:

a: Ta có: ΔDAC vuông cân tại D

=>\(\widehat{DAC}=\widehat{DCA}=45^0\)

Ta có: ΔABC vuông cân tại A

=>\(\widehat{ABC}=\widehat{ACB}=45^0\)

Ta có: \(\widehat{DAC}=\widehat{ACB}\left(=45^0\right)\)

mà hai góc này là hai góc ở vị trí so le trong

nên AD//CB

=>ABCD là hình thang

Hình thang ABCD có AD\(\perp\)DC

nên ABCD là hình thang vuông

b: ABCD là hình thang vuông có hai đáy là AD,CB và AD\(\perp\)DC

=>CB\(\perp\)CD

=>\(\widehat{ADC}=\widehat{DCB}=90^0\)

Ta có: AD//CB

=>\(\widehat{DAB}+\widehat{ABC}=180^0\)

=>\(\widehat{DAB}=180^0-45^0=135^0\)

a) 

\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\\ =\dfrac{8}{9}-\left(\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\right)\\ =\dfrac{8}{9}-\left(\dfrac{1}{8\cdot9}+\dfrac{1}{7\cdot8}+\dfrac{1}{6\cdot7}+\dfrac{1}{6\cdot5}+\dfrac{1}{4\cdot5}+\dfrac{1}{3\cdot4}+\dfrac{1}{2\cdot3}+\dfrac{1}{1\cdot2}\right)\\ =\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)\\ =\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\\ =\dfrac{8}{9}-\dfrac{8}{9}\\ =0\)

b) 

\(\left(-\dfrac{1}{2}\right)-\left(\dfrac{-3}{5}\right)+\left(-\dfrac{1}{9}\right)+\dfrac{1}{127}-\dfrac{7}{18}+\dfrac{4}{35}-\left(\dfrac{-2}{7}\right)\\ =\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\dfrac{1}{127}\\ =\dfrac{-9-2-7}{18}+\dfrac{21+10+4}{35}+\dfrac{1}{127}\\ =-1+1+\dfrac{1}{127}\\ =\dfrac{1}{127}\)

c) (*sửa*)

\(\dfrac{3}{5}+\dfrac{3}{11}-\dfrac{-3}{7}+\dfrac{2}{97}-\dfrac{1}{35}-\dfrac{3}{4}-\dfrac{23}{44}\\ =\dfrac{3}{5}+\dfrac{3}{11}+\dfrac{3}{7}+\dfrac{2}{97}-\dfrac{1}{35}-\dfrac{3}{4}+\dfrac{23}{44}\\ =\left(\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{1}{35}\right)+\left(\dfrac{3}{11}-\dfrac{3}{4}-\dfrac{23}{44}\right)+\dfrac{2}{97}\\ =\dfrac{21+15-1}{35}+\dfrac{12-33-23}{44}+\dfrac{2}{97}\\ =1+\left(-1\right)+\dfrac{2}{97}\\ =\dfrac{2}{97}\)

ABCD là hình vuông

=>AB//CD

mà C\(\in\)DE

nên AB//DE

Ta có: DEFG là hình chữ nhật

=>DE//FG

mà AB//DE

nên AB//FG

a) 

\(32< 2^x< 128\\ =>2^5< 2^x< 2^7\\ =>5< x< 7\\ =>x=6\)

b) 

\(2\cdot16\ge2^x>4\\ =>2\cdot2^4\ge2^x>2^2\\ =>2^5\ge2^x>2^2\\ =>5\ge x>2\\ =>x\in\left\{3;4;5\right\}\)

c) 

\(9\cdot27\le3^x\le243\\ =>3^2\cdot3^3\le3^x\le3^5\\ =>3^5\le3^x\le3^5\\ =>5\le x\le5\\ =>x=5\)

d) 

\(x^{2019}=x\\ =>x^{2019}-x=0\\ =>x\left(x^{2018}-1\right)=0\)

TH1: x = 0

TH2: `x^2018-1=0`

`=>x^2018=1`

`=>x^2018=1^2018`

`=>x=1` hoặc `x=-1`

a: \(32< 2^x< 128\)

=>\(2^5< 2^x< 2^7\)

=>5<x<7

mà x là số tự nhiên

nên x=6

b: \(2\cdot16>=2^x>4\)

=>\(2^5>=2^x>2^2\)

=>2<x<=5

mà x là số tự nhiên

nên \(x\in\left\{3;4;5\right\}\)

c: \(9\cdot27< =3^x< =243\)

=>\(243< =3^x< =243\)

=>\(3^x=243=3^5\)

=>x=5

d: \(x^{2019}=x\)

=>\(x\left(x^{2018}-1\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x^{2018}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^{2018}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

e: \(2^{x+1}+4\cdot2^x=3\cdot2^7\)

=>\(2^x\cdot2+4\cdot2^x=6\cdot2^6\)

=>\(6\cdot2^x=6\cdot2^6\)

=>x=6

f: \(2^{2x}+2^{2x+3}=3^2\cdot8^4\)

=>\(2^{2x}+2^{2x}\cdot8=9\cdot8^4\)

=>\(9\cdot2^{2x}=9\cdot2^{12}\)

=>2x=12

=>x=6

g: \(27^{x+1}=9^{x+5}\)

=>\(3^{3\left(x+1\right)}=3^{2\left(x+5\right)}\)

=>3(x+1)=2(x+5)

=>3x+3=2x+10

=>3x-2x=10-3

=>x=7

h: \(3^{x+2}+5\cdot3^{x+1}=648\)

=>\(3^x\cdot9+5\cdot3^x\cdot3=648\)

=>\(3^x\cdot24=648\)

=>\(3^x=\dfrac{648}{24}=27=3^3\)

=>x=3

\(\left(1:\dfrac{1}{7}\right)^2\left[\left(2^2\right)^3:2^5\right]\cdot\dfrac{1}{49}\\ =7^2\left(2^6:2^5\right)\cdot\dfrac{1}{7^2}\\=\left(7^2\cdot\dfrac{1}{7^2}\right)\cdot2^{6-5}\\ =1\cdot2^1\\ =2\)

\(\left(1:\dfrac{1}{7}\right)^2\left[\left(2^2\right)^3:2^5\right]\cdot\dfrac{1}{49}\)

\(=\dfrac{7^2}{49}\cdot\left(2^6:2^5\right)\)

\(=\dfrac{49}{49}\cdot2=2\)

\(\left[\left(\dfrac{4}{3}\right)^{-3}\cdot\left(\dfrac{3}{4}\right)^5\right]:\left(\dfrac{3}{8}\right)^7\\ =\left[\left(\dfrac{3}{4}\right)^3\cdot\left(\dfrac{3}{4}\right)^5\right]:\left(\dfrac{3}{8}\right)^7\\ =\left(\dfrac{3}{4}\right)^{3+5}:\dfrac{3^7}{8^7}\\ =\left(\dfrac{3}{4}\right)^8\cdot\dfrac{8^7}{3^7}\\ =\dfrac{3^8}{4^8}\cdot\dfrac{8^7}{3^7}\\ =\dfrac{3^8}{2^{16}}\cdot\dfrac{2^{21}}{3^7}=3\cdot2^5=3\cdot32=96\)

\(\left[\left(\dfrac{4}{3}\right)^{-3}\cdot\left(\dfrac{3}{4}\right)^5\right]:\left(\dfrac{3}{8}\right)^7\)

\(=\left[\left(\dfrac{3}{4}\right)^3\cdot\left(\dfrac{3}{4}\right)^5\right]:\dfrac{3^7}{8^7}\)

\(=\left(\dfrac{3}{4}\right)^8\cdot\dfrac{8^7}{3^7}=\dfrac{3^8}{4^8}\cdot\dfrac{8^7}{3^7}=\dfrac{3\cdot2^{21}}{2^{16}}=3\cdot2^5=3\cdot32=96\)

Ta có: xy\(\perp\)AB

x'y'\(\perp\)AB

Do đó: xy//x'y'

a) Ta có: \(\widehat{cNb}+\widehat{MNb}=180^{\circ}\) (hai góc kề bù)

\(\Rightarrow\widehat{MNb}=180^{\circ}-\widehat{cNb}=180^{\circ}-55^{\circ}=125^{\circ}\)

b) Ta có: \(\widehat{MNb}=\widehat{aMN}\left(=125^{\circ}\right)\)

Mà hai góc này đều nằm ở vị trí so le trong

Nên \(Ma//Nb\)

12 We have been taught French by Mr Smith for 2 years

13 The children weren't looked after properly 

14 This street wasn't swept last week

15 A great deal of tea is drunk in England

16 English is spoken all over the world

17 Two poems were being written by Tom

18 Her dog is often taken for a walk by hẻ

19 How many lessons are going to be learnt by you next month?

20 I wasn't introduced to her mother by her

21 Electric lights had been invented before I was born