Refer

1. “Your cousin speaks English very well” Paul told me

Paul said that ___________my cousin spoke English very well____________

2. “The man broke out of prison yesterday” said the policeman

The policeman told us_that the man had broken out of prison the day beforr__

3. “I’ll lend you this book as soon as I finish it” Owen said to me

Owen said __me that he would lend me that book as soon as he finished it___

4. “I think I forgot to turn off the lights this morning” Brenda told Brian

Brenda told Brian ____that he thought he had forgotten to turn off the lights that morning.____

5. “I work eight hours a day, except when the children are on holiday” said Mrs. Wood

Mrs. Wood said me that he worked eight hours a day, excepted when the children were on holiday

6. “You’ve been making good progress this semester” Miss Lynn told me

Miss Lynn said that _____I had been making good progress that semester_________

7. “If you bought all the tickets, you would win the lottery” the man said

The man told me ______that If I had bought all the tickets, I would win the lottery______________

8. “I like swimming but I don’t go very often” Jill said to Pam

Jill said that ______he liked swimming but he didn’t go very often___________________________

9. “I want to buy it, but I haven’t brought any money” said Patrick

Patrick told me _________that he wanted to buy it, but he hadn’t brought any money_______________________

10. “I’m going to visit my aunt in Hue, but I’m not sure when” said Mai

Mai told me _________that she was going to visit her aunt in Hue, but she was not sure when__________________

\(a,\dfrac{x^2+4x+4}{2x^2+4x}=\dfrac{\left(x+2\right)^2}{2x\left(x+2\right)^2}=\dfrac{x+2}{2x}\ne\dfrac{x+2}{2}\\ b,\dfrac{x^2-2}{x^2-1}\ne\dfrac{x+2}{x+1}\\ c,\dfrac{x^3-36x}{x^3+12x^2+36}=\dfrac{x\left(x-6\right)\left(x+6\right)}{x\left(x+6\right)^2}=\dfrac{x-6}{x+6}\ne\dfrac{-\left(x-6\right)}{x+6}=\dfrac{6-x}{x+6}\)

\(=2x^2\left(x-1\right)-4x\left(x-1\right)=\left(x-1\right)\left(2x^2-4x\right)=2x\left(x-2\right)\left(x-1\right)\)

I

1.What type of energy do you use at home?
2.John uses low energy light bulbs to save electricity.
3.Mai can't go to the cinema at present because she is studying for her exam.
4.What book are you currently reading?
5.She is not swimming in the swimming pool at the moment.
6.Scientists are looking for a new energy source to replace coal now

II.

1. Coal is more expensive and more polluting than natural gas.

2. We are looking for cheap and clean and effective sourced of energy..

3.. You should switch off electrical appliances when they aren’t in use.

4. Despite being the most polluting of fossil fuels, coal is still the largest sources of energy worldwide.

5. People will use biogas for fuel in homes and for transport

Bài 5 hình 1: (tự vẽ hình nhé bạn)
a) Xét ΔABD và ΔACB ta có:
\(\widehat{BAD}\)= \(\widehat{BAC}\) (góc chung)
\(\widehat{ABD}\)= \(\widehat{ACB}\) (gt)
=> ΔABD ~ ΔACB (g-g)
=> \(\dfrac{AB}{AC}\) = \(\dfrac{BD}{CB}\) = \(\dfrac{AD}{AB}\) (tsđd)
b) Ta có: \(\dfrac{AB}{AC}\) = \(\dfrac{AD}{AB}\) (cm a)
=> \(AB^2\) = AD.AC
=> \(2^2\) = AD.4
=> AD = 1 (cm)
Ta có: AC = AD + DC (D thuộc AC)
      => 4   =   1   + DC
      => DC = 3 (cm)
c) Xét ΔABH và ΔADE ta có: 
   \(\widehat{AHB}\) = \(\widehat{AED}\) (=\(90^0\))
   \(\widehat{ADB}\) = \(\widehat{ABH}\) (ΔABD ~ ΔACB)
=> ΔABH ~ ΔADE
=> \(\dfrac{AB}{AD}\) = \(\dfrac{AH}{AE}\) = \(\dfrac{BH}{DE}\) (tsdd)
Ta có: \(\dfrac{S_{ABH}}{S_{ADE}}\) = \(\left(\dfrac{AB}{AD}\right)^2\)= \(\left(\dfrac{2}{1}\right)^2\)= 4
=> đpcm

Tiếp bài 5 hình 2 (tự vẽ hình)
a) Xét ΔABC vuông tại A ta có:
\(BC^2\) = \(AB^2\) + \(AC^2\)
\(BC^2\) = \(21^2\) + \(28^2\)
BC = 35 (cm)
b) Xét ΔABC và ΔHBA ta có:
\(\widehat{BAC}\) = \(\widehat{AHB}\) ( =\(90^0\))
\(\widehat{ABC}\) = \(\widehat{ABH}\) (góc chung)
=> ΔABC ~ ΔHBA (g-g)
=> \(\dfrac{AB}{BH}\) = \(\dfrac{BC}{AB}\) (tsdd)
=> \(AB^2\) = BH.BC
=> \(21^2\) = 35.BH
=> BH = 12,6 (cm)
c) Xét ΔABC ta có:
BD là đường p/g (gt)
=> \(\dfrac{AD}{DC}\) = \(\dfrac{AB}{BC}\) (t/c đường p/g)
Xét ΔABH ta có: 
BE là đường p/g (gt)
=> \(\dfrac{HE}{AE}\) = \(\dfrac{BH}{AB}\) (t/c đường p/g)
Mà: \(\dfrac{AB}{BC}\) = \(\dfrac{BH}{AB}\) (cm b)
=> đpcm
d) Ta có: \(\left\{{}\begin{matrix}\widehat{HBE}+\widehat{BEH}=90^0\\\widehat{ABD}+\widehat{ADB=90^0}\\\widehat{HBE}=\widehat{ABD}\end{matrix}\right.\)
=> \(\widehat{BEH}=\widehat{ADB}\)
Mà \(\widehat{BEH}=\widehat{AED}\) (2 góc dd)
Nên \(\widehat{ADB}=\widehat{AED}\)
=> đpcm