g, PT \(\Leftrightarrow\dfrac{x+24}{1996}+1+\dfrac{x+25}{1995}+1+\dfrac{x+26}{1994}+1+\dfrac{x+27}{1993}+1+\dfrac{x+2036}{4}-4=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{1996}+\dfrac{1}{1995}+\dfrac{1}{1994}+\dfrac{1}{1993}+\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

Vậy ...

D

2 You should say you are sorry

3 He ought ask his teacher about it

4 You should start having lessons

5 She ought not to watch TV late

6 You should get some exercise

E

1 can

2 could

3 ought

4 can

5 should

6 couldn't

7 can't

8 ought to

9 can

10 can

F

1 - D

2 - C

3 - B

4 - E

5 - A

2 You should say you are sorry

3 He ought ask his teacher about it

4 You should start having lessons

5 She ought not to watch TV late

6 You should get some exercise

a: \(y=\dfrac{1}{2}x\)

Câu 1: B

Câu 2: A 

Câu 3: C

Câu 4: B

Câu 5: D

Câu 6: B

Câu 7: A

Câu 8: A

Câu 9: C

Câu 10: D

Câu 11: D

Câu 12: B 

Câu 13: B

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

bị động hả em

 7. Nam is not doing his homework now.

→……nam's homework isn't being done by him now.………………

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