1 If you don't finish your homework, you can't go out with your friend

2 They are not sure how to operate the new system

3 I spent 4 hours reading the first chapter of the book

4 Tennis is not as dangerous as snowboarding

X

1 to go to school by bike when they were young

2 teaching her children to play the piano 4 years ago

3 I could cook as well as my mom

4 using Perfume Pagoda as the theme of the presentation

Câu 16: A

Câu 14: C

Câu 12: A

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

Bài em cần đâu em?

Y chứa NaOH, NaAlO₂

Gọi số mol NaOH, NaAlO₂ trong mỗi phần là x, y (mol)

TN1:

\(n_{HCl}=0,1.1=0,1\left(mol\right)\)

PTHH: NaOH + HCl --> NaCl + H₂O

            0,1<----0,1

=> x = 0,1 (mol)

TN3: n_HCl = 0,75.1 = 0,75 (mol)

PTHH: NaOH + HCl --> NaCl + H₂O

             0,1--->0,1

             NaAlO₂ + HCl + H₂O  --> NaCl + Al(OH)₃

                y------>y------------------------>y

             Al(OH)₃ + 3HCl --> AlCl₃ + 3H₂O

           \(\dfrac{0,65-y}{3}\)<-(0,65-y)

=> \(n_{Al\left(OH\right)_3\left(3\right)}=y-\dfrac{0,65-y}{3}=\dfrac{4y-0,65}{3}\left(mol\right)\)

TN2: \(n_{HCl}=1.0,45=0,45\left(mol\right)\)

- Nếu kết tủa không bị hòa tan:

PTHH: NaOH + HCl --> NaCl + H₂O

            0,1--->0,1

            NaAlO₂ + HCl + H₂O  --> NaCl + Al(OH)₃

                0,35<--0,35-------------------->0,35

Điều kiện: y \(\ge\) 0,35

=> \(n_{Al\left(OH\right)_3\left(2\right)}=0,35\left(mol\right)\)

Do \(n_{Al\left(OH\right)_3\left(2\right)}=3.n_{Al\left(OH\right)_3\left(3\right)}\)

=> \(0,35=4y-0,65\)

=> y = 0,25 (Loại)

=> Kết tủa bị hòa tan 1 phần

PTHH: NaOH + HCl --> NaCl + H₂O

            0,1--->0,1

            NaAlO₂ + HCl + H₂O  --> NaCl + Al(OH)₃

                 y---->y------------------------->y

              Al(OH)₃ + 3HCl --> AlCl₃ + 3H₂O

          \(\dfrac{0,35-y}{3}\)<--(0,35-y)

=> \(n_{Al\left(OH\right)_3\left(2\right)}=y-\dfrac{0,35-y}{3}=\dfrac{4y-0,35}{3}\left(mol\right)\)

Do \(n_{Al\left(OH\right)_3\left(2\right)}=3.n_{Al\left(OH\right)_3\left(3\right)}\)

=> \(\dfrac{4y-0,35}{3}=4y-0,65\)

=> y = 0,2 

Vậy trong Y chứa \(\left\{{}\begin{matrix}NaOH:0,3\left(mol\right)\\NaAlO_2:0,6\left(mol\right)\end{matrix}\right.\)

Bảo toàn Na: n_Na = 0,9 (mol)

Bảo toàn Al: n_Al = 0,6 (mol)

=> m = 0,9.23 + 0,6.27 = 36,9 (g)

Y chứa NaOH, NaAlO₂

Gọi số mol NaOH, NaAlO₂ trong mỗi phần là x, y (mol)

TN1:

nHCl=0,1.1=0,1(mol)nHCl=0,1.1=0,1(mol)

PTHH: NaOH + HCl --> NaCl + H₂O

            0,1<----0,1

=> x = 0,1 (mol)

TN3: n_HCl = 0,75.1 = 0,75 (mol)

PTHH: NaOH + HCl --> NaCl + H₂O

             0,1--->0,1

             NaAlO₂ + HCl + H₂O  --> NaCl + Al(OH)₃

                y------>y------------------------>y

             Al(OH)₃ + 3HCl --> AlCl₃ + 3H₂O

           nAl(OH)3(3)=y−0,65−y3=4y−0,653(mol)nAl(OH)3(3)=y−0,65−y3=4y−0,653(mol)

TN2: nHCl=1.0,45=0,45(mol)nHCl=1.0,45=0,45(mol)

- Nếu kết tủa không bị hòa tan:

PTHH: NaOH + HCl --> NaCl + H₂O

            0,1--->0,1

            NaAlO₂ + HCl + H₂O  --> NaCl + Al(OH)₃

                0,35<--0,35-------------------->0,35

Điều kiện: y  0,35

=> nAl(OH)3(2)=0,35(mol)nAl(OH)3(2)=0,35(mol)

Do nAl(OH)3(2)=3.nAl(OH)3(3)nAl(OH)3(2)=3.nAl(OH)3(3)

=> 

=> y = 0,25 (Loại)

=> Kết tủa bị hòa tan 1 phần

PTHH: NaOH + HCl --> NaCl + H₂O

            0,1--->0,1

            NaAlO₂ + HCl + H₂O  --> NaCl + Al(OH)₃

                 y---->y------------------------->y

              Al(OH)₃ + 3HCl --> AlCl₃ + 3H₂O

          nAl(OH)3(2)=y−0,35−y3=4y−0,353(mol)nAl(OH)3(2)=y−0,35−y3=4y−0,353(mol)

Do nAl(OH)3(2)=3.nAl(OH)3(3)nAl(OH)3(2)=3.nAl(OH)3(3)

=> 

lỗi

lỗi

\(a,4y^2+5y^2-2y^2=\left(4+5-2\right)y^2=7y^2\\ b,\dfrac{11}{3}xy^2+\dfrac{2}{3}xy^2+\dfrac{5}{3}xy^2=\left(\dfrac{11}{3}+\dfrac{2}{3}+\dfrac{5}{3}\right)xy^2=6xy^2\)

I. Write sentences with the cues given

1. Mai / usually / listen / K - pop music / free time.

→ Mai usually listens to K-pop music in her free time.

2. when / I / be / a child / I / enjoy / play / computer games

→ When I was a child, I enjoyed playing computer games.

3. my father / spend / most / spare time / look after / the garden

→ My father spends most of his spare time looking after the garden.

4. watching TV / most / popular / leisure activity / Britain ?

→ Is watching TV the most popular leisure activity in Britain?

5. many teenagers / addicted / the Internet / computer games 

→ Many teenagers are addicted to the Internet and computer games.

6. she / get / hooked / the medical drama / after / watch / the first people 

→ She got hooked on the medical drama after watching the first episode.

7. most / my friends / prefer / play sports / to / surf the net

→ Most of my friends prefer to play sports rather than surf the net.

8. today's world  / teenagers / rely / technology / more / the past 

→ In today's world, teenagers rely on technology more than in the past

II. Write the second sentences so that it has a similar meaning to the first one

1. It takes us more than two hours to see the film " Avatar "

→ The film " Avatar " requires more than two hours of our time to watch.

2. She likes to hang out with friends on Saturday evening

→ She's interested in socializing with friends on Saturday evening