16 funny

17 furious

18 awful

19 hilarious

20 bad

21 listen

22 to

23 speak

24 tell

25 about

26 say

27 pronounce

28 understand

29 mean

30 repeat

31 realize

32 allow

33 choose

34 panic

35 recommends

#\(yGLinh\)

16 funny

17 furious

18 awful

19 hilarious

20 bad

1 were traveling - saw

2 bought - were shopping

3 were watching - fell

4 hated - was

5 fell - was playing

6 C

7 I

8 C

9 C

10 I 

11 Walking is good for your health.

12 John is fed up with getting up early.

13 Doing yoga is relaxing.

14 I like spending time with my family.

15 I'm planning on having a birthday party.

#\(yGLinh\)

Task 1

1 were travelling - saw

2 bought - were shopping

3 were watching - fell

4 hated - was

5 fell - was playing

a) Căc cặp góc so le trong là:

\(\widehat{R_1}\) và \(\widehat{S_3}\)

\(\widehat{R_2}\) và \(\widehat{S_4}\)

Các cặp góc đồng vị là: 

\(\widehat{S_1}\) và \(\widehat{R_1}\)

\(\widehat{S_2}\) và \(\widehat{R_2}\)

\(\widehat{S_3}\) và \(\widehat{R_3}\)

\(\widehat{S_4}\) và \(\widehat{R_4}\)

Các cặp góc trong cùng phía là:

\(\widehat{S_4}\) và \(\widehat{R_1}\)

\(\widehat{S_3}\) và \(\widehat{R_2}\)

b) Ta có: 

\(\widehat{R_2}=120^o=>\widehat{S_2}=120^o\) (đồng vị)

\(\widehat{R_2}=120^o=>\widehat{S_3}=180^o-\widehat{R_2}=180^o-120^o=60^o\) (trong cùng phía) 

\(\widehat{S_4}=120^o=>\widehat{R_4}=120^o\) (đồng vị) 

\(\widehat{S_4}=120^o=>\widehat{R_1}=180^o-\widehat{S_4}=180^o-120^o=60^o\) (trong cùng phía)  

\(\widehat{R_1}=60^o=>\widehat{S_1}=60^o\) (đồng vị)  

\(\widehat{S_3}=60^o=>\widehat{R_3}=60^o\) (đồng vị) 

a) Ta có:

\(\left\{{}\begin{matrix}AB\perp AC\\KH\perp AC\end{matrix}\right.=>AB//KH\) 

b) Ta có: 

\(\widehat{ABK}=\widehat{BKI}\left(=60^o\right)\)

Mà hai góc này ở vị trí so le trong 

=> AB//KI 

c) AB//HK = > \(\widehat{ABK}+\widehat{HKB}=180^o\)

Mà: \(\widehat{ABK}=\widehat{BKI}\) 

\(=>\widehat{BKI}+\widehat{HKB}=180^o\)

=> \(\widehat{HKI}\) là góc bẹt hay H, K, I thẳng hàng

a) Ta có:

\(\widehat{ADE}=\widehat{ABC}\left(=45^o\right)\)

Mà hai góc này ở vị trí đồng vị

=> DE//BC

b) Ta có: 

\(\widehat{FEC}=\widehat{ECB}\left(gt\right)\)

Mà hai góc này ở vị trí so le trong 

=> EF//BC

c) Ta có: DE//BC

=> \(\widehat{DEC}+\widehat{ECB}=180^o\) (trong cùng phía) 

Mà: \(\widehat{FEC}=\widehat{ECB}\left(gt\right)\)

\(=>\widehat{FEC}+\widehat{ECB}=180^o\)

\(=>\widehat{DEF}\) là góc bẹt

=> D, E, F thẳng hàng

a) Ta có: 

\(\widehat{MAB}=\widehat{ABC}\left(=55^o\right)\)

Mà hai góc này ở vị trí so le trong 

=> AM//BC 

b) Ta có:
\(\widehat{NAC}=\widehat{ACB}\left(=40^o\right)\)

Mà hai góc này ở vị trí so le trong

=> AN//BC 

c) Xét tam giác ABC có:

\(\widehat{BAC}+\widehat{ABC}+\widehat{ACB}=180^o\\ =>\widehat{BAC}=180^o-\widehat{ABC}-\widehat{ACB}\\ =>\widehat{BAC}=180^o-55^o-40^o=85^o\) 

\(\widehat{MAB}+\widehat{BAC}+\widehat{NAC}=55^o+85^o+40^o=180^o\) 

=> \(\widehat{MAN}\) là góc bẹt => M, A, N thẳng hàng

\(\left(8+2\dfrac{1}{3}-\dfrac{3}{5}\right):\left(5-\dfrac{1}{4}-\dfrac{5}{8}\right)\\ =\left(8+2+\dfrac{1}{3}-\dfrac{3}{5}\right):\left(5-\dfrac{2}{8}-\dfrac{5}{8}\right)\\ =\left(10+\dfrac{1}{3}-\dfrac{3}{5}\right):\left(5-\dfrac{7}{8}\right)\\ =\left(\dfrac{150}{15}+\dfrac{5}{15}-\dfrac{9}{15}\right):\left(\dfrac{40}{8}-\dfrac{7}{8}\right)\\ =\dfrac{146}{15}:\dfrac{33}{8}\\ =\dfrac{146}{15}\cdot\dfrac{8}{33}\\ =\dfrac{1168}{495}\)

\(f\left(x\right)+g\left(x\right)=x^2h\left(x\right)\)

\(2x^3-3x^2+3x+1+g\left(x\right)=x^2\left(2x+1\right)\)

\(g\left(x\right)=2x^3+x^2-2x^3+3x^2-3x-1=4x^2-3x-1\)

chọn C

`#3107.101107`

Ta có:

`f(x) + g(x) = x^2h(x)`

`\Rightarrow g(x) = x^2h(x) - f(x)`

`g(x) = x^2 * (2x + 1) - (2x^3 - 3x^2 + 3x + 1)`

`= 2x^3 + x^2 - 2x^3 + 3x^2 - 3x - 1`

`= 4x^2 - 3x - 1`

Chọn C.

1C

2B

3B

4A

5D

6D

7A

8B

9D

10D

11B

12C 

13B 

14B

15D

16A

17A

18D

19D

 Xét đường thẳng BC, có AH, AB lần lượt là đường vuông góc và đường xiên kẻ từ A đến BC. Do đó \(AH< AB\).

 Chứng minh tương tự, ta được \(BK< BC\) và \(CL< CA\)

 Cộng theo vế 3 BĐT vừa tìm được, ta có:

 \(AH+BK+CL< AB+BC+CA\) (đpcm)