Bài 3: 

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{9}=\dfrac{b}{8}=\dfrac{c}{7}=\dfrac{d}{6}=\dfrac{b-d}{8-6}=\dfrac{70}{2}=35\)

Do đó: a=315; b=280; c=245; d=210

Ta có: \(6x+1=2\left(3x-1\right)+3\)

Vì \(2\left(3x-1\right)⋮\left(3x-1\right)\Rightarrow3⋮\left(3x-1\right)\)

\(\Rightarrow3x-1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow3x=\left\{2;0;4;-2\right\}\)

\(\Rightarrow x=\left\{\frac{2}{3};0;\frac{4}{3};\frac{-2}{3}\right\}\)

Vì biểu thức là số nguyên 

Vậy x = 0

Đầu bài có thiếu không nhỉ?

\(6x+1=6x-2+3.\)

\(\left(6x-2\right)+3=2\left(3x-1\right)+3⋮3x-1\)

1 C

2 A

3 C

4 B

5 D

23.

Gọi I là trung điểm MN \(\Rightarrow I\left(3;3\right)\)

\(\Rightarrow\overrightarrow{IN}=\left(2;-1\right)\Rightarrow IN=\sqrt{5}\)

Phương trình đường tròn đường kính MN, nhận I là tâm và có bán kính \(R=IN\) là:

\(\left(x-3\right)^2+\left(y-3\right)^2=5\)

Thay tọa độ E vào pt ta được:

\(\left(x-3\right)^2+4=5\Rightarrow\left(x-3\right)^2=1\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)  \(\Rightarrow x_1x_2=8\)

Cả 4 đáp án của câu này đều sai

24.

Gọi \(M\left(x;y\right)\) là 1 điểm bất kì thuộc \(\Delta\)

Do \(\Delta\) là đường phân giác của góc tạo bởi d và k nên:

\(d\left(M;d\right)=d\left(M;k\right)\Leftrightarrow\dfrac{\left|2x+y\right|}{\sqrt{2^2+1^2}}=\dfrac{\left|x+2y-3\right|}{\sqrt{1^2+2^2}}\)

\(\Leftrightarrow\left|2x+y\right|=\left|x+2y-3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+y=x+2y-3\\2x+y=-x-2y+3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-y+3=0\\x+y-1=0\end{matrix}\right.\)

- Với \(x-y+3=0\), ta có: 

\(\left(x_E-y_E+3\right)\left(x_F-y_F+3\right)=2.1=2>0\Rightarrow E;F\) nằm cùng phía so với \(x-y+3=0\) (thỏa mãn)

- Với \(x+y-1=0\) ta có:

\(\left(x_E+y_E-1\right)\left(x_F+y_F-1\right)=2.7=14>0\Rightarrow E;F\) nằm cùng phía so với \(x+y-1=0\) (thỏa mãn)

Vậy cả đáp án A và D đều đúng

Tương tự như câu 23, câu 24 đề bài tiếp tục sai

1 Japanese is too difficult for me to learn

2 If he felt well, he wouldn't go to bed early

3 I wish I hadn't gone out in the rain

4 They had me carry te box upstairs

5 Your sister is too young to enjoy this film

6 THey had this letter posted by me

7 The morning was too cold for them to go out

8 If the tickets were cheap enough, we would go there

9 The pupils collected these old clothes

10 The coffee is so excellent so I can drink it

11 Her house was sold 2 years ago

12 Some tickets was bought there by him

13 Waste paper was recycled to save money and labour

14 The TV set was so heavy that the girl couldn't move it

15 I will have someone type the report for you

16 There was so much fog that the driver couldn't see far

17 THe day was fine enoigh for them to enjoy sunbathing on the beach

18 I wish your were here now

19 Susan offeref me to lend her some money

20 They asked me is I could fill in the form

21 The acrobat whom we saw on TV last night is his uncle

22 The traffic jam prevented us from coming late

23 It is sucha wide gate that the struck can't get through it

24 Thet novel was written by Charales Dickens

25 Your answer is too difficult for me to answer

26 We've had so much homework that we can't find time to go out

27 He can't borrow the book because he don't have the library card

28 If the buses weren't full up, they would stop

29 She is not strong enough to carry her luggage

30 The dog which is playing with a ball is a big one

II

1 B

2 A

3 A

4 D

5 B

6 B

7 A

8 D

9 D

10 D

1 B

2 C

3 A

4 D

5 D

6 B

7 D

8 C

9 A

10 B

1 The meeting was canceled 3 days ago

2 She told me she was watching a film with her sister then

3 I admire the guitarist who is perfroming on the stage

4 Had it not been for Pauline's interest, the project would have been abandoned 

IV. 

1. The meeting was canceled 3 days ago.

2. She told me shewas watching a film with her sister then

3. I admire the guitarist who is performing on the stage

4. Had it not been for Pauline's interest, the project was abandoned. (chắc thế)

71.

\(\left\{{}\begin{matrix}BB'\perp\left(ABCD\right)\\BB'\in\left(ABB'A'\right)\end{matrix}\right.\) \(\Rightarrow\left(ABCD\right)\perp\left(ABB'A'\right)\)

74.

\(\left\{{}\begin{matrix}DD'\perp\left(ABCD\right)\\DD'\in\left(CDD'C'\right)\end{matrix}\right.\) \(\Rightarrow\left(ABCD\right)\perp\left(CDD'C'\right)\)