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Bài 2:

a. 3x(x - 6) - 2x² = x² + 6

<=> 3x² - 18x - 2x² - x² - 6 = 0

<=> 3x² - 2x² - x² - 18x - 6 = 0

<=> -18x - 6 = 0

<=> -18x = 6

<=> x = \(\dfrac{6}{-18}=\dfrac{-1}{3}\)

b. (x - 3)(x - 2) - 5 = x² - 4x

<=> x² - 2x - 3x + 6 - 5 - x² + 4x = 0

<=> x² - x² - 2x - 3x + 4x + 6 - 5 = 0

<=> -x + 1 = 0

<=> -x = -1

<=> x = 1

c. (x + 5)² - 8x = x² + 15

<=> x² + 10x + 25 - 8x - x² - 15 = 0

<=> x² - x² + 10x - 8x + 25 - 15 = 0

<=> 2x + 10 = 0

<=> 2x = -10

<=> x = -5

d. x² - 4x + 4 = 0

<=> x² - 2.2.x + 2² = 0

<=> (x - 2)² = 0

<=> x - 2 = 0

<=> x = 2

e. x² + 8x + 16 = 0

<=> x² + 2.x.4 + 4² = 0

<=> (x + 4)² = 0

<=> x + 4 = 0

<=> x = -4

f. x² - 36 = 0

<=> x² - 6² = 0

<=> (x - 6)(x + 6) = 0

<=> \(\left[{}\begin{matrix}x-6-0\\x+6=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

g. (x + 3)² - 16 = 0

<=> (x + 3)² - 4² = 0

<=> (x + 3 + 4)(x + 3 - 4) = 0

<=> (x + 7)(x - 1) = 0

<=> \(\left[{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\)

k: Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-2x^3+8\)

\(=x^3-8-2x^3+8\)

\(=-x^3\)

g: Ta có: \(\left(x-3\right)\left(x+3\right)-6x^2-x-10\)

\(=x^2-9-6x^2-x-10\)

\(=-5x^2-x-19\)

h: Ta có: \(\left(x+2\right)^3-2x^2\left(x-5\right)\)

\(=x^3+6x^2+12x+8-2x^3+10x^2\)

\(=-x^3+16x^2+12x+8\)

a: k=3/8

b: y=3/8x

4 going to buy a lot of things for the party tomorrow

5 is go out somewhere for a drink

6 Jane going to travel around the world

7 short and fair

8 a good football player

9 a fast runner

10 hard workers

11 swim well

12 worked busily

1 A

2 D

3 C

4 A

5 A

1 C

2 A

3 D

4 A

5 B

6 D

7 C

8 C

II

1 A

2 B

III

1 deaf

2 countless

3 inventor

4 delivery

IV

1 He worked with deaf-mutes at Boston University in America

2 I am going to see the movie Dream city at 6.45 this evening

V

1 going to visit my friends next Sunday

2 go to the movie tonight