1.A= 1.2.3+2.3.4+...+29.30.31+x=15

\(4A=1.2.3.4+2.3.4.\left(5-1\right)+...+29.30.31.\left(32-28\right)+4x=60\)

\(\Rightarrow4A=1.2.3.4+2.3.4.5-1.2.3.4+...+29.30.31.32-28.29.30.31+4x=60\)

Từ đó suy ra nha bạn

2.\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(=\frac{2}{2\left(2+1\right)}+\frac{2}{3.\left(3+1\right)}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(=2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\\ =1-\frac{2}{\left(x+1\right)}=\frac{2007}{2009}\)

\(\Rightarrow\frac{2}{x+1}=\frac{2}{2009}\Rightarrow x+1=2009\Rightarrow x=2008\)

\(=\dfrac{-8}{27}\cdot81+\dfrac{9}{16}\cdot32\)

=-24+18

=-6

1.

1.2 +2.3 +...+97.98

=1/3.(1.2.3 +2.3.3 +3.4.3 +...+97.98.3)

=1/3.(1.2.3 - 0.1.2+ 2.3.4 -1.2.3 + 3.4.5 -2.3.4 + ... +97.98.99 -96.97.98)

=1/3 . 97.98.99

= 313698

=>1.2 +2.3 +...+97.98-x=16

=>313698-x=16

=> x=313682

4. 

\(\left[\left(\frac{36}{x}-x\right):x-x\right]:x-x=-x\)

\(\left[\left(\frac{36}{x}-x\right):x-x\right]:x=-x+x\)

\(\left[\left(\frac{36}{x}-x\right):x-x\right]:x=0\)

\(\left[\left(\frac{36}{x}-x\right):x-x\right]=0\)

\(\left(\frac{36}{x}-x\right):x=x\Rightarrow\frac{36}{x}-x=x^2\)

\(\frac{36}{x}=x^2+x=x\left(x+1\right)\Rightarrow36=x^2\left(x+1\right)\)

Mà Ư(36)={1;2;3;4;6;9;12;18;36}; 9 là số chính phương duy nhất bé hơn 36=> x² = 9 => x=3

2 câu kia thì đợi một lúc.

Ta có: \(x^2+5y^2+2xy-4x-8y+2015\)

\(=\left(x^2+2xy+y^2\right)-\left(4x+8y\right)+4+\left(4y^2-4y+1\right)+2010\)

\(=[\left(x+y\right)^2-4\left(x+2y\right)+4]+\left(4y^2-4y+1\right)+2010\)

\(=\left(x+y-2\right)^2+\left(2y-1\right)^2+2010\)

mà \(\left(x+y-2\right)^2,\left(2y-1\right)^2\ge0\)

nên \(x^2+5y^2+2xy-4x-8y+2015\ge2010\)

Vậy MIN= 2010 \(\Leftrightarrow x=\frac{3}{2},y=\frac{1}{2}.\)

Tờ 1

41 It's very important to use body language in communication

42 Despite her age, she still leads an active life

43 My mother said that you had to decorate the room carefully

44 People recycle old cans to make new ones

45 Tim is always forgetting his homework

46 T

47 F

48 T

49 T

50 F

Tờ 2

17 C => hard

18 do => make

19 D => has

20 to go => going

21 A => At

22 B => to

23 C => beautifully

24 D => five-star

25 is => was

V

26 would travel

27 be

28 to buy

29 has spoken

30 Has - just been finished

VI

31 for

32 as

33 about

34 with

35 than

VII

36 development

37 exploration

38 behavior

39 deforestation

40 specialness

Tờ 3

IV

1 sat

2 Was - published

3 would help

4 had left

5 cleaning

V

21 belonging

22 arrangement

23 household

24 cruelty

25 unacceptable

VI

26 C => with

27 A => the prettier

28 C => not smoke

29 C => going

30 A => Despite

VII

31 in

32 of

33 up

34 of

35 for

VIII

36 When we were children, we used to flight our kites in the afternoons

37 She prectise playing the guitar more regularly than her sister

38 He will fail the next exam unless he study hard

A = 0,5 - | x - 3,5 |

Vì | x - 3,5 | >= 0

=> A = 0,5 - | x - 3,5 | < = 0,5

Dấu ( = ) xảy ra khi : | x - 3,5 | = 0

                                  x  - 3,5  = 0

                                           x  = 3,5

Vậy A đạt GTLN là 0,5 khi x = 3,5

B = - | 1,4 - x | - 2

Vì | 1,4 - x  | > = 0

=> B = - | 1,4 - x | - 2 < = - 2

Dấu ( = ) xảy ra khi : | 1,4 - x | = 0

                                    1,4 - x = 0

                                           x  = 1,4

Vậy B đạt GTLN là -2 khi x = 1,4

A vì cái trị tuyệt đối ý nó luôn lớn hơn hoặc bằng 0 ý nên A luôn bé hơn hoặc bằng 0,5 ý vậy GTLN của A là 0,5 ý

B vì âm trị tuyệt đối luôn bé hơn hoặc bằng 0 ý nên B luôn bé hơn hoặc bằng -2 ý vậy GTLN của A là -2 ý

16x⁴ - 64 = 16(x⁴ - 4) = 16[(x²)² - 2²] = 16(x² - 2)(x² + 2) = 16[x² -\(\left(\sqrt{2}\right)^2\)](x² + 2) = 16\(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x^2+2\right)\)

\(16x^4-64\)

\(=16\left(x^4-4\right)\)

\(=16\left(x^2-2\right)\left(x^2+2\right)\)

\(=16\left(x^2-\left(\sqrt{2}\right)^2\right)\left(x^2+2\right)\)

\(=16\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x^2+2\right)\)

Bài này ra kết quả trên là lớp 9 . Còn lớp 8 là : \(16\left(x^2-2\right)\left(x^2+2\right)\)

\(16x^4-64=16\left(x^4-4\right)=16\left[\left(x^2\right)^2-2^2\right]=16\left(x^2+2\right)\left(x^2-2\right)=16\left(x^2+2\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)

a)\(\left|x+\dfrac{2}{3}\right|=\dfrac{5}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=\dfrac{-5}{6}\\x+\dfrac{2}{3}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)

b) \(\left(x-\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{2}{3}\\x-\dfrac{1}{3}=\dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)

a) Ta có: \(\left|x+\dfrac{2}{3}\right|=\dfrac{5}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=-\dfrac{5}{6}\\x+\dfrac{2}{3}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)

b) Ta có: \(\left(x-\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{2}{3}\\x-\dfrac{1}{3}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)

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