13. False
14. True
15. True
16. False
17. D
18. A

Câu 28

a. \(A=\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right).\dfrac{x+2\sqrt{x}}{\sqrt{x}}\)

\(=\left[\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\)

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

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

\(=\dfrac{4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}\)

\(=\dfrac{4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

1/ \(A=6+5^2+5^3+5^4+...+5^{1996}+5^{1997}\)

\(=1+5+5^2+5^3+5^4+...+5^{1996}+5^{1997}\)

\(\Rightarrow5A=5+5^2+5^3+5^4+5^5+...+5^{1997}+5^{1998}\)

Khi đó: 

\(5A-A=\left(5+5^2+5^3+5^4+5^5+...+5^{1997}+5^{1998}\right)-\left(1+5+5^2+5^3+5^4+...+5^{1996}+5^{1997}\right)\)

\(4A=5+5^2+5^3+5^4+5^5+...+5^{1997}+5^{1998}-1-5-5^2-5^3-5^4-...-5^{1996}-5^{1997}\)

\(4A=5^{1998}-1\)

\(\Rightarrow A=\dfrac{5^{1998}-1}{4}\)

2/ \(B=10+9^2+9^3+9^4+...+9^{2004}+9^{2005}\)

\(=1+9+9^2+9^3+9^4+...+9^{2004}+9^{2005}\)

\(\Rightarrow9B=9+9^2+9^3+9^4+9^5+...+9^{2005}+9^{2006}\)

Khi đó:

\(9B-B=\left(9+9^2+9^3+9^4+9^5+...+9^{2005}+9^{2006}\right)-\left(1+9+9^2+9^3+9^4+...+9^{2004}+9^{2005}\right)\)

\(8B=9+9^2+9^3+9^4+9^5+...+9^{2005}+9^{2006}-1-9-9^2-9^3-9^4-...-9^{2004}-9^{2005}\)

\(8B=9^{2006}-1\)

\(\Rightarrow B=\dfrac{9^{2006}-1}{8}\)

3/ \(C=\dfrac{5^{997}\left(5^{100}+2\right)-10.5^{996}-1}{4}\)

\(=\dfrac{5^{997}.5^{100}+5^{997}.2-10.5^{996}-1}{4}\)

\(=\dfrac{5^{1097}+5^{996}.5.2-10.5^{996}-1}{4}\)

\(=\dfrac{5^{1097}+10.5^{996}-10.5^{996}-1}{4}\)

\(=\dfrac{5^{1097}-1}{4}\)

4/ Theo giả thiết, ta có: x = 2005 => x - 2005 = 0

Do đó:

\(D=x^{20}-2006x^{19}+2006x^{18}-2006x^{17}+...+2006x^2-2006x+2006\)

\(=\left(x^{20}-2005x^{19}\right)-\left(x^{19}-2005x^{18}\right)+\left(x^{18}-2005x^{17}\right)+...+\left(x^2-2005x\right)-\left(x-2005\right)+1\)

\(=x^{19}\left(x-2005\right)-x^{18}\left(x-2005\right)+x^{17}\left(x-2005\right)+...+x\left(x-2005\right)-\left(x-2005\right)+1\)

\(=1\)

1. A. chose B. decided C. selected D. picked
2. A. transport B. carry C. tour D. travel
3. A. said B. spoke C. told D. explained
4. A. because B. so C. but D. and
5. A. forget B. lose C. leave D. miss
6. A. further B. after C. next D. later
7. A. down B. along C. from D. away
8. A. shall B. could C. must D. would
9. A. nervous B. disappointed C. angry D. bored
10. A. knew B. found C. made D. met

5. C

6. C

uh

Phân tích đa thức thành nhân tử: 

a) x² - 2x + 2xy - 4y

= (x² - 2x) + (2xy - 4y)

= x(x - 2) + 2y(x - 2)

= (x + 2y)(x - 2)

b) (x + 1)(x + 3)(x + 5)(x + 7) + 15

= [(x + 1)(x + 7)][(x + 3)(x + 5)] + 15

= (x² + 8x + 7)(x² + 8x + 15) + 15

= (x² + 8x + 7)[(x² + 8x + 7) + 8] + 15

Đặt t = x² + 8x + 7, khi đó ta có:

t(t + 8) + 15

= t² + 8t + 15

= (t² + 3t) + (5t + 15)

= t(t + 3) + 5(t + 3)

= (t + 5)(t + 3)

Thay t = x² + 8x + 7, ta được:

(x² + 8x + 7 + 5)(x² + 8x + 7 + 3)

= (x² + 8x + 12)(x² + 8x + 10)

= [(x² + 2x) + (6x + 12)](x² + 8x + 10)

= [x(x + 2) + 6(x + 2)](x² + 8x + 10)

= (x + 6)(x + 2)(x² + 8x + 10)

Tìm x:

x⁴ + 2x³ - 4x - 4 = 0

<=> x⁴ + 2x³ + x² - x² - 4x - 4 = 0

<=> (x⁴ + 2x³ + x²) - (x²​ + 4x + 4) = 0

<=> x²(x² + 2x + 1) - (x²​ + 4x + 4) = 0

<=> x²(x + 1)² - (x + 2)² = 0

<=> [x(x + 1) - (x + 2)][x(x + 1) + (x + 2)] = 0

<=> x(x + 1) - (x + 2) = 0 hoặc x(x + 1) + (x + 2) = 0

<=> x² + x - x - 2 = 0 hoặc x² + x + x + 2 = 0

<=> x² - 2 = 0 hoặc (x² + 2x + 1) + 1 = 0

<=> x² = 2 hoặc (x + 1)² + 1 = 0 (vô lí)

<=> \(x=\sqrt{2}\) hoặc \(x=-\sqrt{2}\)

Vậy \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}.\)

Tìm cặp số nguyên x, y:

2x + 5 = xy + y

<=> (x + 1)y = 2x + 5

Với x = -1, ta có: 0y = 3 (vô lí)

Với \(x\ne-1,\) khi đó phương trình tương đương:

\(y=\dfrac{2x+5}{x+1}\)

\(\Leftrightarrow y=\dfrac{\left(2x+2\right)+3}{x+1}\)

\(\Leftrightarrow y=\dfrac{2\left(x+1\right)+3}{x+1}\)

\(\Leftrightarrow y=2+\dfrac{3}{x+1}\)

Để y nguyên \(\Rightarrow\dfrac{3}{x+1}\) nguyên \(\Rightarrow3⋮\left(x+1\right)\Rightarrow\left(x+1\right)\inƯ\left(3\right)\)

mà \(Ư\left(3\right)=\left\{-3;-1;1;3\right\}.\)

\(\Rightarrow\left[{}\begin{matrix}x+1=-3\\x+1=-1\\x+1=1\\x+1=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-2\\x=0\\x=2\end{matrix}\right.\)

So với yêu cầu đề bài thấy thỏa mãn.

Với \(x=-4\Rightarrow y=1\)

Với \(x=-2\Rightarrow y=-1\)

Với \(x=0\Rightarrow y=5\)

Với \(x=2\Rightarrow y=3\)

Vậy nghiệm của phương trình là: \(\left(x;y\right)=\left(-4;1\right),\left(-2;-1\right),\left(0;5\right),\left(2;3\right).\)

cụ thể là định lí 2

định lí đường trung bình của tam giác đó

VIII.
2. My father has not smoked for 10 years.
3. Annie asked me if I could possibly tell her the time.
4. She used to be a skillful golf player.
IX.
1. Nowadays, teenagers rely on technology more than in the past.