\(f'\left(x\right)=x^2+2x\)

a.

\(f'\left(-3\right)=3\) ; \(f\left(-3\right)=-2\)

Phương trình tiếp tuyến:

\(y=3\left(x+3\right)-2\Leftrightarrow y=3x+7\)

b.

Gọi \(x_0\) là hoành độ tiếp điểm, do hệ số góc tiếp tuyến bằng 3

\(\Rightarrow f'\left(x_0\right)=3\Rightarrow x_0^2+2x_0=3\Rightarrow x_0^2+2x_0-3=0\)

\(\Rightarrow\left[{}\begin{matrix}x_0=1\Rightarrow y_0=-\dfrac{2}{3}\\x_0=-3\Rightarrow y_0=-2\end{matrix}\right.\)

Có 2 tiếp tuyến thỏa mãn:

\(\left[{}\begin{matrix}y=3\left(x-1\right)-\dfrac{2}{3}=3x-\dfrac{11}{3}\\y=3\left(x+3\right)-2=3x+7\end{matrix}\right.\)

c. Tiếp tuyến song song (d) nên có hệ số góc bằng 8

Gọi \(x_0\) là hoành độ tiếp điểm \(\Rightarrow x_0^2+2x_0=8\)

\(\Rightarrow\left[{}\begin{matrix}x_0=2\Rightarrow y_0=\dfrac{14}{3}\\x_0=-4\Rightarrow y_0=-\dfrac{22}{3}\end{matrix}\right.\)

Có 2 tiếp tuyến thỏa mãn:

\(\left[{}\begin{matrix}y=8\left(x-2\right)+\dfrac{14}{3}=...\\y=8\left(x+4\right)-\dfrac{22}{3}=...\end{matrix}\right.\)

Gọi số học sinh khối 6 là x

Theo đề, ta có: \(x-3\in BC\left(8;12;15\right)\)

\(\Leftrightarrow x-3\in\left\{120;240;360;...\right\}\)

\(\Leftrightarrow x\in\left\{123;243;363\right\}\)

mà 200<=x<=300

nên x=243

Gọi số học sinh khối 6 là a

a + 3 \(⋮8;12;15\)

\(\Rightarrow\) \(a+3\in BC\left(8;12;15\right)\)

8 = 2 . 3

12 = 2² . 3

15 = 3 . 5

\(\Rightarrow\) BCNN (8; 12; 15) = 2² . 3 . 5 = 60

Mà 203 < a + 3 < 303 học sinh​​

\(\Rightarrow\) a + 3 \(\in\) {240; 300}

\(\Rightarrow\) a \(\in\) {237; 207}

1 It took me 10 minutes to walk to my office

2 After she had written a letter, she went to bed

3 Before he bought a radio, he had checked the price

4 After they had argued, they fought

5 Before she met a close friend,she had gone out for a walk

6 Before she decided to go away, she had faced the matter

7 After she had watched the film, she wrote a report

8 I saw her cross the road

9 He heard them sing a song

10 They let him use their car

11 After the train had left, he arrived at the station

12 Peter and Maria decided to go to the cinema

Bài 8:

a: Thay x=16 vào P, ta được:

\(P=\dfrac{4+2}{4-1}=\dfrac{6}{3}=2\)

b: Thay \(x=3+2\sqrt{2}\) vào P, ta được:

\(P=\dfrac{\sqrt{2}+1+2}{\sqrt{2}+1-1}=\dfrac{3+\sqrt{2}}{\sqrt{2}}=3\sqrt{2}+2\)

1 A

2 C

3 D

4 A

5 B

6 A

7 D

8 B

9 A

10 D

11 A

12 C

13 B

14 C

15 B

16 A

17 C

18 B

19 B

20 B

Bài 2:

\(10M=\dfrac{10^{12}+10}{10^{12}+1}=1+\dfrac{9}{10^{12}+1}\)

\(10N=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)

Ta có: \(10^{12}+1>10^{11}+1\)

=>\(\dfrac{9}{10^{12}+1}< \dfrac{9}{10^{11}+1}\)

=>\(\dfrac{9}{10^{12}+1}+1< \dfrac{9}{10^{11}+1}+1\)

=>10M<10N

=>M<N

Bài 1:

\(A=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{900}\right)\)

\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{30}\right)\cdot\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{30}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{29}{30}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{31}{30}\)

\(=\dfrac{1}{30}\cdot\dfrac{31}{2}=\dfrac{31}{60}\)