8.

Hàm có 1 điểm cực đại \(\left(x=-1\right)\)

9. 

Hàm có 1 điểm cực tiểu (\(x=-1\))

14.

\(y'=\dfrac{2x\left(x+1\right)-\left(x^2+3\right)}{\left(x+1\right)^2}=\dfrac{x^2+2x-3}{\left(x+1\right)^2}\)

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

Xét dấu y' trên trục số:

Từ dấu của y' ta thấy \(x=1\) là điểm cực tiểu

\(\Rightarrow y_{CT}=y\left(1\right)=2\)

1.
There are both many good things and many potential dangers in social network 
It's important for parents to teach their children how to use social media wisely
Social network is the start of bad things like cyberbullying

2.

A recent report has said that many teenagers have contacted online with strangers and it makes them feel scared or uncomfortable.

Others receive onine advertising that is inapproriate for their age.

It is important that parents should be aware of what children are doing online.

Parents should make their children understand that they respect children's privacy.
However, parents want to make sure that their children are safe.

1,C

2, A

3, D

 4, B

5, D

6, A

7, C

8, B

9, D

10, C

16. advices

17. Homelessness

18. disabled

19. natural

Giúp mình câu 14 và15 với ạ

Câu 14)

\(a,\\ =-\dfrac{3}{8}+\dfrac{8}{17}+\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{9}{17}\\ =\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{8}{17}+\dfrac{9}{17}\right)-\dfrac{3}{5}\\ =\left(-1\right)+1-\dfrac{3}{5}=0-\dfrac{3}{5}=\dfrac{-3}{5}\\ b,\\ =\dfrac{7}{15}.\dfrac{-15}{14}+\left(\dfrac{27}{16}-\dfrac{1}{8}\right):\dfrac{5}{8}\) 

\(=\dfrac{-1}{2}+\dfrac{25}{16}.\dfrac{8}{5}=\dfrac{-1}{2}+\dfrac{5}{2}=2\\ c,\\ =\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+.....+\dfrac{2}{99}-\dfrac{2}{100}\\ =1-\dfrac{1}{50}=\dfrac{49}{50}\) 

Câu 15

\(a,2x+\dfrac{-1}{4}=\dfrac{3}{2}\\ 2x=\dfrac{3}{2}-\dfrac{-1}{4}=\dfrac{7}{4}\\ x=\dfrac{7}{4}:2=\dfrac{7}{8}\\ b,\dfrac{15}{x}=\dfrac{-3}{4}\\ x=\dfrac{15.4}{-3}=-20\)

\(14.a)A=\dfrac{-3}{8}+\dfrac{8}{17}+\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{9}{17}\)

        \(A=\dfrac{-3}{8}+\dfrac{8}{17}+\dfrac{-5}{8}+\dfrac{-3}{5}+\dfrac{9}{17}\)

        \(A=\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{8}{17}+\dfrac{9}{17}\right)+\dfrac{-3}{5}\)

        \(A=\left(-1\right)+1+\dfrac{-3}{5}\)

        \(A=0+\dfrac{-3}{5}=\dfrac{-3}{5}\)

\(b)B=\dfrac{7}{5}.\dfrac{-15}{14}+\left(\dfrac{27}{16}-\dfrac{1}{8}\right)\div\dfrac{5}{8}\)

   \(B=\dfrac{7}{5}.\dfrac{-15}{14}+\dfrac{25}{16}\div\dfrac{5}{8}\)

   \(B=\left(\dfrac{-3}{2}\right)+\dfrac{5}{2}=1\)

\(c)C=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{99.100}\)

   \(C=1-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+\dfrac{2}{4}-...-\dfrac{2}{99}+\dfrac{2}{99}-\dfrac{2}{100}\)

   \(C=1-\dfrac{2}{100}=1-\dfrac{1}{50}=\dfrac{49}{50}\)

\(15.a)2x+\dfrac{-1}{4}=\dfrac{3}{2}\)

        \(2x\)            \(=\dfrac{3}{2}-\left(\dfrac{-1}{4}\right)=\dfrac{3}{2}+\dfrac{1}{4}=\dfrac{7}{4}\)

          \(x\)            \(=\dfrac{7}{4}\div2=\dfrac{7}{8}\)

\(b)\dfrac{15}{x}=\dfrac{-3}{4}\)

\(\Rightarrow x=\dfrac{15.4}{-3}=-20\)

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