bài 7

Bài 8

Bài 9

1. go / her / we / birthday / will / to / party

=> We will go to her birthday party

2. be / on / May / Lan / will / Sunday / 20th / thirteen.

=> Lan will be thirteen on Sunday 20th

3. will / her / call / six / after / I / again / o'clock

=> I will call her again after six o'clock

4. speak / can / to / I / Mai / please?

=> Can I speak to Mai please?

5. telephone / your / is / number / what?

=> What is your telephone number?

6. morning / will / free / you / be / tomorrow?

=> Will you be free tomorrow morning?

7. weekend / think / will / the / free / on / I / I / be

=> I think I will be free on the weekend.

8. of / invite / friends / some / will / she / her.

=> She will invite some of her friends.

Bài 1:
b) \(B=A.\dfrac{-10}{x-4}=\dfrac{x-4}{x+5}.\dfrac{-10}{x-4}=\dfrac{-10}{x+5}\)

Để B nguyên <=> x+5 nguyên mà \(x\in Z\Rightarrow x+5\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

\(\Leftrightarrow x\in\left\{-6;-4;-3;-7;0;-10;-15;5\right\}\) kết hợp với điều kiện của x

\(\Rightarrow x\in\left\{-15;-10;-6;-7;-3;0;5\right\}\)

Bài 5:

Có \(\left|x-2018\right|+\left|2x-2019\right|+\left|3x-2020\right|\ge0\) \(\forall\)x

\(\Rightarrow x-2021\ge0\) \(\Leftrightarrow x\ge2021\)

\(\Rightarrow x-2018>0,2x-2019>0,3x-2020>0\)

PT \(\Leftrightarrow x-2018+2x-2019+3x-2020=x-2021\)

\(\Leftrightarrow5x=4036\) \(\Leftrightarrow x=\dfrac{4036}{5}< 2021\) (L)

Vậy pt vô nghiệm

\(\sqrt{51-7\sqrt{8}}=\sqrt{7^2-7.2\sqrt{2}+\left(\sqrt{2}\right)^2}=\sqrt{\left(7-\sqrt{2}\right)^2}=7-\sqrt{2}\)

(vì\(7=\sqrt{49}>\sqrt{2}\Rightarrow7-\sqrt{2}>0\))

pt sao có 1 vé vậy bạn

Câu 3:

a)xe máy,xe đạp

b)hoa lan,hoa cúc

c)theo lũ bạn lên đê thả diều,đuổi bắt

d)trong mát,chảy hiền hòa.

TICK GIÚP MIK NHA CẢM ƠN^^

Bài 5 ở đâu vậy ạ?

a) Ta có:

 \(H=\left(\dfrac{x}{x^2-4}+\dfrac{1}{x+2}+\dfrac{2}{2-x}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\\ =\left(\dfrac{x}{x^2-4}+\dfrac{x-2}{x^2-4}-\dfrac{2\left(x+2\right)}{x^2-4}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\\ =\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}\\ =\dfrac{-6}{x-2}\cdot\dfrac{1}{6}=\dfrac{1}{2-x}\)

b) Để H < 0 thì \(\dfrac{1}{2-x}\) < 0 hay 2 - x < 0 ( do 1 > 0) suy ra x > 2

Vậy với x > 2 thì H < 0.

c) Ta có:

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

+) Với x = 3 thì:

H = \(\dfrac{1}{2-3}=-1\)

+) Với x = -3 thì:

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

Vậy với |x| = 3 thì H = -1 hoặc H = 1/5

a: Ta có: \(H=\left(\dfrac{x}{x^2-4}+\dfrac{1}{x+2}+\dfrac{2}{2-x}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

\(=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x^2-4+10-x^2}{x+2}\)

\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}\)

\(=\dfrac{-1}{x-2}\)

b: Để H<0 thì x-2<0

hay x<2

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x< 2\\x\ne-2\end{matrix}\right.\)

c: Ta có: |x|=3

nên \(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Thay x=3 vào H, ta được:

\(H=\dfrac{-1}{3-2}=-1\)

Thay x=-3 vào H, ta được:

\(H=\dfrac{-1}{-3-2}=\dfrac{-1}{-5}=\dfrac{1}{5}\)

lớp mấy

ê c*c tau đi đâu cũng gặp mày hết á

chửi cail*z đy học hay đy chữi z đ*  😏