There is......... girl in my class.

A. not     B. no     C. aren't     D. isn't

mik cảm ơn bạn nhé

\(A=-\left(x^2-4x+4\right)-\left(y^2+4y+4\right)+10\\ A=-\left(x-2\right)^2-\left(y+2\right)^2+10\le10\\ A_{max}=10\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

1, What would he like to have for breakfast?

He would like to have a sandwich

2,Who would you like to go fishing with?

I would like to go fishing with my father

3,What would her children like to do in the summer?

They would like to go swimming

4,When would Mrs Tam like to go shopping?

She would like to go shopping at weekends

5,Where would Hung and Tung like to study

They would like to study in the library

1 What would he like for breakfast? 

He'd like a sandwich

2 Who would you like to go fishing with?

I would like to go with my father

3 What would her children like to do in summer?

They would like to swim inpool

4 When would Mrs Tam like to go shopping?

She would like to go shopping on the weekends

5 Where would Tung and Hung like to study ?

THey would like to study in the school library

1 isn't done

2 is asked

DẠ CẢM ƠN NHIỀU Ạ ^^❤

bạn đăng tách ra nhé 

Bài 3 : 

Ta có :\(1+\dfrac{1}{2+x}=\dfrac{12}{x^3+8}\)

đk : x khác -2 

\(\Rightarrow x^3+8+x^2-2x+4=12\Leftrightarrow x^3+x^2-2x=0\)

\(\Leftrightarrow x\left(x^2+x-2\right)=0\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\Leftrightarrow x=0;x=1;x=-2\left(ktm\right)\)

Bài 2:

a,ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)

\(\dfrac{1}{x}+\dfrac{2}{x-2}=0\\ \Leftrightarrow\dfrac{x-2}{x\left(x-2\right)}+\dfrac{2x}{x\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x-2+2x}{x\left(x-2\right)}=0\\ \Rightarrow3x-2=0\\ \Leftrightarrow x=\dfrac{2}{3}\left(tm\right)\)

b, ĐKXĐ:\(x\ne\pm2\)

\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2-5x}{x^2-4}\\ \Leftrightarrow\dfrac{x^2-3x+2-x^2-2x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2-5x}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow\dfrac{-5x+2-2+5x}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow0=0\left(tm\right)\)

em chịu_{em thua}

cái gì đây?

3x.(x-2)-x²+2x=0

⇔3x²-6x-x²+2x=0

⇔2x²-4x=0

⇔2x(x-2)=0

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

vậy x=0 và x=2

3x(x-2)-x^2+2x=0

<=>3x(x-2)-x(x-2)=0

<=>(3x-x)(x-2)=0

<=>2x(x-2)=0

<=>2x=0 hoặc x-2=0

<=>x=0 hoặc x=2