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Những câu hỏi liên quan
LH
Lê Huy Tường
VIP
16 tháng 4 2021
There is......... girl in my class.
A. not B. no C. aren't D. isn't
13 tháng 11 2021
\(A=-\left(x^2-4x+4\right)-\left(y^2+4y+4\right)+10\\ A=-\left(x-2\right)^2-\left(y+2\right)^2+10\le10\\ A_{max}=10\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)
TN
2
9 tháng 8 2021
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
DT
Đỗ Thanh Hải
CTVVIP
9 tháng 8 2021
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
TN
2
QN
2
18 tháng 2 2022
bạn đăng tách ra nhé
Bài 3 :
Ta có :\(1+\dfrac{1}{2+x}=\dfrac{12}{x^3+8}\)
đk : x khác -2
\(\Rightarrow x^3+8+x^2-2x+4=12\Leftrightarrow x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(x^2+x-2\right)=0\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\Leftrightarrow x=0;x=1;x=-2\left(ktm\right)\)
18 tháng 2 2022
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:\(x\ne\pm2\)
\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2-5x}{x^2-4}\\ \Leftrightarrow\dfrac{x^2-3x+2-x^2-2x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2-5x}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow\dfrac{-5x+2-2+5x}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow0=0\left(tm\right)\)
HT
4
LP
19 tháng 8 2021
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
19 tháng 8 2021
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
6. \(\left(3+7\right)+\left(15+25\right)+30+\left(18+2\right)+\left(6+4\right)\)
\(=10+40+30+20+10=110\)
7. 120 : 2 +120 : 5 - 120 : 10 = 60 + 24 -12= 72
8. (23 - 8 ) + ( 15 -5 ) + (10 - 4) = 15 + 10 + 6 = 31
Cảm ơn cô giáo O(∩_∩)O