giúp e với,e đg cần gấp
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![](https://rs.olm.vn/images/avt/0.png?1311)
b: Gọi A là Tọa độ giao điểm của hàm số với trục tung
=>Tọa độ của A là:
\(\left\{{}\begin{matrix}x=0\\y=\left(m-2\right)\cdot0+m+3=m+3\end{matrix}\right.\)
Gọi B là Tọa độ giao điểm của hàm số với trục hoành
=>Tọa độ của B là:
\(\left\{{}\begin{matrix}\left(m-2\right)\cdot x+m+3=0\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-m-3}{m-2}\\y=0\end{matrix}\right.\)
Theo đề, ta có: \(\left|m+3\right|=\left|\dfrac{-m-3}{m-2}\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}m+3=\dfrac{-m-3}{m-2}\\m+3=\dfrac{m+3}{m-2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m^2+m-6+m+3=0\\m^2+m-6-m-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(m+3\right)\left(m-1\right)=0\\\left(m+3\right)\left(m-3\right)=0\end{matrix}\right.\Leftrightarrow m\in\left\{-3;1;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
c: Ta có: \(\left(x-3\right)^3-\left(x^3-27\right)+9\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-9x^2+27x+9x^2+18x+9=15\)
\(\Leftrightarrow45x=6\)
hay \(x=\dfrac{2}{15}\)
d: Ta có: \(x\left(x-5\right)\left(x+5\right)-\left(x^3+8\right)=3\)
\(\Leftrightarrow x^3-25x-x^3-8=3\)
\(\Leftrightarrow-25x=11\)
hay \(x=-\dfrac{11}{25}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
my - your
my
I - my
her
Nam and Lan's teacher
is
Her
Those students' school bags
your
He
is - your - She
its
her
His
their
hers
him
her
![](https://rs.olm.vn/images/avt/0.png?1311)
1.Yes, they do
2..Yes, it is
3.People buy fruits and flowers from the market and decorate their house
4.People visit their family and friends
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\dfrac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x-2\right)}{x+2}\)
Thay \(x=\dfrac{1}{2}\) vào A ta được:
\(A=\dfrac{2\cdot\left(\dfrac{1}{2}-2\right)}{\dfrac{1}{2}+2}=\dfrac{-3}{\dfrac{5}{2}}=-\dfrac{6}{5}\)
b) \(B=\dfrac{x^3-x^2y+xy^2}{x^3+y^3}=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\)
Thay \(x=-5,y=10\) vào B ta đc:
\(B=\dfrac{-5}{-5+10}=-1\)
fewer => less
are => do
two => three
cook => cooking
since => for
taking => take
listening => to listen
palace => palaces
is come => comes
hundred => hundreds