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: \(\left|m+3\right|=\left|\dfrac{-m-3}{m-2}\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}m+3=\dfrac{-m-3}{m-2}\\m+3=\dfrac{m+3}{m-2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m^2+m-6+m+3=0\\m^2+m-6-m-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(m+3\right)\left(m-1\right)=0\\\left(m+3\right)\left(m-3\right)=0\end{matrix}\right.\Leftrightarrow m\in\left\{-3;1;3\right\}\)

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 \(x=-\dfrac{11}{25}\)

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

1 my/your

2 my

3I/my

4 her

5 Nam and Lan's teacher

6 is./Her

7 Those students' school bags

8 your/He

9 is/your/she

10 its

11 Her

12 His

13 their

14 hers

15 him

16 her

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

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:

     \(B=\dfrac{-5}{-5+10}=-1\)