B

b

B đó bạn

a: Xét ΔADC có 

E là tđiểm của AD

I là tđiểm của AC

Do đó: EI là đường trung bình

=>EI//DC

Xét ΔBCA có 

F là tđiểm của BC

I là tđiểm của AC

Do đó: IF là đường trung bình

=>IF//AB

\(x^3=9x\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

Thang

Hình thang.

Bài gần giống b tham khảo nka 

1.Although she disliked coffee, she drank it to keep herself warm

2.In spite of Mary’s dislike of flying, she will take a plane

3.Although Marcy was sad because of losing the contest, she managed to smile

4.Despite the cloudy weather, we took many pictures

5.Even though the old woman had a poor memory, she told interesting stories to the children

6.In spite of his being absent frequently, he has managed to pass the test

7.Though Nancy had promised not to tell anyone the secret, she told me that

8.Even though we know we will not win a prize, we plan to buy a ticket for the drawing

9.Even though going to the movies is costly, my daughters insist on doing that every Saturday

10.In spite of his being on a diet, he ate the chocolate cake

11: \(=x^2-2x+1-x^2+4\)

=-2x+5

13: \(=\left(6x+1-6x+1\right)^2\)

=4

\(x^2+\left(x+3\right)\left(x-9\right)=-27\\ \Rightarrow x^2+x^2+3x-9x-27=-27\\ \Rightarrow2x^2-6x=0\\ \Rightarrow2x\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(x^2+\left(x+3\right)\left(x-9\right)=-27\)

\(\Rightarrow2x^2-6x=0\)

\(\Rightarrow2x\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(\Rightarrow x^2+x^2-6x-27=-27\\ \Rightarrow2x^2-6x=0\\ \Rightarrow2x\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Cj gãy cổ rồi e ơi =))

a) Để biểu thức A có giá trị nguyên thì :

\(a+1⋮2a\)

Mà \(2a⋮2a\)

\(\Leftrightarrow\left\{{}\begin{matrix}2a+2⋮2a\\2a⋮2a\end{matrix}\right.\)

\(\Leftrightarrow2⋮2a\)

Vì \(a\in Z;2⋮2a\Leftrightarrow2a\inƯ\left(2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2a=1\Leftrightarrow a=\dfrac{1}{2}\left(loại\right)\\2a=2\Leftrightarrow a=1\left(tm\right)\\2a=-1\Leftrightarrow a=\dfrac{-1}{2}\left(loại\right)\\2a=-2\Leftrightarrow a=-1\left(tm\right)\end{matrix}\right.\)

Vậy ..............

b) Để phân số \(\dfrac{a-2}{7}\in Z\) thì :

\(a-2⋮7\)

Vì \(a\in Z\Leftrightarrow a-7\in Z;a-7\inƯ\left(7\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}a-7=1\Leftrightarrow a=8\left(tm\right)\\a-7=7\Leftrightarrow a=14\left(tm\right)\\a-7=-1\Leftrightarrow a=6\left(tm\right)\\a-7=-7\Leftrightarrow0\left(tm\right)\end{matrix}\right.\)

Vậy .............

\(a+1⋮2a\)

\(\Rightarrow2a+2⋮2a\)

\(\Rightarrow2⋮2a\)

\(\Rightarrow1⋮a\Rightarrow a=\pm1\)

\(2a+1⋮a-3\)

\(\Rightarrow2a-6+7⋮a-3\)

\(\Rightarrow2\left(a-3\right)+7⋮a-3\)

Xét ước

\(a-2⋮7\)

\(\Rightarrow a-2\in B\left(7\right)\)

\(B\left(7\right)=\left\{0;7;14;....\right\}\)

\(a\in\left\{-2;5;12;....\right\}\)