a) Vì AB = AC (gt)

\(\Rightarrow\) \(\Delta ABC\) cân tại A

\(\Rightarrow\) AM là đường trung tuyến đồng thời là đường cao

Vậy AM \(\perp\) BC.

b) Xét hai tam giác ABM và DCM có:

MA = MD (gt)

\(\widehat{AMB}=\widehat{DMC}\) (đối đỉnh)

MB = MC (gt)

Vậy \(\Delta ABM=\Delta DCM\left(c-g-c\right)\)

Suy ra: \(\widehat{BAM}=\widehat{CDM}\) (hai góc tương ứng)

Mà hai góc này ở vị trí so le trong

Do đó AB // DC (đpcm).

1 for

2 been

3 than

4 so

5 writes

6 more

7 at

8 take

9 the

10 of

11 but

12 not

chữ đẹp thế

3. Tin's grandma told him not to go out then because it was snowing severely.
4. My brother said to me water boiled at 100 degrees Celsius.
5. Louis said watching cartoons had been his number one hobby years before, but he were into horor movies then.
6. The librarian told Arinna the book she was looking couldn't be there.
7. Mom said her friend had invited her to a wedding ceremony so she would have to buy a new dress.
8. Jennie said she had to finish that assignment that day and send it to the teacher via e-mail the next day.
9. The girl said those dry flowers looked even more gorgeous than the ones she had received.
10. Mr. Thomas always says his children are well-behaved.

ko nhìn thấy j đâu e

pls help me ;-;

\(a,3\left(x-1\right)-4=2\left(x+1\right)-7\\ \Leftrightarrow3x-3-4=2x+2-7\\ \Leftrightarrow3x-7=2x-5\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ b,\left(x+1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

c, ĐKXĐ:\(x\ne\pm3\)

\(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{3}{x^2-9}\\ \Leftrightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{\left(x-3\right)\left(x+3\right)}=0\\ \Leftrightarrow\dfrac{\left(x^2+6x+9\right)-\left(x^2-6x+9\right)-3}{\left(x-3\right)\left(x+3\right)}=0\\ \Rightarrow x^2+6x+9-x^2+6x-9-3=0\\ \Leftrightarrow12x-3=0\\ \Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

d, \(d,\dfrac{x-3}{4}-\dfrac{2x+3}{3}=\dfrac{5x-1}{3}+\dfrac{2x+9}{12}\\ \Leftrightarrow d,\dfrac{3\left(x-3\right)}{12}-\dfrac{4\left(2x+3\right)}{12}=\dfrac{4\left(5x-1\right)}{12}+\dfrac{2x+9}{12}\\ \Leftrightarrow3x-9-8x-12=20x-4+2x+9\\ \Leftrightarrow-5x-21=22x+5\\ \Leftrightarrow27x+26=0\\ \Leftrightarrow x=-\dfrac{26}{27}\left(tm\right)\)

a: \(2+\dfrac{3\left(x+5\right)}{8}>\dfrac{x-1}{4}\)

=>\(2+\dfrac{3}{8}x+\dfrac{15}{8}>\dfrac{1}{4}x-\dfrac{1}{4}\)

=>\(\dfrac{3}{8}x+\dfrac{31}{8}>\dfrac{1}{4}x-\dfrac{1}{4}\)

=>\(\dfrac{1}{8}x>-\dfrac{1}{4}-\dfrac{31}{8}=\dfrac{-33}{8}\)

=>x>-33

b: \(2x-x\left(2x+1\right)< =15-2x\left(x+2\right)\)

=>\(2x-2x^2-x< =15-2x^2-4x\)

=>x<=15-4x

=>5x<=15

=>x<=3

a: \(\Leftrightarrow2x^2-2x+3x+6-2x^2+3=0\)

=>x+9=0

hay x=-9

b: \(\Leftrightarrow10x-30-4x=10x-5\)

=>-4x=-5+30=25

=>x=-25/4