Bài 1

ta có góc A1=góc B1=55 độ

suy ra a//b (slt)                        (1)

Ta có CE vg góc vs b

         CE vg góc vs c

suy ra b//c ( tc từ vg góc đến ss )    (2)

Từ (1) và (2) suy ra a//b//c

bài 2

ta có góc LJK=110 độ

  Mà góc JKM=70 độ 

suy ra JL//KM( hai góc trong cùng phía bù nhau)   (1)

Ta lại có LM vg góc vs JL                                         (2)

Suy ra LM//KM ( tc từ vg góc đến ss)

Bài 3

Ta có góc HFG + góc FGH= 105+75=180 độ

suy ra FH//GI ( hai góc tcp bù nhau)

Vì FH// GI suy ra x+HIG=180 độ

Mà HIG=90 độ suy ra x=90 độ

vì x =90 độ suy ra FH vg góc vs HI

didn't do - was - does

felt - went - is

waters - forgot 

will you see

did you come - had

brushes - are

do you have - have - had

was - did not have

1.didn't do - was - does

2.felt - went - is

3.waters - forgot 

4.will you see

5.did you come - had

6.brushes - are

7.do you have - have - had

8.was - did not have

c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)

\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\) 

Ở nơi x=9/4-1/2 là x-9/4-1/2 nha

a. -1,5 + 2x = 2,5

<=> 2x = 2,5 + 1,5

<=> 2x = 4

<=> x = 2

b. \(\dfrac{3}{2}\left(x+5\right)-\dfrac{1}{2}=\dfrac{4}{3}\)

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

<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)

<=> 9x + 45 - 3 = 8

<=> 9x = 8 + 3 - 45

<=> 9x = -34

<=> x = \(\dfrac{-34}{9}\)