1. Let's collect old textbooks to give to children in mountainous areas.

2. My sister is fond of making vlogs about animals.

3. Last week , I had a sore throat , so I stayed at home .

1 . He had a stomachache .

2. Why don't we organize a talent show to raise money .

3. You shouldn't eat junk food every day .

4. I prefer listening to country music more than classical music .

1. I drink a glass of milk before going to bed to help me sleep .

2. Volunteering at the food bank is the great way to help people in need .

3. The dancer moved gracefully acoss the stage during the performance .

Em chx biết làm bài này

 Tổng số tiền của 3 quyển sách là:

    3. 120 000 = 360 000 ( đồng )

 Số tiền bạn Lan được giảm là :

        360 000 . 10% = 36 000 ( đồng )

Số tiền bạn phải trả là :

     360 000 - 36 000 = 324 000 ( đồng )

Cửa hàng phải trả lại bạn Lan số tiền là:

    350 000 - 324 000 = 26 000 ( đồng )

a) Ta có: \(\widehat{BAC}\) +  \(\widehat{CAx}\) = 180^o ( hai góc kề bù )

               100^o+ \(\widehat{CAx}\) = 180^o

                          \(\widehat{CAx}\) = 180^o-100^o

                         \(\widehat{CAx}\) = 80^o

b) Ta có: \(\widehat{xAy}\) = \(\widehat{yAC}\)= \(\dfrac{\widehat{CAx}}{2}\)=\(\dfrac{80^o}{2}=40^o\)

        ( Vì tia Ay là tia phân giác của góc CAx )

   Ta có: \(\widehat{BCA}\) = \(\widehat{CAy}\)= 40^o mà hai góc này ở vị trí so le trong nên Ay//BC

                 ( Dấu hiệu nhận biết hai đường thẳng song song )

c) Xét tam giác ABC ta có: \(\widehat{A}\) + \(\widehat{B}\) + \(\widehat{C}\) = 180^o( tổng ba góc trong 1 tam giác)

                                           100^o+ \(\widehat{B}\) + 40^o = 180^o

                                                     \(\widehat{B}\)           = 180^o-100^o-40^o

                                                      \(\widehat{B}\)          = 40^o

       ^{Suy ra \(\widehat{ABC}\)} = 40^o

a) \(x-\dfrac{2}{3}=\dfrac{1}{6}\)

    x=\(\dfrac{1}{6}+\dfrac{2}{3}\)

    x= \(\dfrac{5}{6}\)

b) \(2x+\dfrac{1}{2}=\dfrac{-5}{3}\)

    \(2x\)= \(\dfrac{-5}{3}-\dfrac{1}{2}\)

     2x = \(\dfrac{-13}{6}\)

     x= \(\dfrac{-13}{6}:2\)

   x= \(\dfrac{-13}{12}\)

c) \(3x+\dfrac{3}{2}=x-\dfrac{5}{3}\)

    3x - x = \(\dfrac{-5}{3}-\dfrac{3}{2}\)

    2x= \(\dfrac{-19}{6}\)

    x= \(\dfrac{-19}{6}:2\)

   x = \(\dfrac{-19}{12}\)

a) \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)

     =\(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+\dfrac{1}{2}-\dfrac{36}{41}\)

     = \(\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}\)

    = \(1-1+\dfrac{1}{2}\)

  =  \(\dfrac{1}{2}\)

b) \(\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{1}{2}.\dfrac{1}{4}+\dfrac{1}{2}\)

 = \(\dfrac{1}{2}.\left(\dfrac{3}{4}+\dfrac{1}{4}+1\right)\)

= \(\dfrac{1}{2}.2\)

=\(\)1

c) \(\left(\dfrac{-3}{4}\right)^2:\left(\dfrac{-1}{4}\right)^2+9.\left(\dfrac{-1}{9}\right)+\left(\dfrac{-3}{2}\right)\)

=\(\dfrac{9}{16}:\dfrac{1}{16}+\left(-1\right)+\left(\dfrac{-3}{2}\right)\)

= \(9+\left(-1\right)+\left(\dfrac{-3}{2}\right)\)

= \(\dfrac{18}{2}-\dfrac{2}{2}-\dfrac{3}{2}\)

= \(\dfrac{13}{2}\)

d) \(\sqrt{0,25}.\left(-3\right)^3-\sqrt{\dfrac{1}{81}}:\left(\dfrac{-1}{3}\right)^3\)

=\(\dfrac{1}{2}.\left(-27\right)-\dfrac{1}{9}.\left(-27\right)\)

= \(\left(-27\right).\left(\dfrac{1}{2}-\dfrac{1}{9}\right)\)

= \(\left(-27\right).\left(\dfrac{9}{18}-\dfrac{2}{18}\right)\)

= (-27) . \(\dfrac{7}{18}\)

=\(\dfrac{-21}{2}\)

a) \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)

     =\(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+\dfrac{1}{2}-\dfrac{36}{41}\)

     = \(\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}\)

    = \(1-1+\dfrac{1}{2}\)

  =  \(\dfrac{1}{2}\)

b) \(\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{1}{2}.\dfrac{1}{4}+\dfrac{1}{2}\)

 = \(\dfrac{1}{2}.\left(\dfrac{3}{4}+\dfrac{1}{4}+1\right)\)

= \(\dfrac{1}{2}.2\)

=\(\)1

c) \(\left(\dfrac{-3}{4}\right)^2:\left(\dfrac{-1}{4}\right)^2+9.\left(\dfrac{-1}{9}\right)+\left(\dfrac{-3}{2}\right)\)

=\(\dfrac{9}{16}:\dfrac{1}{16}+\left(-1\right)+\left(\dfrac{-3}{2}\right)\)

= \(9+\left(-1\right)+\left(\dfrac{-3}{2}\right)\)

= \(\dfrac{18}{2}-\dfrac{2}{2}-\dfrac{3}{2}\)

= \(\dfrac{13}{2}\)

d) \(\sqrt{0,25}.\left(-3\right)^3-\sqrt{\dfrac{1}{81}}:\left(\dfrac{-1}{3}\right)^3\)

=\(\dfrac{1}{2}.\left(-27\right)-\dfrac{1}{9}.\left(-27\right)\)

= \(\left(-27\right).\left(\dfrac{1}{2}-\dfrac{1}{9}\right)\)

= \(\left(-27\right).\left(\dfrac{9}{18}-\dfrac{2}{18}\right)\)

= (-27) . \(\dfrac{7}{18}\)

=\(\dfrac{-21}{2}\)

a) Ta có : \(\widehat{mOx}\) + \(\widehat{xOn}\) = 180^o  ( hai góc kề bù )

                30^o +  \(\widehat{xOn}\) = 180^o

                                ^{\(\widehat{xOn}\)}   = 180^o - 30^o

                                ^{\(\widehat{xOn}\)}   = 150^o

   Vì Ot là tia phân giác của  \(\widehat{nOx}\)   nên  \(\widehat{nOt}\) =  \(\widehat{tOx}\) = ^{\(\dfrac{\widehat{nOx}}{2}\)} = \(\dfrac{150}{2}^{ }\) = 75^o

    ^{Vậy \(\widehat{nOt}\)} = 75^o

^{b) Ta có: a // b nên  \(\widehat{A^{ }_{4^{ }}}\) =   \(\widehat{B_2}\) =} 65^o ( hai góc so le trong)

    ^{Ta có : \(\widehat{B_2}\)}  +  \(^{\widehat{B_3}}\) = 180^o ( hai góc kề bù)

             65^o + \(\widehat{B_3}\) = 180^o

                            ^{\(\widehat{B_3}\)} = 180^o - 65^o

                       \(\widehat{B_3}\) = 115^o

^{Vậy \(\widehat{B_3}\)} = 115^o