đau?

các bạn làm giúp mk nhé

tính

a, 11/9+ 7/6 . ( 2-1/3)

b,20,17 . 43 + 57 . 20,17

2:

Chiều cao là: 26*2:8=6,5cm

S2-S1=1/2*6,5*11-1/2*6,5*8=9,75cm²

VII.

1. Anna used to have long hair when she was at school

2. Mary didn't use to listen when her teachers were speaking

3. Ricardo used to get up at 6:00 when he was training for the Olympics

4. What did you use to usually do on saturday evenings?

5. Sophie used to be afraid of dogs when she was a litlle girl

6. We used to give our teachers presents at the end of term

7. Did you use to live next door to Mrs Harrison?

8. My brother used to wear glasses when he was young

VI.

1. Viet Nam's anthem Tien Quan Ca was composed by Van Cao.

2. If you are not careful, you’ll hurt yourself.

3. How much does a bicycle cost?

4. There aren't any carrots in the market.

5. No one in the school plays guitar better than you

VIII.

1. I used to stay up late to watch football matches last year.

2. There used to be some trees in the field.

3. Anna used to live with her parents.

4. He used to be a poor man, but now he becomes a rich businessman

5. They didn’t use to go to cinema every Sunday last year.

6. In the past my hair used to be shorter.

7. I used to have much time to collect stamps when I was in primary school.

8. Did you use to go to the beach when you lived in Nha Trang?

Câu 1.

\(CH_3COOH+NaOH\rightarrow CH_4\uparrow+Na_2CO_3\)

\(2CH_4\xrightarrow[LLN]{1500^oC}C_2H_2+2H_2\uparrow\)

\(C_2H_2+H_2\underrightarrow{t^o,xtPd}C_2H_4\)

a)\(C_2H_4+HBr\rightarrow C_2H_5Br\)

b)\(C_2H_4+HCl\underrightarrow{as}C_2H_5Cl\)

\(A^3_n+5A^2_n=2\left(n+15\right)\)

ĐK: n ≥ 3 (n∈N)

<=> \(\dfrac{n!}{\left(n-3\right)!}+\dfrac{5.n!}{\left(n-2\right)!}=2\left(n+15\right)\)

<=> \(\dfrac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}{\left(n-3\right)!}+\dfrac{5n\left(n-1\right)\left(n-2\right)!}{\left(n-2\right)!}=2\left(n+15\right)\)

<=> \(n\left(n-1\right)\left(n-2\right)+5\left(n-1\right)n-2n-30=0\)

<=> \(n^3+2n^2-5n-30=0\) <=> n=3

Bài 4: 

Ta có: \(A=x^2+4x+y^2-5y+20\)

\(=x^2+4x+4+y^2-5y+\dfrac{25}{4}+\dfrac{39}{4}\)

\(=\left(x+2\right)^2+\left(y-\dfrac{5}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}\forall x,y\)

Dấu '=' xảy ra khi x=-2 và \(y=\dfrac{5}{2}\)

Qua O, kẻ tia Oz//Aa

Oz//Aa

Aa//BC

Do đó: Oz//BC

Oz//Aa

=>\(\widehat{zOA}=\widehat{OAa}\)(hai góc so le trong)

=>\(\widehat{zOA}=30^0\)

\(\widehat{zOA}+\widehat{zOB}=\widehat{AOB}=90^0\)

=>\(\widehat{zOB}=90^0-30^0=60^0\)

Oz//BC

=>\(\widehat{zOB}=\widehat{OBC}\)(hai góc so le trong)

=>\(x=60^0\)

Đặt tên các điểm, các tên đường thẳng như trên hình vẽ.

Qua O, kẻ tia Oz//Bb(Oz và Bb là hai tia nằm ở hai phía khác nhau)

Oz//Bb 

=>\(\widehat{zOB}=\widehat{bBO}\)(hai góc so le trong)

=>\(\widehat{zOB}=40^0\)

Oz//Bb

Bb//Ac

=>Oz//Ac

=>\(\widehat{zOA}+\widehat{OAc}=180^0\)(hai góc trong cùng phía)

=>\(\widehat{zOA}=180^0-120^0=60^0\)

\(\widehat{BOA}=\widehat{zOB}+\widehat{zOA}=60^0+40^0=100^0\)

Vẽ đường thẳng c//a//b

Ta có: c//a

=> x₁ = 40^o (hai góc so le trong bằng nhau)     (1)

Ta lại có:

=> x₂ = 180^o - 120^o = 60^o (hai góc phụ nhau)   (2)

Từ (1) và (2) => x = x₁ + x₂ = 40 + 60 = 100^o