1. I (meet) met them at the movie theater last night.

2. He (go) goes to the English club every Sunday. But last Sunday he (not go) doesn't go there, he (go) went to the City's Museum.

3. Don't talk to the children. They (do) are doing their homework at the moment.

4. Two years ago, she (teach) taught at a village school.

5. What did you (do) do yesterday morning?- I (write) wrote my essay.

6. Next week, we (have) will have dinner together.

7. My sister (live) lives in a flat in London but she (travel) travels in Europe at this time.

8. The children (be) are in class now, but ten minutes ago they (be) were in the school yard.

\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)

\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)

\(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)

\(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}\)

\(\frac{1}{x+1}=\frac{1}{30}\)

\(\Rightarrow x+1=30\)

\(x=30-1=29\)

1. a. father                b. afternoon                  c. family

2 a. ten                   b he                    c twenty

3 a. five               b nice                   c miss

Ta có:

\(BE=\frac{AD+CH}{2}\) (Đường trung bình của hình thang bằng một nửa tổng độ dài 2 đáy)

\(32=\frac{24+x}{2}\) (\(BE=32cm\)(gt); \(AD=24cm\left(gt\right)\);\(CH=x\left(gt\right)\))

\(\Rightarrow24+x=32\cdot2\)

\(24+x=64\)

\(x=64-24\)

\(x=40\)

Vậy \(x=40\)

\(\frac{89}{16}+25\cdot\left(\frac{9}{16}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)

\(=\frac{89}{16}+25\cdot\frac{36}{125}:\left(-\frac{27}{8}\right)\)

\(=\frac{89}{16}+7\frac{1}{5}:\left(-\frac{27}{8}\right)\)

\(=\frac{89}{16}-2\frac{2}{15}\)

\(=3\frac{103}{240}\)

a) \(\left(19\cdot x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)

\(\left(19\cdot x+50\right):14=9\)

\(19\cdot x+50=9\cdot14\)

\(19\cdot x+50=126\)

\(19\cdot x=126-50\)

\(19\cdot x=76\)

\(x=76:19\)

\(x=4\)

Vậy \(x=4\)

\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

\(=1-\frac{1}{11}\)

\(=\frac{10}{11}\)

bìa này trên olm đúng không! đây là bài tuần (the bottom surface)

Đổi 120%= 120/100= 6/5   .  

Số tiền mua cá là:

88000 /( 6+5) *6 =48 000( đồng)

Đáp số : 48 000 đồng

a) Ta có:

\(\widehat{xOy} + \widehat{yOz} = 180^O\) (kề bù)

\(70^O + \widehat{yOz} = 180^O (\widehat{xOy} = 70^O(gt))\)

\(\widehat{yOz} = 180^O - 70^O\)

\(\widehat{yOz} = 110^O\)

Vậy \(\widehat{yOz} = 110^O\)

b) Ta có:

\(\widehat{aOy} = \frac{\widehat{xOy}}{2}\) (Oa là tia phân giác của \(\widehat{xOy}\)(gt))

\(\widehat{aOy} = \frac{70^O}{2} (\widehat{xOy} = 70^O(gt))\)

\(\Rightarrow\)\(\widehat{aOy} = 35^O\)

\(\widehat{aOy} + \widehat{yOz} = \widehat{aOz}\) (Oy nằm giữa hai tia Oa và Oz)

\(35^O + 110^O = \widehat{aOz} (\widehat{aOy}=35^O (cmt); \widehat{yOz} = 110^O (cmt))\)

\(\Rightarrow\)\(\widehat{aOz} = 145^O\)

Vậy \(\widehat{aOz} = 145^O\)

c) Ta có:

\(\widehat{yOb} = \frac{\widehat{yOz}}{2}\) (Ob là tia phân giác của \(\widehat{yOz}\)(gt))

\(\widehat{yOb} = \frac{110^O}{2} (\widehat{yOz} = 110^O(cmt))\)

\(\Rightarrow\)\(\widehat{yOb} = 55^O\)

\(\widehat{yOb} + \widehat{aOy} = \widehat{aOb}\) (Oy nằm giữa Oa và Ob)

\(55^O + 35^O = \widehat{aOb}(\widehat{yOb}=55^O (cmt); \widehat{aOy} = 35^O (cmt))\)

\(\Rightarrow\)\(\widehat{aOb} = 90^O\)

Vậy \(\widehat{aOb} = 90^O\)