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1. I like comedies because they are very funny.
2. My mother never watches a football match.
3. There are a lot of music programmes on TV nowadays.
4. I am interested in watching TV every day.
5. I spend more time watching TV than doing outdoor activities.
\(#WendyDang\)
1. I/ comedies/very funny/are/ because/ like / they /.
I like comedies because they are very funny
2. My/football match/watches/ mother / never/a/.
My mother never watches a football match
3. There/a lot of /TV/on / music programmes/ nowadays/ are.
There are a lot of music programmes on TV nowadays
4. am/ I/ interested in/ every day/ watching TV/.
I am interested in watching TV every day
5. more time/ I/ spend/ watching TV /than doing/ outdoor activities.
I spend more time watching TV than doing outdoor activities
- \(\dfrac{5}{6}\).(\(\dfrac{1}{3}\) - \(\dfrac{7}{9}\)).\(\dfrac{12}{25}\) - \(\dfrac{5}{4}\)
- \(\dfrac{5}{6}\).(-\(\dfrac{4}{9}\)).\(\dfrac{12}{25}\) - \(\dfrac{5}{4}\)
= \(\dfrac{5.4.2.6}{6.9.5.5}\) - \(\dfrac{5}{4}\)
= \(\dfrac{8}{45}\) - \(\dfrac{5}{4}\)
= -\(\dfrac{193}{180}\)
\(\dfrac{1}{5}.\left(\dfrac{1}{2}-\dfrac{7}{4}\right)-\dfrac{3}{4}.\left(\dfrac{1}{3}-\dfrac{11}{12}\right)\)
\(\dfrac{1}{5}.\left(-\dfrac{5}{4}\right)-\dfrac{3}{4}.\left(-\dfrac{7}{12}\right)\)
\(-\dfrac{1}{4}-\left(-\dfrac{7}{16}\right)\)
\(-\dfrac{1}{4}+\dfrac{7}{16}\)
\(\dfrac{3}{16}\)
Đây là dạng toán nâng cao chuyên đề điểm và đoạn thẳng cấu trúc thi hsg, hôm nay olm.vn sẽ hướng dẫn em làm dạng này như sau:
Vì O;A; C thẳng hàng nên O \(\in\) AC;
Vì O;B;D thẳng hàng nên O \(\in\) DB
Vậy O là giao điểm của AC và BD.
Kết luận vị trí của điểm O sao cho ba điểm A; O; C và ba điểm; B;O;D thẳng hàng là O là giao điểm của AC và BD.
. Circle the word (A, B, C or D) with an underlined part pronounced differently from the others.
1. A. nothing B. then C. them D. those
2. A. bathroom B. sixth C. thumb D. father
3. A. please B. feature C. bear D. beat
4. A. fear B. near C. hear D. treat
5. A. pea B. leather C. tea D. peanut
II. Choose the best answer.
1. The weather was nice , _______ it was a bit cold.
A. and B. but C. so D. because
2. It was raining,_______ I took my umbrella.
A. or B. but C. so D. although
3. We often play soccer with our friends in the park _______ Sunday afternoon.
A. in B. at C. with D. on
4. _______ do Thuy and Hoa go to the cinema? - Once a month.
A. How long B. When C. How often D. How many
5. The Red River is _______ river in the North of Vietnam.
A. longer B. the longest C. longest D. the longer
6. Mika _________ return to America last week.
A. doesn’t B. isn’t C. didn’t D. wasn’t
7. “How often are the Olympic games held?” _______
A. Every four years. B. Four years. C. In 2009 D. Every year.
8. Children should participate in many _________ activities at school.
A. sport B. outdoor C. games D. A or B
9. _______ is your favorite program?
A. Where B. Which C. What D. Who
10. Last summer, we _______ a good time in Ha Long Bay.
A. go B. have C. had D. were
11.__________run too fast. You may fall.
A. Don’t B. Do C. Didn’t D. Did
12.The football match on TV last night great. Our team lost.
A. is B. wasn't C. isn't D. was
13.What’s your favourite ? - I like swimming.
A. subject B. game C. drink D. sport
14.This small is used to play table tennis.
A. racket B. stick C. bat D. hoop
15.Last weekend Phong and Lan their grandparents.
A. visit B. is visiting C. will visit D. visited
Lời giải:
a.
$\frac{3n+2}{3}=n+\frac{2}{3}> n+\frac{1}{2}=\frac{2n+1}{2}$
$\Rightarrow \frac{3}{3n+2}< \frac{2}{2n+1}$
b.
$\frac{2n+1}{2n}=1+\frac{1}{2n}> 1+\frac{1}{3n}=\frac{1+3n}{3n}$
$\Rightarrow \frac{2n}{2n+1}< \frac{3n}{3n+1}$
\(\dfrac{3}{7}-\dfrac{1}{2}x=\dfrac{5}{3}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{3}{7}-\dfrac{5}{3}\)
\(\Rightarrow x=2\cdot-\dfrac{26}{21}\)
\(\Rightarrow x=\dfrac{-52}{21}\)
______
\(2x-\dfrac{3}{4}=\dfrac{-5}{8}\)
\(\Rightarrow2x=\dfrac{-5}{8}+\dfrac{3}{4}\)
\(\Rightarrow2x=\dfrac{1}{8}\)
\(\Rightarrow x=\dfrac{1}{8}:2\)
\(\Rightarrow x=\dfrac{1}{16}\)
________
\(\dfrac{1}{4}x-\left(-\dfrac{7}{5}\right)=\dfrac{-5}{3}\)
\(\Rightarrow\dfrac{1}{4}x+\dfrac{7}{5}=\dfrac{-5}{3}\)
\(\Rightarrow\dfrac{1}{4}x=\dfrac{-5}{3}-\dfrac{7}{5}\)
\(\Rightarrow x=4\cdot-\dfrac{46}{15}\)
\(\Rightarrow x=-\dfrac{184}{15}\)
______
\(2\dfrac{1}{3}x-\dfrac{3}{4}=1\dfrac{1}{6}\)
\(\Rightarrow\dfrac{7}{3}x-\dfrac{3}{4}=\dfrac{7}{6}\)
\(\Rightarrow\dfrac{7}{3}x=\dfrac{7}{6}+\dfrac{3}{4}\)
\(\Rightarrow\dfrac{7}{3}x=\dfrac{23}{12}\)
\(\Rightarrow x=\dfrac{23}{12}:\dfrac{7}{3}\)
\(\Rightarrow x=\dfrac{23}{28}\)
\(\dfrac{-5}{8}:\dfrac{15}{4}=\dfrac{-5}{8}\cdot\dfrac{4}{15}=\dfrac{-1\cdot1}{2\cdot3}=\dfrac{-1}{6}\)
\(\dfrac{-15}{17}:\dfrac{25}{-34}=\dfrac{-15}{17}\cdot\dfrac{-34}{25}=\dfrac{-3\cdot-2}{1\cdot5}=\dfrac{-6}{5}\)
\(-12:\dfrac{8}{3}=-12\cdot\dfrac{3}{8}=\dfrac{-12\cdot3}{8}=\dfrac{-3\cdot3}{2}=\dfrac{-9}{2}\)
\(\dfrac{-15}{14}:\dfrac{20}{-21}=\dfrac{-15}{14}\cdot\dfrac{-21}{20}=\dfrac{-3\cdot-3}{2\cdot4}=\dfrac{-9}{8}\)
\(-48:\dfrac{-24}{5}=-48\cdot\dfrac{5}{-24}=\dfrac{-48\cdot5}{-24}=2\cdot5=10\)
\(\dfrac{-30}{7}:\dfrac{-5}{-14}=\dfrac{-30}{7}:\dfrac{5}{14}=\dfrac{-30}{7}\cdot\dfrac{14}{5}=\dfrac{-6\cdot2}{1\cdot1}=-18\)
a) Gọi \(ƯCLN\left(a^2,a+b\right)=d\) với \(d\inℕ^∗\)
\(\Rightarrow\left\{{}\begin{matrix}a^2⋮d\\a+b⋮d\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2⋮d\\a^2+ab⋮d\end{matrix}\right.\)
\(\Rightarrow ab⋮d\)
Vì \(a,b\) nguyên tố cùng nhau \(\Rightarrow\left[{}\begin{matrix}a⋮d\\b⋮d\end{matrix}\right.\)
Hơn nữa, vì \(a+b⋮d\) nên nếu \(a⋮d\) thì \(b⋮d\). Nếu \(b⋮d\) thì \(a⋮d\). Như vậy \(a,b⋮d\).
Nhưng do \(a,b\) nguyên tố cùng nhau nên \(d=1\) \(\RightarrowƯCLN\left(a^2,a+b\right)=1\) hay \(a^2,a+b\) nguyên tố cùng nhau.
b) Gọi \(ƯCLN\left(ab,a+b\right)=d\)
\(\Rightarrow\left\{{}\begin{matrix}ab⋮d\\a+b⋮d\end{matrix}\right.\)
Vì a và b nguyên tố cùng nhau nên từ \(ab⋮d\Rightarrow\left[{}\begin{matrix}a⋮d\\b⋮d\end{matrix}\right.\). Đến đây kết hợp với \(a+b⋮d\) và lập luận tương tự như câu a), sẽ chứng minh được \(d=1\)
\(\dfrac{1}{5}\cdot\dfrac{4}{7}+\dfrac{1}{5}\cdot\dfrac{3}{7}-\dfrac{1}{5}=\dfrac{1}{5}\cdot\left(\dfrac{4}{7}+\dfrac{3}{7}-1\right)=\dfrac{1}{5}\cdot\left(1-1\right)=0\)
________
\(\dfrac{3}{7}\cdot\dfrac{5}{8}+\dfrac{3}{7}\cdot\dfrac{11}{8}-\dfrac{3}{7}=\dfrac{3}{7}\cdot\left(\dfrac{5}{8}+\dfrac{11}{8}-1\right)=\dfrac{3}{7}\cdot\left(\dfrac{16}{8}-1\right)=\dfrac{3}{7}\cdot\left(2-1\right)=\dfrac{3}{7}\cdot1=\dfrac{3}{7}\)
_______
\(12\left(\dfrac{7}{6}-\dfrac{8}{12}+\dfrac{29}{4}\right)=12\cdot\dfrac{7}{6}-12\cdot\dfrac{8}{12}+12\cdot\dfrac{29}{4}=14-8+87=93\)
__________
\(58\cdot\left(3\dfrac{1}{29}-2\dfrac{1}{58}\right)=58\cdot\left(\dfrac{88}{29}-\dfrac{117}{58}\right)=58\cdot\dfrac{88}{29}-58\cdot\dfrac{117}{58}=176-117=59\)