1.

Để ý rằng \(\dfrac{36}{4}=9\) nên 4 đỉnh tạo thành hình vuông khi chúng lần lượt cách nhau 9 đỉnh

Do đó ta có các bộ (1;10;19;28), (2;11;20;29),... (9; 18; 27, 36), tổng cộng 9 bộ hay 9 hình vuông

Xác suất: \(P=\dfrac{9}{C_{36}^4}=...\)

2.

Trong mp (ABCD), nối BM kéo dài cắt AD tại E

\(\Rightarrow SE=\left(SAD\right)\cap\left(SBM\right)\)

b. Gọi N là trung điểm SC \(\Rightarrow\dfrac{DG}{DN}=\dfrac{2}{3}\) (t/c trọng tâm)

Do \(AD||BC\) , áp dụng Talet:

\(\dfrac{IB}{ID}=\dfrac{BC}{AD}=\dfrac{1}{2}\Rightarrow\dfrac{IB}{ID}=\dfrac{1}{2}\Rightarrow\dfrac{ID}{BD}=\dfrac{2}{3}\)

\(\Rightarrow\dfrac{DG}{DN}=\dfrac{ID}{IB}\Rightarrow IG||BN\Rightarrow IG||\left(SBC\right)\)

c. Trong mp (SAD), nối QE cắt SD tại P

Talet: \(\dfrac{BC}{DE}=\dfrac{MC}{MD}=1\Rightarrow BC=DE\Rightarrow DE=\dfrac{1}{3}AE\)

Áp dụng Menelaus cho tam giác SAE:

\(\dfrac{QS}{QA}.\dfrac{AE}{ED}.\dfrac{DP}{PS}=1\) \(\Leftrightarrow1.3.\dfrac{DP}{PS}=1\Leftrightarrow SP=3DP\)

\(\Rightarrow\dfrac{SP}{SD}=\dfrac{3}{4}\)

3.

\(2sinx.cosx-4sinx+mcosx-2m=0\)

\(\Leftrightarrow2sinx\left(cosx-2\right)+m\left(cosx-2\right)=0\)

\(\Leftrightarrow\left(2sinx+m\right)\left(cosx-2\right)=0\)

\(\Leftrightarrow sinx=-\dfrac{m}{2}\)

Phương trình có nghiệm khi và chỉ khi:

\(-1\le-\dfrac{m}{2}\le1\Leftrightarrow-2\le m\le2\)

4.

\(cot\dfrac{A}{2}+cot\dfrac{C}{2}=2cot\dfrac{B}{2}\Leftrightarrow\dfrac{cos\dfrac{A}{2}}{sin\dfrac{A}{2}}+\dfrac{cos\dfrac{C}{2}}{sin\dfrac{C}{2}}=\dfrac{2cos\dfrac{B}{2}}{sin\dfrac{B}{2}}\)

\(\Leftrightarrow\dfrac{cos\dfrac{A}{2}sin\dfrac{C}{2}+cos\dfrac{C}{2}sin\dfrac{A}{2}}{sin\dfrac{A}{2}sin\dfrac{C}{2}}=\dfrac{2cos\dfrac{B}{2}}{sin\dfrac{B}{2}}\)

\(\Leftrightarrow\dfrac{sin\left(\dfrac{A+C}{2}\right)}{sin\dfrac{A}{2}sin\dfrac{C}{2}}=\dfrac{2cos\dfrac{B}{2}}{sin\dfrac{B}{2}}\Leftrightarrow\dfrac{cos\dfrac{B}{2}}{sin\dfrac{A}{2}sin\dfrac{C}{2}}=\dfrac{2cos\dfrac{B}{2}}{sin\dfrac{B}{2}}\)

\(\Leftrightarrow sin\dfrac{B}{2}=2sin\dfrac{A}{2}sin\dfrac{C}{2}\)

\(\Leftrightarrow sin\dfrac{B}{2}=cos\left(\dfrac{A-C}{2}\right)-cos\left(\dfrac{A+C}{2}\right)\)

\(\Leftrightarrow sin\dfrac{B}{2}=cos\left(\dfrac{A-C}{2}\right)-sin\dfrac{B}{2}\)

\(\Leftrightarrow2sin\dfrac{B}{2}=cos\left(\dfrac{A-C}{2}\right)\Leftrightarrow2sin\dfrac{B}{2}cos\dfrac{B}{2}=cos\dfrac{B}{2}.cos\left(\dfrac{A-C}{2}\right)\)

\(\Leftrightarrow2sinB=cos\left(\dfrac{A+B-C}{2}\right)+cos\left(\dfrac{B+C-A}{2}\right)\)

\(\Leftrightarrow2sinB=sinC+sinA\)

\(\Leftrightarrow\dfrac{2b}{R}=\dfrac{c}{R}+\dfrac{a}{R}\Leftrightarrow2b=a+c\)

Đáp án:

Ba chìm bảy nổi chín lênh đênh

Trả lời:

ba ..chìm.. bảy ..nổi.. chín lênh đênh.

CHÚC BẠN HỌC TỐT!

Ba chìm bảy nổi chín lênh đênh.

                      Nhớ đúng !

1 The distance  from my home to  school is about 3 km

2 My mum used to live in a small village when she was small
3 Despite being a millionaire , he lives in a small flat 
4 when does the festive take place ?
5 It is about two kilometres from my home to school
6 he didn't use to ride his bike to school 
7 Despite having a test tomrrow , they are still watching TV now 

\(\overrightarrow{AB}.\overrightarrow{AC}=AB.ACcos\widehat{BAC}=a.a\sqrt{2}.cos45^0=a^2\)

 Mọi người ơi giúp em với ạ

Tham khảo:

In a spell of dry weather, when the Birds could find very little to drink, a thirsty Crow found a pitcher with a little water in it.

But the pitcher was high and had a narrow neck, and no matter how he tried, the Crow could not reach the water. The poor thing felt as if he must die of thirst.

Then an idea came to him. Picking up some small pebbles, he dropped them into the pitcher one by one. With each pebble the water rose a little higher until at last it was near enough so he could drink.

“In a pinch a good use of our wits may help us out.”