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.
\(cot\dfrac{A}{2}+cot\dfrac{C}{2}=2cot\dfrac{B}{2}\Leftrightarrow\dfrac{cos\dfrac{A}{2}}{sin\dfrac{A}{2}}+\dfrac{cos\dfrac{C}{2}}{sin\dfrac{C}{2}}=\dfrac{2cos\dfrac{B}{2}}{sin\dfrac{B}{2}}\)
\(\Leftrightarrow\dfrac{cos\dfrac{A}{2}sin\dfrac{C}{2}+cos\dfrac{C}{2}sin\dfrac{A}{2}}{sin\dfrac{A}{2}sin\dfrac{C}{2}}=\dfrac{2cos\dfrac{B}{2}}{sin\dfrac{B}{2}}\)
\(\Leftrightarrow\dfrac{sin\left(\dfrac{A+C}{2}\right)}{sin\dfrac{A}{2}sin\dfrac{C}{2}}=\dfrac{2cos\dfrac{B}{2}}{sin\dfrac{B}{2}}\Leftrightarrow\dfrac{cos\dfrac{B}{2}}{sin\dfrac{A}{2}sin\dfrac{C}{2}}=\dfrac{2cos\dfrac{B}{2}}{sin\dfrac{B}{2}}\)
\(\Leftrightarrow sin\dfrac{B}{2}=2sin\dfrac{A}{2}sin\dfrac{C}{2}\)
\(\Leftrightarrow sin\dfrac{B}{2}=cos\left(\dfrac{A-C}{2}\right)-cos\left(\dfrac{A+C}{2}\right)\)
\(\Leftrightarrow sin\dfrac{B}{2}=cos\left(\dfrac{A-C}{2}\right)-sin\dfrac{B}{2}\)
\(\Leftrightarrow2sin\dfrac{B}{2}=cos\left(\dfrac{A-C}{2}\right)\Leftrightarrow2sin\dfrac{B}{2}cos\dfrac{B}{2}=cos\dfrac{B}{2}.cos\left(\dfrac{A-C}{2}\right)\)
\(\Leftrightarrow2sinB=cos\left(\dfrac{A+B-C}{2}\right)+cos\left(\dfrac{B+C-A}{2}\right)\)
\(\Leftrightarrow2sinB=sinC+sinA\)
\(\Leftrightarrow\dfrac{2b}{R}=\dfrac{c}{R}+\dfrac{a}{R}\Leftrightarrow2b=a+c\)
câu 3 5c mn giúp em cần gấp lắm ạ
