-Lúc bắt đầu gieo, chuẩn bị mặt ruộng tốt, giống tốt, ngâm ủ nảy mầm trên 90%.

-Cần tháo nước trước lúc thu hoạch để thúc đẩy quá trình chín của ruộng lúa, tạo điều kiện cho lúa chuyển sang vàng.

chúc bn học tốt ^^

Câu 16.

Hình vẽ tương đối thôi nha!!!

Bảo toàn động lương ta có:

\(\overrightarrow{p}=\overrightarrow{p_1}+\overrightarrow{p_2}\)

\(\Rightarrow p^2=p_2^2-p_1^2\)\(\Rightarrow p_2=\sqrt{p^2+p_1^2}\)

\(\Rightarrow m_2\cdot v_2=\sqrt{\left(m_1+m_2\right)\cdot v+m_1\cdot v_1}\)

\(\Rightarrow0,3\cdot v_2=\sqrt{[\left(0,5+0,3\right)\cdot3]^2+(0,5\cdot4)^2}=3,124\)

\(\Rightarrow v_2=10,41\)m/s

C3: Hệ bpt trở thành: \(\left\{{}\begin{matrix}x\ge1-m\\mx\ge2-m\end{matrix}\right.\)

a, Để hệ phương trình vô nghiệm thì \(m=0\)

b, Để hệ có nghiệm duy nhất thì \(\left\{{}\begin{matrix}m\ne0\\\dfrac{m-2}{m}=1-m\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}m\ne0\\m=\pm\sqrt{2}\end{matrix}\right.\) \(\Leftrightarrow\) \(m=\pm\sqrt{2}\)

c, \(x\in\left[-1;2\right]\) \(\Leftrightarrow\) \(-1\le x\le2\)

Để mọi \(x\in\left[-1;2\right]\) là nghiệm của hệ bpt trên thì

\(\left\{{}\begin{matrix}-1\le1-m\le2\\-1\le\dfrac{2-m}{m}\le2\end{matrix}\right.\) với \(m\ne0\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2\ge m\ge-1\\m\ge\dfrac{2}{3}\end{matrix}\right.\) \(\left(m\ne0\right)\)

\(\Leftrightarrow\) \(2\ge m\ge\dfrac{2}{3}\)

Vậy \(m\in\left[\dfrac{2}{3};2\right]\) thì mọi \(x\in\left[-1;2\right]\) là nghiệm của hệ bpt

Chúc bn học tốt!

27.

Bán kính mặt cầu ngoại tiếp tứ diện vuông được tính bằng:

\(R=\sqrt{\dfrac{OA^2+OB^2+OC^2}{4}}=\sqrt{\dfrac{1^2+2^2+3^2}{4}}=\dfrac{\sqrt{14}}{2}\)

28.

Từ giả thiết suy ra \(A\left(2;2;2\right)\)

Gọi điểm thuộc mặt Oxz có tọa độ dạng \(D\left(x;0;z\right)\)

\(\Rightarrow\overrightarrow{AD}=\left(x-2;-2;z-2\right)\)

\(\overrightarrow{BD}=\left(x+2;-2;z\right)\) ; \(\overrightarrow{CD}=\left(x-4;-1;z+1\right)\)

D cách đều A, B, C \(\Rightarrow\left\{{}\begin{matrix}AD=BD\\AD=CD\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2+4+\left(z-2\right)^2=\left(x+2\right)^2+4+z^2\\\left(x-2\right)^2+4+\left(z-2\right)^2=\left(x-4\right)^2+1+\left(z+1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+z=1\\2x-3z=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}z=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow P\left(\dfrac{3}{4};0;-\dfrac{1}{2}\right)\)

29.

Do tâm I mặt cầu thuộc Oz nên tọa độ có dạng: \(I\left(0;0;z\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AI}=\left(-3;1;z-2\right)\\\overrightarrow{BI}=\left(-1;-1;z+2\right)\end{matrix}\right.\)

Mặt cầu qua A, B nên \(AI=BI\)

\(\Leftrightarrow3^2+1^2+\left(z-2\right)^2=1^2+1^2+\left(z+2\right)^2\)

\(\Leftrightarrow8z=8\Rightarrow z=1\)

\(\Rightarrow I\left(0;0;1\right)\Rightarrow R=IB=\sqrt{1^2+1^1+3^2}=\sqrt{11}\)

Phương trình mặt cầu:

\(x^2+y^2+\left(z-1\right)^2=11\)

30.

Từ phương trình mặt cầu ta có:

\(R=\sqrt{1^2+\left(-2\right)^2+2^2-\left(-m\right)}=\sqrt{m+9}\)

\(\Rightarrow\sqrt{m+9}=5\Rightarrow m=16\)

31.

Khoảng cách giữa điểm M và điểm đối xứng với nó qua Ox là \(2\sqrt{y_M^2+z_M^2}=2\sqrt{65}\)

32.

Gọi \(I\left(x;y;z\right)\) là tâm mặt cầu

\(\overrightarrow{AI}=\left(x-1;y;z\right)\) ; \(\overrightarrow{BI}=\left(x;y-1;z\right)\) ; \(\overrightarrow{CI}=\left(x;y;z+1\right)\); \(\overrightarrow{DI}=\left(x-1;y;z-3\right)\)

Do I là tâm mặt cầu

\(\Rightarrow\left\{{}\begin{matrix}AI=BI\\AI=CI\\AI=DI\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)^2+y^2+z^2=x^2+\left(y-1\right)^2+z^2\\\left(x-1\right)^2+y^2+z^2=x^2+y^2+\left(z-1\right)^2\\\left(x-1\right)^2+y^2+z^2=\left(x-1\right)^2+y^2+\left(z-3\right)^2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}-x+y=0\\-x+z=0\\-6z+9=0\end{matrix}\right.\) \(\Rightarrow x=y=z=\dfrac{3}{2}\)

Hay \(I\left(\dfrac{3}{2};\dfrac{3}{2};\dfrac{3}{2}\right)\) \(\Rightarrow D\) đúng

\(\left(2x+1\right)^2-\frac{4}{9}=0\)

\(\Leftrightarrow\left(2x+1\right)^2=\frac{4}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x+1\right)^2=\left(\frac{2}{3}\right)^2\\\left(2x+1\right)^2=\left(-\frac{2}{3}\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\frac{2}{3}\\2x+1=-\frac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-\frac{1}{3}\\2x=-\frac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{6}\\x=-\frac{5}{6}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{6};-\frac{5}{6}\right\}\)

refer

Viet Nam is a beautiful country with not only a lot of interesting places and historic places but also many heritage site. one of the most beautiful sites is hlb

Firstly, in 1994, UNESCO recognized the heart of Ha Long Bay as being a site of Natural World Heritage and stated that “Apart from Ha Long Bay there are no equivalent sites on the World Heritage List…”. In 2012, Ha Long Bay is officially recognized as one of the new Wonders of Nature.

Secondly, Ha Long Bay’s greatest attraction is its natural scenery. The towering limestone outcrops jutting skywards from the emerald-blue waters. The sheer numbers and size of the islands, islets and karsts within the bay serve to humble even the most jaded traveler.
Swimming, kayaking, hiking and of course photography are favorite pastimes for visitors to Ha Long Bay – one of the most popular tourist attractions in all of Vietnam. Abundant fresh local seafood is another delight.Floating fishing villages are the most popular human attraction. Entire village populations live, work and die on these ’floating islands’ in Ha Long Bay. The villagers are mainly fisherfolk and ply the waters around their villages selling their fresh produce to the passing tourist trade.

Thirdly,Historical research surveys have shown the presence of prehistorical human beings in this area tens of thousands years ago. The successive ancient cultures are the Soi Nhụ culture around 18,000–7000 BC, the Cai Beo culture 7000–5000 BC and the Ha Long culture 5,000–3,500 years ago. Hạ Long Bay also marked important events in the history of Vietnam with many artifacts found in Bài Thơ Mount, Đau Go Cave, Bai Chay.500 years ago, Nguyen Trai praised the beauty of Ha Long Bay in his verse "Lo nhap Van Đon", in which he called it "rock wonder in the sky". In 1962, the Ministry of Culture, Sports and Tourism of North Vietnam listed Hạ Long Bay in the National Relics and Landscapes publication. In 1994, the core zone of Hạ Long Bay was listed by UNESCO as a World Heritage Site according to criterion , and listed for a second time according to criterion.

For these reasons,a trip to Ha Long will make you feel unforgettable even only one day!

Phương trình hoành độ giao điểm (d) và (P):

\(x^2-2x-3=ax-a-3\)

\(\Leftrightarrow x^2-\left(a+2\right)x+a=0\) 

\(\Delta=\left(a+2\right)^2-4a=a^2+4>0;\forall a\Rightarrow\) (d) luôn cắt (P) tại 2 điểm pb

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_A+x_B=a+2\\x_Ax_B=a\end{matrix}\right.\)

Mặt khác do A, B thuộc (d) nên: \(\left\{{}\begin{matrix}y_A=ax_A-a-3\\y_B=ax_B-a-3\end{matrix}\right.\)

\(y_A+y_B=0\)

\(\Leftrightarrow a\left(x_A+x_B\right)-2a-6=0\)

\(\Leftrightarrow a\left(a+2\right)-2a-6=0\)

\(\Leftrightarrow a^2-6=0\)

\(\Leftrightarrow a=\pm\sqrt{6}\)