Tìm GTLN của:A=-x^2-2y^2+2xy+2x-4y+100
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\(A=-10x^2-y^2+6xy-4x+20\)
\(=-9x^2+6xy-y^2-x^2-4x-4+24\)
\(=-\left(3x-y\right)^2-\left(x+2\right)^2+24\le24\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}3x-y=0\\ x+2=0\end{cases}\Rightarrow\begin{cases}x=-2\\ y=3x=3\cdot\left(-2\right)=-6\end{cases}\)
Góc bẹt luôn có số đo là 180 độ vì nó được định nghĩa là một góc mà hai cạnh của nó nằm trên cùng một đường thẳng và kéo dài về hai phía đối diện nhau.
góc bẹt gấp đôi góc vuông, mà góc vuông là 90 độ,góc bẹt là 90 độ nhân 2[vì đôi cũng đồng nghĩa với 2] =180 độ
Olm chào em, cảm ơn đánh giá của em về chất lượng bài giảng của Olm, cảm ơn em đã đồng hành cùng Olm trên hành trình tri thức. Chúc em học tập hiệu quả và vui vẻ cùng Olm em nhé!
137.25 + 25 - 38.25
= 137.25 + 25.1 - 38.25
= 25.(137 + 1 - 38)
= 25.(138 - 38)
= 25.100
= 2500
137. 25 + 25 - 38.25
= 137. 25 + 25.1 - 38.25
= 25 . (137 + 1 - 38)
= 25 . (138 - 38)
= 25 . 100
= 2500
10+9-8+7-6+5-4+3-2+1-0
= 10 +[(9 - 8) + (7 - 6) + (5 - 4) + (3 - 2) + (1 - 0)]
= 10 + [1 + 1 + 1 + 1 + 1]
= 10 + 1 x 5
= 10 + 5
= 15
Số số hạng của dãy số là:
10-0+1=11(số)
Tổng của dãy số là: \(\left(10+0\right)\times\frac{11}{2}=10\times\frac{11}{2}=5\times11=55\)
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1. Gom nhóm và sắp xếp lại
\(A = - x^{2} - 2 y^{2} + 2 x y + 2 x - 4 y + 100\)
Nhóm thành:
\(A = - \left(\right. x^{2} - 2 x y + 2 y^{2} \left.\right) + 2 x - 4 y + 100\)
2. Nhận dạng hằng đẳng thức
\(x^{2} - 2 x y + 2 y^{2} = \left(\right. x - y \left.\right)^{2} + y^{2}\)
Suy ra:
\(A = - \left(\right. \left(\right. x - y \left.\right)^{2} + y^{2} \left.\right) + 2 x - 4 y + 100\) \(A = - \left(\right. x - y \left.\right)^{2} - y^{2} + 2 x - 4 y + 100\)
3. Đặt ẩn phụ
Đặt \(u = x - y \textrm{ }\textrm{ } \Rightarrow \textrm{ }\textrm{ } x = u + y\).
Thay vào:
\(A = - u^{2} - y^{2} + 2 \left(\right. u + y \left.\right) - 4 y + 100\) \(A = - u^{2} - y^{2} + 2 u + 2 y - 4 y + 100\) \(A = - u^{2} - y^{2} + 2 u - 2 y + 100\)
4. Phân tích theo từng biến
\(A \left(\right. u , y \left.\right) = - \left(\right. u^{2} - 2 u \left.\right) - \left(\right. y^{2} + 2 y \left.\right) + 100\) \(= - \left(\right. u^{2} - 2 u + 1 \left.\right) + 1 - \left(\right. y^{2} + 2 y + 1 \left.\right) + 1 + 100\) \(= - \left(\right. u - 1 \left.\right)^{2} - \left(\right. y + 1 \left.\right)^{2} + 102\)
5. Tìm giá trị lớn nhất
\(u - 1 = 0 \text{v} \overset{ˋ}{\text{a}} y + 1 = 0\)
Tức là \(u = 1 , y = - 1\).
Amax=102A_{\max} = 102Amax=102
✅ Đáp số:
Amax=102A_{\max} = 102Amax=102
(Đạt được khi \(x = u + y = 1 + \left(\right. - 1 \left.\right) = 0 , \textrm{ }\textrm{ } y = - 1\))