Nguyễn Thị Huyền
Giới thiệu về bản thân
Kích thước cả khung:
\(\left(\right. 17 + 2 x \left.\right) \left(\right. 25 + 2 x \left.\right) = 513\) \(\Rightarrow 425 + 84 x + 4 x^{2} = 513\) \(\Leftrightarrow 4 x^{2} + 84 x - 88 = 0\) \(\Leftrightarrow x^{2} + 21 x - 22 = 0\) \(\Delta = 21^{2} + 88 = 529 \Rightarrow \sqrt{\Delta} = 23\) \(x = \frac{- 21 \pm 23}{2} \Rightarrow x = 1 \&\text{nbsp}; \left(\right. \text{nh}ậ\text{n} \left.\right) , \&\text{nbsp}; x = - 22 \&\text{nbsp}; \left(\right. \text{lo}ạ\text{i} \left.\right)\) \(\boxed{x = 1 \&\text{nbsp};\text{cm}}\)Câu 18
a)
\(cos \alpha = \frac{\mid 3 \cdot 5 + 4 \cdot \left(\right. - 12 \left.\right) \mid}{\sqrt{3^{2} + 4^{2}} \cdot \sqrt{5^{2} + \left(\right. - 12 \left.\right)^{2}}} = \frac{\mid 15 - 48 \mid}{5 \cdot 13} = \frac{33}{65}\)b)
Đường thẳng vuông góc với \(\Delta : 3 x + 4 y + 7 = 0\) có dạng:
Tâm \(C \left(\right. 3 , - 2 \left.\right)\), bán kính \(R = 6\)
\(\frac{\mid 4 \cdot 3 - 3 \left(\right. - 2 \left.\right) + c \mid}{\sqrt{4^{2} + \left(\right. - 3 \left.\right)^{2}}} = 6\) \(\frac{\mid 12 + 6 + c \mid}{5} = 6 \Rightarrow \mid 18 + c \mid = 30\) \(\Rightarrow c = 12 \&\text{nbsp};\text{ho}ặ\text{c}\&\text{nbsp}; c = - 48\)Vậy các đường thẳng cần tìm:
\(4 x - 3 y + 12 = 0 \text{ho}ặ\text{c} 4 x - 3 y - 48 = 0\)Câu 17
a)
\(\Delta = \left(\right. m - 1 \left.\right)^{2} - 4 \left(\right. m + 5 \left.\right) = m^{2} - 6 m - 19\) \(\Delta < 0 \Rightarrow m^{2} - 6 m - 19 < 0\) \(\Rightarrow 3 - 2 \sqrt{7} < m < 3 + 2 \sqrt{7}\)
b)
\(2 x^{2} - 8 x + 4 = x - 2\) \(\Leftrightarrow 2 x^{2} - 9 x + 6 = 0\) \(\Delta = 33\) \(x = \frac{9 \pm \sqrt{33}}{4}\)
Câu 19. (1 điểm)
Gọi \(x\) (km) là độ dài \(A B\) ⇒ \(B C = \sqrt{\left(\right. 5 - x \left.\right)^{2} + 1^{2}}\)
Tổng chi phí:
\(2 x + 3 \sqrt{\left(\right. 5 - x \left.\right)^{2} + 1} = 13\) \(\Rightarrow 3 \sqrt{\left(\right. 5 - x \left.\right)^{2} + 1} = 13 - 2 x\)
Bình phương:
\(9 \left[\right. \left(\right. 5 - x \left.\right)^{2} + 1 \left]\right. = \left(\right. 13 - 2 x \left.\right)^{2}\) \(9 \left(\right. x^{2} - 10 x + 26 \left.\right) = 169 - 52 x + 4 x^{2}\) \(9 x^{2} - 90 x + 234 = 169 - 52 x + 4 x^{2}\) \(\Rightarrow 5 x^{2} - 38 x + 65 = 0\) \(\Delta = 1444 - 1300 = 144 \Rightarrow \sqrt{\Delta} = 12\) \(x = \frac{38 \pm 12}{10} \Rightarrow x = 5 \&\text{nbsp};\text{ho}ặ\text{c}\&\text{nbsp}; x = \frac{13}{5}\)
Chọn \(x = 5\) ⇒ \(B C = 1\)
Tổng chiều dài:
\(A C = A B + B C = 5 + 1 = 6 \&\text{nbsp};\text{km}\) \(\boxed{6 \&\text{nbsp};\text{km}}\)
a)
\(cos \alpha = \frac{\mid 3 \cdot 12 + \left(\right. - 4 \left.\right) \cdot \left(\right. - 5 \left.\right) \mid}{\sqrt{3^{2} + \left(\right. - 4 \left.\right)^{2}} \cdot \sqrt{12^{2} + \left(\right. - 5 \left.\right)^{2}}} = \frac{\mid 36 + 20 \mid}{5 \cdot 13} = \frac{56}{65}\)
b)
Đường thẳng \(d\) song song với \(\Delta : 3 x - 4 y + 7 = 0\) có dạng:
\(3 x - 4 y + c = 0\)
Tâm \(C \left(\right. - 3 , 2 \left.\right)\), bán kính \(R = 6\)
\(\frac{\mid 3 \left(\right. - 3 \left.\right) - 4 \cdot 2 + c \mid}{\sqrt{3^{2} + \left(\right. - 4 \left.\right)^{2}}} = 6\) \(\frac{\mid - 9 - 8 + c \mid}{5} = 6 \Rightarrow \mid c - 17 \mid = 30\) \(\Rightarrow c = 47 \&\text{nbsp};\text{ho}ặ\text{c}\&\text{nbsp}; c = - 13\)
Vậy:
\(\boxed{d : 3 x - 4 y + 47 = 0 \text{ho}ặ\text{c} d : 3 x - 4 y - 13 = 0}\)
Câu 17
a)
\(- 2 x^{2} + 18 x + 20 \geq 0\) \(\Leftrightarrow 2 x^{2} - 18 x - 20 \leq 0\) \(\Leftrightarrow x^{2} - 9 x - 10 \leq 0\) \(\Delta = 121 \Rightarrow \sqrt{\Delta} = 11\) \(x = \frac{9 \pm 11}{2} \Rightarrow x = - 1 , \&\text{nbsp}; 10\) \(\Rightarrow - 1 \leq x \leq 10\)
b)
\(2 x^{2} - 8 x + 4 = x - 2\) \(\Leftrightarrow 2 x^{2} - 9 x + 6 = 0\) \(\Delta = 33\) \(x = \frac{9 \pm \sqrt{33}}{4}\)