

Võ Khánh Linh
Giới thiệu về bản thân



































a) Xét tứ giác ABCD ta có:
\(\hat{A} + \hat{B} + \hat{C} + \hat{D} = 36 0^{o}\)
\(\Rightarrow \hat{D} = 36 0^{o} - 10 2^{o} - 10 2^{o} - 10 2^{o}\)
\(\Rightarrow \hat{D} = 5 4^{o}\)
b) Xét tam giác vuông AOD ta có:
\(A D^{2} = O D^{2} + O A^{2}\)
\(\Rightarrow O A = \sqrt{A D^{2} - O D^{2}}\)
\(\Rightarrow O A = \sqrt{3 0^{2} - 26 , 7^{2}} \approx 13 , 7 \left(\right. c m \left.\right)\)
Xét tam giác vuông AOB ta có:
\(A B^{2} = O A^{2} + O B^{2}\)
\(\Rightarrow O B = \sqrt{A B^{2} - O A^{2}}\)
\(\Rightarrow O B = \sqrt{17 , 5^{2} - 13 , 7^{2}} \approx 10 , 9 \left(\right. c m \left.\right)\)
Độ dài đường chéo BD là:
\(B D = O B + O D = 26 , 7 + 10 , 9 \approx 37 , 6 \left(\right. c m \left.\right)\)
a) \(x y + y^{2} - x - y\)
\(= \left(\right. x y + y^{2} \left.\right) - \left(\right. x + y \left.\right)\)
\(= y \left(\right. x + y \left.\right) - \left(\right. x + y \left.\right)\)
\(= \left(\right. y - 1 \left.\right) \left(\right. x + y \left.\right)\)
b) \(\left(\left(\right. x^{2} y^{2} - 8 \left.\right)\right)^{2} - 1\)
\(= \left(\left(\right. x^{2} y^{2} - 8 \left.\right)\right)^{2} - 1^{2}\)
\(= \left(\right. x^{2} y^{2} - 8 - 1 \left.\right) \left(\right. x^{2} y^{2} - 8 + 1 \left.\right)\)
\(= \left(\right. x^{2} y^{2} - 9 \left.\right) \left(\right. x^{2} y^{2} - 7 \left.\right)\)
\(= \left(\right. x y + 3 \left.\right) \left(\right. x y - 3 \left.\right) \left(\right. x^{2} y^{2} - 7 \left.\right)\)
a) \(\left(\right. - 12 x^{13} y^{15} + 6 x^{10} y^{14} \left.\right) : \left(\right. - 3 x^{10} y^{14} \left.\right)\)
\(= - 12 x^{13} y^{15} : - 3 x^{10} y^{14} + 6 x^{10} y^{14} : - 3 x^{10} y^{14}\)
\(= 4 x^{3} y - 2\)
b) \(\left(\right. x - y \left.\right) \left(\right. x^{2} - 2 x + y \left.\right) - x^{3} + x^{2} y\)
\(= x^{3} - 2 x^{2} + x y - x^{2} y + 2 x y - y^{2} - x^{3} + x^{2} y\)
\(= - 2 x^{2} + 3 x y - y^{2}\)