Kết quả tìm kiếm

Chiếm sạch

phải high

2a. $ 3,7 < x < 5\frac{1}{10} = 5,1 \\ \Rightarrow 3,7 < x < 5,1 $ 1. Gọi 3 số cần tìm là $a, b, c$ Theo đề bài ta có: $ \frac{a + b + c}{3} = 34 \Rightarrow a + b + c = 34 \times 3 = 102 \\ a + 8 = b + c \Rightarrow a + b + c = a + a + 8 = 2a + 8 = 102 \\\Rightarrow 2a = 102 - 8 = 94 \\...

$ \frac{a}{c} = \frac{c}{b} \\ \Rightarrow ab = c^2 \\ \frac{b - a}{a} = \frac{(b - a)(b + a)}{a(b + a)} = \frac{b^2 - a^2}{ab + a^2} = \frac{b^2 - a^2}{c^2 + a^2} $

I: Match the questions (1;2;3;4;5;6;7;8) with the correct answers (A;B;C;D;E;F;G;H) (4ms) 1. How do you go to school? 2. How far is it from here to your school? 3. How old are you? 4. When is your birthday? 5. What is your middle name? 6. Where do you live? 7. Who are you talking to? 8. What...

$A = 1^0 + 2^1 + 2^2 + ... + 2^{2014} \\ = 2^0 + 2^1 + 2^2 + ... + 2^{2014} + 2^{2015} - 2^{2015} \\ = (2^0 + 2^1 + 2^2) + (2^3 + 2^4 + 2^5) + ... + (2^{2013} + 2^{2014} + 2^{2015}) - 2^{2015} \\ = 2^0(2^0 + 2^1 + 2^2) + 2^3(2^0 + 2^1 + 2^2) + ... + 2^{2013}(2^0 + 2^1 + 2^2) - 2^{2015} \\ = (2^0...

thế hệ false

Câu 3 tạo ra 4 góc vuông mà?

$ a^3 + b^3 + c^3 = 3abc \\ \Leftrightarrow a^3 + 3a^2b + 3ab^2 + b^3 + c^3 = 3abc + 3a^2b + 3ab^2 \\ \Leftrightarrow (a + b)^3 + c^3 = 3ab(a + b + c) \\ \Leftrightarrow (a + b)^3 + c^3 - 3ab(a + b + c) = 0 \\ \Leftrightarrow (a + b + c)(a^2 + 2ab + b^2 - ac - bc + c^2) - 3ab(a + b + c) = 0 \\...

$ x^2 - x + 6 \\ = x^2 - x + \frac{1}{4} + \frac{23}{4} \\ = x^2 - 2 . \frac{1}{2} . x + \left (\frac{1}{2} \right )^2 + \frac{23}{4} \\ = \left (x - \frac{1}{2} \right )^2 + \frac{23}{4} \\ \left (x - \frac{1}{2} \right )^2 \geq 0 \\ \Rightarrow \left (x - \frac{1}{2} \right )^2 +...

$ 3^{x + 2} + 3^{x} = 9^{2} . 10 \\ \Leftrightarrow 3^{x}(3^{2} + 1) = (3^{2})^{2} . 10 \\ \Leftrightarrow 3^{x} . 10 = 3^{4} . 10 \\ \Leftrightarrow 3^{x} = 3^{4} \\ Vì\; 3 \neq 0; 3 \neq \pm 1 \Rightarrow x = 4 $