10.
Vì M thuộc $ y = 5x \Rightarrow y = 5x $
a)
$ x + y = 0 \\\Leftrightarrow x + 5x = 0 \\\Leftrightarrow 6x = 0 \\\Leftrightarrow x = 0 \\\Rightarrow y = 5x = 5 . 0 = 0 $
b)
$ x - 3y = -12 \\\Leftrightarrow x - 3 . 5x = -12 \\\Leftrightarrow x - 15x = -12 \\\Leftrightarrow -14x = -12 \\\Leftrightarrow x = \frac{6}{7} \\\Rightarrow y = 5 . \frac{6}{7} = \frac{30}{7} $
c)
$ |2x - y| = 9 \\\Leftrightarrow |2x - 5x| = 9 \\\Leftrightarrow |-3x| = 9 \\\Leftrightarrow -3x = 9 \; or \; -3x = -9 \\\Leftrightarrow x = -3 \; or \; x = 3 \\\Rightarrow y = 5 . 3 = 15 \; or \; y = 5 . (-3) = -15 $
11.
a) Tự thay
b)
$ y = f(x) = -2 \\\Leftrightarrow 5x^2 - 2 = -2 \\\Leftrightarrow 5x^2 = 0 \\\Leftrightarrow x^2 = 0 \\\Leftrightarrow x = 0 $
$ y = f(x) = 3 \\\Leftrightarrow 5x^2 - 2 = 3 \\\Leftrightarrow 5x^2 = 5 \\\Leftrightarrow x^2 = 1 \\\Leftrightarrow x = 1 \; or \; x = -1 $
c)
Ta có
$ x^2 = (-x) ^2 \\\Leftrightarrow 5x^2 = 5(-x) ^2
\\\Leftrightarrow 5x^2 - 2 = 5(-x)^2 - 2 \\\Leftrightarrow f(x) = f(-x) $
12.
$ f(0) = a. 0^2 + b. 0 + c = c \\\Leftrightarrow 2010 = c $
$ f(1) = a. 1^2 + b. 1 + c = a + b + 2010 \\\Leftrightarrow 2011 = a + b + 2010 \\\Leftrightarrow 1 = a + b $
$ f(1) = a. (-1)^2 + b. (- 1) + c = a - b + 2010 \\\Leftrightarrow 2012 = a - b + 2010 \\\Leftrightarrow 2 = a - b $
$ a + b + a - b = 1 + 2 \\\Leftrightarrow 2a = 3 \\\Leftrightarrow a = 1,5 \\\Leftrightarrow b = -0,5 $
$ \Rightarrow f(x) = 1,5x^2 - 0,5x + 2010 $
Thay $ x = -2 $ vào