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\(f\left(x\right)-g\left(x\right)=5x^2-2x+5-\left(5x^2-6x-\frac{1}{3}\right)\)
= \(5x^2-2x+5-5x^2+6x+\frac{1}{3}\)
=\(4x+\frac{16}{3}\)
Bài 1:
|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}
A(-1) = 2(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)) + 5
A(-1) = \(\dfrac{2}{9}\) + 1 + 5
A (-1) = \(\dfrac{56}{9}\)
A(1) = 2.(\(\dfrac{1}{3}\) )2- \(\dfrac{1}{3}\).3 + 5
A(1) = \(\dfrac{2}{9}\) - 1 + 5
A(1) = \(\dfrac{38}{9}\)
|y| = 1 ⇒ y \(\in\) {-1; 1}
⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))
B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)2
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)
B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).1 + 12
B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1
B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)2
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).1 + (1)2
B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1
B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)
a, 3 : ( 1 - 3/2x ) = 4 : ( 2 - x )
<=> \(\frac{3}{1-\frac{3}{2}x}=\frac{4}{2-x}\)
<=> 3 ( 2 - x ) = 4 ( 1 - 3/2x )
<=> 6 - 3x = 4 - 6x
<=> -3x + 6x = 4 - 6
<=> 3x = -2
<=> x = -2/3
b, 2.3x + 3x-1 = 7( 32 + 2.62 )
b, 2.3x + 3x-1 = 7( 32 + 2.62 )
<=> 2.3x + 3x-1 = 7.81
<=> 3x-1(2.3 + 1) = 7.81
<=> 3x-1.7 = 7.81
<=> 3x-1=81
<=> 3x-1 = 34
=> x - 1 = 4 => x = 5
ét o ét các bn lm giúp mk ik,mk đang cak gấp ạ
Bài 2:
5x-(-2x-0,3)=2,4
=>5x+2x+0,3=2,4
=>7x=2,1
hay x=0,3