2X & # 178; (2 / 1xy & # 178; - y) - (X & # 178; Y & # 178;) × (- 3x) - 2Y & # 178; (X & # 178; - 5xy + 2 / 1y) - (2XY) &# 178;

2X & # 178; (2 / 1xy & # 178; - y) - (X & # 178; Y & # 178;) × (- 3x) - 2Y & # 178; (X & # 178; - 5xy + 2 / 1y) - (2XY) &# 178;

[(3x-2 / 1y & # 178;) + 3Y (X-12 / y)] / [(2x + Y & # 178;) - 4Y (x + 4 / 1y)], where x = - 7.8, y = 8.7
Given that point C is the golden section of line AB, and AC = 1cm, then the length of line AB is (). What else is the golden section
AB = 2 ^ (√ 5 - 1) or 2 ^ (3 - √ 5)
Golden section point: given the line AB, if a point C on AB can make AC △ AB = BC △ ac.. Then C is the golden section point of line ab
That's what our teacher said
If f (x) is an odd function defined on R, when x is less than 0, f (x) = x (2-x) to find the analytic expression of F (x)
Why is f (x) = x (2 + x) when x > 0
f(-x)=-f(x）=-x(2-x)
Because X0
When x > 0, f (x) = x (2 + x) is obtained by replacing - x with X
2 / 2 (x-3 / 1y & # 178;) + (3 / 1y & # 178; - 2 / 3x), where x = 2 / 3, y = - 2
2 / 2 (x-3 / 1y & # 178;) + (3 / 1y & # 178; - 2 / 3x)
=1/2x-2x+2/3y²+1/3y²-3/2x
=-3x+y²
=-3x2/3+(-2)²
=-2+4
=2
one hundred and twenty-one thousand two hundred and twelve
If point C is the golden section of line AB, and AC = 2, then AB = what
Come on?!
Seek complete process, two kinds of answers!
1) Near a
Because C is the dividing point
So AC / AB = 3-radical 5 / 2
SO 2 / AB = 3-radical 5 / 2
So AB = 3-radical 5
2) Near B
Because C is the dividing point
So AC / AB = root 5-1 / 2
SO 2 / AB = root 5-1 / 2
So AB = root 5 + 1
Let f (x) be an odd function defined on R, and when x > 0, f (x) = - 2x ^ 2 + 3x + 1, find the analytic expression of F (x)
F (x) is an odd function defined on R
f(-x)=-f(x)
When x > 0, f (x) = - 2x ^ 2 + 3x + 1
X0
f(-x)=-2x^2-3x+1=-f(x)
=>f(x)=2x^2+3x-1
therefore
The analytic expression of F (x) is piecewise
When x > 0, f (x) = - 2x ^ 2 + 3x + 1
When x
① X & # 178; - 1 / 3x-1 + 1-x / x = 2 ② X & # 178; - 4 / 2x + 2-x / 1 = 1 / 3
③ X & # 178; - 1 / x + 1-3x / 1 = 3x-3 / 1
(3x-1) / (x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\/ 3, X2 = 1
If point C is the golden section of line AB, AC = 2cm, then the length of AB is ()
A.4cm
B. (1 + root 5) cm
C. (1-radical 5) cm
D. (3 + root 5) cm
By the way, how does it count
BD first of all, what is the golden section? 1. Let the known line segment be AB, BC ⊥ AB through point B, and BC = AB / 2; 2. Connect AC; 3. Take C as the center of the circle, CB as the radius of the arc, intersect AC at d; 4. Take a as the center of the circle, ad as the radius of the arc, intersect AB at P, then point P is the golden section point of ab
B
Let f (x) be an odd function defined on R, and if x is greater than 0, f (x) = minus 2x square + 3x + 1,
Finding the analytic expression of function f (x)
Monotone interval of function f (x)
When x is larger than 0, we will take x 0 {\ (x) (x) (x) (x) (x) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ (x >...)
When ﹣ x = ﹣ x + 1
If x0, then f (- x) = - 2 (- x) &# 178; + 3 (- x) + 1
=-2x²-3x+1
Because Let f (x) be an odd function defined on R
So f (- x) = - f (x)
So f (x) = - [- 2x & # 178; - 3x + 1... Expansion
When x > 0, f (x) = - 2x & # 178; + 3x + 1
If x0, then f (- x) = - 2 (- x) &# 178; + 3 (- x) + 1
=-2x²-3x+1
Because Let f (x) be an odd function defined on R
So f (- x) = - f (x)
So f (x) = - [- 2x & # 178; - 3x + 1]
=2x²+3x-1
That is, when x0, f (x) = - 2x & # 178; + 3x + 1
When x > 0, f (x) = - 2x & # 178; + 3x + 1, x = 3 / 4
Monotone increasing interval is (0,3 / 4) monotone decreasing interval is (3 / 4, positive infinity)
Because f (x) is an odd function, when x0, f (x) = - 2x ^ 2 + 3x + 1 x0, f (- x) = - 2x ^ 2-3x+
When x = 0, f (- 0) = - f (0) = > 2F (0) = 0 = = > F (0) = 0
When x0;
f(-x)= -2(-x)^2+3(-x)+1
=-2x^2-3x+1
Because f (x) is an odd function, so
f(-x)= -f(x)= -2x^2-3x+1
f(x)= 2x^2+3x-1
therefore
F (x) = {- 2x ^... Expansion
When x = 0, f (- 0) = - f (0) = > 2F (0) = 0 = = > F (0) = 0
When x0;
f(-x)= -2(-x)^2+3(-x)+1
=-2x^2-3x+1
Because f (x) is an odd function, so
f(-x)= -f(x)= -2x^2-3x+1
f(x)= 2x^2+3x-1
therefore
f(x)={-2x^2+3x+1 (x>0)
{0 (x=0)
When {2x ^ 2 + 3x-1 (x0,
F (x) = - 2x ^ 2 + 3x + 1, the axis of symmetry is: x = 3 / 4; the opening is downward,
The monotone increasing interval on [0, + ∞) is: [0,3 / 4],
The monotone decreasing interval is: (3 / 4, + ∞)
The monotone interval of odd function is symmetric about the origin,
So the monotone increasing interval of the original function is: [- 3 / 4,3 / 4]
The monotone decreasing interval is: (- ∞, - 3 / 4); (3 / 4, + ∞) close
[calculation] 1 / 2 X-2 (x-3 / 1y & # 178;) + (- 2 / 3x + 3 / 1y & # 178;)
x/2-2*（x-y²/3)+(-3x/2+y²/3)
=x/2-2x+2y²/3-3x/2+y²/3
=x/2-2x-3x/2+2y²/3+y²/3
=-3x+y²
Where. / is preceded by a numerator and followed by a denominator