When k takes what value, the equation 3x2-2 (3K + 1) x + 3k2-1 = 0 has two real roots which are opposite to each other

When k takes what value, the equation 3x2-2 (3K + 1) x + 3k2-1 = 0 has two real roots which are opposite to each other

3x2-2(3k+1)x+3k2-1=0
Two numbers are opposite to each other
There must be: - 2 (3K + 1) = 0
k=-1/3

Proof: in the common domain, the product of odd function and odd function is even function
2. The product of odd function and even function is odd function
3. The product of even function and even function is even function

1, let two odd functions F1 (x), F2 (x), and f (x) = F1 (x) * F2 (x)
f1(-x)=-f1(x),f2(-x)=-f2(x)
F(-x)=f1(-x)*f2(-x)=[-f1(x)]*[-f2(x)]=f1(x)*f2(x)=F(x)
So f (x) is an even function
2 and 3 are the same as above

What are the characteristics of odd and even functions

Algebraic characteristics: odd function: if f (- x) = - f (x) exists for any X in the domain of F (x), then f (x) is called odd function. Even function: if f (- x) = f (x) exists for any X in the domain of F (x), then f (x) is called even function

It is known that f (x) is an even function defined on (- ∞, + ∞) and an increasing function on (- ∞, 0),
Let a = f (log with 4 as the bottom 7) B = f (log with 1 / 2 as the bottom 3) C = f (0.2 ^ - 0.6), the size relationship of a, B, C

a＝f(log4 7)=f(log2 √7),
b＝f(log1/2 3)=f(-log2 3)=f(log2 3)
c＝f(0.2^-0.6)=f(5^0.6),
√7

F (x) is an even function and f (x) is an increasing function at (0, + ∞)
If x ∈ [0.5,1] f (AX + 1) ≤ f (X-2) is constant, then the value range of real number a is. (note, the answer is [- 2,0]), otherwise, it must be scribbled

If f (x) is an even function, then f (x) = f (| x |)
F (AX + 1) ≤ f (X-2), that is, f (| ax + 1 |)

Compare the size of x2 + Y2 and X + y + XY-1?

(x2+y2)-(x+y+xy-1)
=(x^2-xy+y^2)-(x+y-1)
=[(x^2-2xy+y^2)+(x^2-2x+1)+(y^2-2y+1)]/2
=[(x-y)^2+(x-1)^2+(y-1)^2]/2≥0,
When X-Y = X-1 = Y-1,
That is, when x = y = 1, the equal sign is taken
So (x2 + Y2) ≥ (x + y + XY-1)

Compare the sizes x2 + Y2 and XY

（x-y)^2>=0
x^2-2xy+y^2>=0
x^2+y^2>=2xy
So, x ^ 2 + y ^ 2 is greater than (equal to) XY

The size of (x2 + Y2) 2 and XY (x + y) 2

(x2 + Y2) 2-xy (x + y) 2 = x ^ 4 + y ^ 4 + 2x ^ 2Y ^ 2-x ^ 3y-2x ^ 2Y ^ 2-y ^ 3x = x ^ 3 (X-Y) - y ^ 3 (X-Y) = (X-Y) ^ 2 (x ^ 2 + XY + y ^ 2) because (X-Y) ^ 2 > = 0, mainly see (x ^ 2 + XY + y ^ 2) = (x + Y / 2) ^ 2 + 3Y ^ 2 / 4 > = 0, so (x2 + Y2) 2-xy (x + y) 2 > = 0 when x = y equation holds, so (x2 + Y2) 2 > = XY (

If x and y are arbitrary real numbers, compare the size of x 2 + y 2 and XY, and prove that

It is proved that: ∵ (X-Y) 2 ≥ 0, ∵ x2 + Y2 ≥ 2XY, ∵ x2 + Y2 ≥ XY

2y-8 = - x 4x + 3Y = 7 for XY

Multiply 1 by 2 to get 4y-16 = - 2x
Use formula 1-2 and then formula 2-1,
The solution is: x = - 2, y = 5
xy=-10