The number of real roots of equation 3x ^ 2 + 6x-1 / x = 0 is

The number of real roots of equation 3x ^ 2 + 6x-1 / x = 0 is

The number of real roots of equation (3x & sup2; + 6x-1) / x = 0 is
By (3x & sup2; + 6x-1) / x = [3 (X & sup2; + 2x) + 1] / x = [3 (x + 1) & sup2; - 2] / x = 0
We obtain 3 (x + 1) & sup2; = 2 and x = - 1 ± √ (2 / 3)

The number of real roots of equation 3x ^ 2 + 6x-1 / x = 0 is

Because 3x ^ 2 + 6x-1 / x = 0, we can let Y1 = 3x ^ 2 + 6x, y2 = 1 / x, then Y1 = Y2, and make the image of these two functions as parabola Y1 = 3x ^ 2 + 6x = 3 (x ^ 2 + 2x + 1) - 3 = 3 (x + 1) ^ 2-3 and hyperbola y2 = 1 / X. obviously, the vertex of parabola is in (- 1, - 3) opening upward, so the vertex is in the third quadrant, and double

The area of a classroom is 50 square meters. How many classrooms can be built in 1 hectare of land?

200

If the power function y = f (x) is known, the analytic expression of the image passing through the point (root 2 of 8,2) can be obtained

Let y = x ^ n
Bring in point
2 ^ radical 2 = 8 ^ n
That is, root 8 = 8 ^ n
n=1/2

Looking for English words,
Find the words ending with OO, OI and Io,

too boi radio

Let the sum of the first n terms of the arithmetic sequence an be Sn, if A6 = S3 = 12, then LIM (Sn / n square)=

a1+5d=3a1+3d=12.
A1 + 5D = 12, a1 + D = 4, the solution is: A1 = D = 2
Sn=2n+n(n-1)*2/2=n^2+n.
LIM (n > - infinite) Sn / N ^ 2 = LIM (n > - infinite) (1 + 1 / N) = 1

If the function y = f (x) is differentiable, it is proved that there must be zeros of F (x) + F '(x) between two different zeros of F (x)

Let a and B be two zero points of F (x)
That is, f (a) = f (b) = 0
Let f (x) = e ^ x * f (x)
Then f (a) = f (b) = 0
By Rolle's theorem, there exists C between a and B such that
F'(c)=0
That is, e ^ C f (c) + e ^ c * f '(c) = 0
If e ^ C is reduced, f (c) + F '(c) = 0
C is the zero point of that function between a and B
It's over

Simple calculation of 3.45-8.7 + 16.55-1.3
Master, help me!

3.45-8.7+16.55-1.3
=3.45+16.55-（8.7+1.3）
=20-10
=10
Answer: 10
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1's cube + 2's cube = 9 = 3's Square 1's cube + 2's cube + 3's cube = 36 = 6's Square 1's cube + 2's cube + 3's cube + 4's cube = 100 = 10
The square of
(1) Please show the law of the above equation with the formula of non-zero natural number n: cube of 1 + cube of 2 + cube of 3 +... + cube of n = ()
(2) Use the law you found to calculate: 1's cube + 2's cube + 3's cube + 4's cube +... + 10's cube
(3) Use the law you found to calculate: 11 cube + 12 cube + 13 cube +... + 20 cube. (write the necessary process) speed... Please answer before 8:30, thank you

(1) Write the square of (1 + 2 + 3 +... + n) in space
(2) 55 square 3025
(3) 1 cube to 20 cube = 210 square minus 1 to 10 cubes and 3025 is 41075

Find the extremum of function f (x, y) = the square of X - XY + the square of Y + 3x

X=1.5
Y=0
The partial derivatives of X and y are obtained respectively, and the reciprocal is zero