Solve the equation m × x square + X + 2m = 0

Solve the equation m × x square + X + 2m = 0

mx²+x+2m=0
X = [- 1 ± root (1-8m & # 178;)] / 2m

(m-1) x square + (2m-1) x + M-3 = 0
You are the best

(1) When m = 1, the original equation = X-2 = 0, x = 2 (2) when m is not equal to 1 △ = (2m-1) - 4 (m-1) (M-3) = 12m-11, when △

If a is the opposite number equal to itself, B is the number of cube equal to itself, and C is the number of square equal to itself, then a + B + C=

a=0
B = ± 1 or 0
C = 0 or 1
therefore
A + B + C = - 1 or 1 or 0 or 2

4. 4, 4, 4 use these four numbers with addition, subtraction, multiplication, division and brackets to make their calculation result equal to 3

(4×4-4)÷4=3

Given the first-order function Y1 = - 2 / 3x + 2 / 3a, y2 = 2 / 3x-a / 2-1. Try to find that when a =? The intersection of two function images is in the fourth quadrant

a=-2

A cuboid pool, covering an area of 6 square meters, with a depth of 1.7 meters, can hold up to () liters of water

A cuboid pool, covering an area of 6 square meters, with a depth of 1.7 meters, can hold up to 10200 liters of water
6x1.7x1000=10200

It's 24:7 13 4 5
How about 7, 13, 4 and 5

（13-5）*（7-4)=24
Do you know?

The general picture of the function f (x) = lnx-12x2 is ()
A. B. C. D.

∵ f (x) = lnx-12x2, its definition domain is (0, + ∞) ∵ f ′ (x) = 1x-x = (1 − x) (1 + x) x, from F ′ (x) > 0, 0 < x < 1; F ′ (x) < 0, x > 1; ∵ f (x) = lnx-12x2, monotonically increasing on (0,1), monotonically decreasing on (1, + ∞); ∵ when x = 1, f (x) takes

The air density is 1.29kg/m3. Can you estimate the buoyancy of a person with a mass of 50kg in the air? (it is estimated that the density of people is similar to that of water)

If the front view and side view of a space geometry are two squares with side length of 1, and the top view is an isosceles right triangle with right angle side length of 1, then the surface area of the geometry is equal to ()
A. 2+2B. 3+2C. 4+2D. 6

According to the three views, the geometry is a straight triangular prism with an isosceles right triangle at the bottom. The length of the right side at the bottom is 1; the height of the prism is 1. So the surface area of the triangular prism is 2S bottom + s side = 2 × 12 × 1 × 1 + 1 × (1 + 1 + 2) = 3 + 2