3x minus 16x plus 5 = 0

3x minus 16x plus 5 = 0

3x^2-16x+5=0
(3x-1)(x-5)=0
x=1/3 ,x=5

The solution equation is: x ^ 2 + 3x = 0,

Factorization
The original equation can be written as
x(x+3)=0
It can be seen that as long as one of X or x + 3 is 0, the equation can be established
So X1 = 0, X2 = - 3

X-1.2-0.3x = 0.2 to solve the equation

x-1.2-0.3X=0.2
0.7x=1.2+0.2
x=1.4/0.7
x=2

If the tangent of circle (x-1) 2 + (y + 3) 2 = 1 is made through point P (2,4), then the tangent equation is______ ．

When the tangent slope does not exist, the tangent equation is x = 2. When the tangent slope exists, let the tangent equation be y-4 = K (X-2), that is, kx-y + 4-2k = 0, and then according to the distance from the center of the circle (1, - 3) to the tangent equal to the radius, we can get | K + 3 + 4 − 2K | K2 + 1 = 1 In conclusion, the tangent equation of a circle is 24x-7y-20 = 0, or x = 2, so the answer is: 24x-7y-20 = 0, or x = 2

Let m = 1 / x + 1 / y, n = 4 / (x + y), and try to compare the size of M and n

M=(x＋y)/(xy)
N=4/(x＋y)
Because m and N are both positive numbers, then:
M/N=(x＋y)²/(4xy)
=(x²＋2xy＋y²)/(4xy)
=(1/4)[(x/y)＋(y/x)]＋(1/2)
Because (x / y) + (Y / x) is greater than or equal to 2, then:
M / N ≥ 1
That is, M is greater than or equal to n

In the pyramid p-abcd, PD is perpendicular to the plane ABCD, PD = DC = BC = 1, ab = 2, AB is parallel to DC, and BCD is 90 degrees

It should be PD vertical plane ABCD,
BC is a straight line on ABCD,
So, PD vertical BC

It is known that y = - Sin & sup2; θ - 2msin θ - 2m-1 is always less than 0 for any angle. Try to find the value range of real number M

y=-sin²θ-2msinθ-2m-1
y=-sin²θ-2msinθ+m²-m²-2m-1
=-（sin²θ+2msinθ+m²)+m²-2m+1-2
=-(sinθ+m)²+(m-1)²-2
∵-(sinθ+m)²≤0
∴(m-1)²-2＜0
∴(1-√2)＜m＜(1+√2)

Solve the plane equation where the point m (- 1,0,4) is perpendicular to the plane: 3x-4y + Z-10 = 0 and parallel to the line (x + 1) / 3 = (Y-3) / 1 = Z / 2?

If the n power of 2 plus 1 is prime, then n is the power of 2

The fourth or the fifth is not prime

In the cube abcd-a'b'c'd ', if M is the midpoint of a'B' and N is the midpoint of BB ', then the cosine value of the angle between AM and cn is equal to
The lengths of the three sides are root 5 / 2, root 5 / 2 and root 6 / 2 respectively. How to find the root 6 / 2

Parallel lines of AM and CN through B '
Are you OK with the first two? The third one √ 6 / 2
√（BC²+（½AB）²+（½CC'）²）√（1²+½²+½²）=√6/2