Solve X & # 179; - 7x & # 178; - 17x-9 = 0,

Solve X & # 179; - 7x & # 178; - 17x-9 = 0,

y(x)=x^3-7x^2-17x-9=0 （1）
y=x^3-7x^2-17x-9
＝（x+1）(x^2-8x-9)
=(x-9)(x+1)^2 ＝0 （2）
Namely:
X1,2 = - 1 (2 multiple roots) X3 = 9 (3)

It is known that y-13 and 7x are in positive proportion. When x = - 1 and x = - 6 / 7, their Y values are opposite. Find the value of X when y-29

y-13=k/7x
y=7kx+13
So - 6K + 13 = - (- 7K + 13)
-6k+13=7k-13
k=2
y=14x+13
So y = - 29
x=-3

3 (7x-5) - 1 / 3 (5-7x) + 1 / 7 (7x-50) = 7 (5-7x)

3 (7x-5) - 1 / 3 (5-7x) + 1 / 7 (7x-5) = 7 (5-7x)
3 (7x-5) + 1 / 3 (7x-5) + 1 / 7 (7x-5) + 7 (7x-5) = 0
(7x-5)(3+1/3+1/7+7)=0
7x-5=0
x=5/7

Given that a and B are two of the equations x2 + X-2 = 0, then 2A2 + 2A + B=______ ．

∵ A and B are two of the equations x2 + X-2 = 0, ∵ A2 + A-2 = 0, a + B = - 1, ∵ A2 + a = 2, ∵ 2A2 + 2A + B = 2A2 + A + A + B = 22 − 1 = 2

Does exponential function x = 0 make sense

Yes
a^0=1
So it intersects with the y-axis at (0,1)

Given that x = 2 is the solution of the equation 2x & # 178; + 3ax-2a = 0 about X, find the solution of the equation M & # 178; - 5 = a about M

Let's take x = 2
8+6a-2a=0
4a=-8
a=-2
m^2-5=-2
m^2=3
M = ± root 3
So m = root 3 or M = - root 3

Exponential function and equation
If f (x) = a ^ x-x-a (a > 0, a is not equal to 1) has two zeros, then the value range of real number a?

Break it down into two functions
F1=a^x
F2=x+a
F (x) = 0, that is F1 = F2
There is obviously one point of intersection between F 1 and F 2 (a)
It's too much trouble to ask for extra points for the rest

It is known that x = 2 / 1 is the solution of the equation 2 / 5x + a = 1-3ax about X, and the value of a is obtained

Is two out of one half? If so, a = - one out of two

Exponential function equation problem
1+3^(-x)
-------- =3
1+3^x

1+3^(-x)
-------- =3
1+3^x
Let 3 ^ x = y
Then (1 + 1 / y) = 3 + 3 * y
3y^2+2y-1=0
Y1 = - 1
y2=1/3
3^x＝1/3
x=-1

Given that the equation B / 2 (x + 1) + 1 = 3ax / 2 has innumerable solutions, find the value of a and B
I don't know if it's 3 / 2 or 2 / 3. Please help me

(b/2)(x+1)+1=(3/2)ax
bx+b+2=3ax
(3a-b)x=b+2
3a-b=b+2=0
b=-2
a=-2/3