If f (x-1) = 2x squared-3x, then f (x)= The square is over X

If f (x-1) = 2x squared-3x, then f (x)= The square is over X

The steps are as follows:
Let t = X-1, then x = t + 1, replace the original formula to get f (T) = 2 (T + 1) ^ 2-3 (T + 1) = 2T ^ 2 + T-1
So f (x) = 2x ^ 2 + X-1

If f (3x-2) = 3x square - 2x + 1, then f (x)=

Let 3x-2 = t
x=(t+2)/3
f(t)=3*[(t+2)/3]^2-2(t+2)/3+1
=(t+2)^2/3-(2t+4)/3+1
=(t^2+4t+4-2t-4+3)/3
=(t^2+2t+3)/3
That is, f (x) = (x ^ 2 + 2x + 3) / 3

It is known that f (1-3x) = 2x square - x + 1
The expression of (1) f (2) value (2) f (x)

(1) 1-3x = 2 x = - 1 / 3 F (1-3 * 1 / 3) = 2 * (1 / 3) - 1 / 3 + 1 = 8 / 9 F (1-3x) = 2x-x + 1 1 1-3x = t x = (1-T) / 3 F (T) = 2 * [(1-T) / 3] - (1-T) / 3 + 1 f (T) = (2t-1-t) / 9 + 1, that is, f (x) = (2t-1-t) / 9 + 1

（2x-10）：（3x+10）=3:

（2x-10）：（3x+10）=3:5
5（2x-10)=3(3x+10）
10x-50=9x+30
10x-9x=30+50
x=80

-X ten (2x-2) one (3x-5)

Original formula = (- 1 + 2-3) x-2-5
=-2x-7

3x + 2x of (x + 1) = 5 of 2

Let a = x / (x + 1)
Then (x + 1) / x = 1 / A
3a+1/(2a)=5/2
Two years on both sides
6a²-5a+1=0
(3a-1)(2a-1)=0
a=1/3,a=1/2
x/(x+1)=1/3
x+1=3x
x=1/2
x/(x+1)=1/2
x+1=2x
x=1
So x = 1 / 2, x = 1

The remainder of F (x) = x ^ 9 + 3x ^ 2 + 2 divided by 2x-3 is?
[I think it's a bit of a problem, because if we use the ordinary case when x = 3 / 2 times, the answer is too big. If we can't find the remainder, we can also find the remainder!]

Let t = 2x-3
x=(t+3)/2
f(x)=1/2^9*(t+3)^9+3/4*(t+3)^2+2
So take t = 0
Remainder = (3 / 2) ^ 9 + 27 / 4 + 2

The maximum value of function f (x) = 2x (5-3x), X ∈ (0, 53) ()
A. 2B. 4C. 256D. 5

∵ function f (x) = 2x (5-3x) = - 6 (x − 56) 2 + 256, X ∈ (0, 53), ∵ when x = 56, the maximum value of function f (x) is 256, so choose: C

The square of (2x-1) = x (3x + 2) - 7

The square of (2x-1) = x (3x + 2) - 7
4x²-4x+1=3x²+2x-7
x²-6x+8=0
(x-2)(x-4)=0
x1=2,x2=4

If the solutions of 2x-1 and - 2 / 7 (3x + 1) are equal, find X

If we simplify the whole formula, we can get 14x-7 = - 6x-2 - > x = 1 / 4