Given f {(3x + 4) / (2x-1)} = x + 5, find y = f (x)?

Given f {(3x + 4) / (2x-1)} = x + 5, find y = f (x)?

f{(3x+4)/（2x-1)}=x+5
Let (3x + 4) / (2x-1) = M
∴x=(m+4)/(2m-3)
∴f(m)=(m+4)/(2m-3)+5
=11(m-1)/(2m-3).
So f (x) = 11 (x-1) / (2x-3)
Let B = (3x + 4) / (2x-1)
Then x = - 4 / (5-2B)
f（b）=x+5=-4/（5-2b）
That is, y = f (x) = - 4 / (5-2x)
In addition, a = (3x + 4) / (2x-1), the solution is x = (4 + a) / (2a-3), because f (a) = x + 5, so f (a) = (11a-11) / (2a-3)
Let t = (3x + 4) / (2x-1), then x = (T + 4) / (2t-3), f (T) = x + 5
∴f(t)=(t+4)/(2t-3)+5=(11t-11)/(2t-3)
This shows that the above function relation is satisfied when the unknown number is t. Both T and X can represent unknowns
∴y=f（x）=(11x-11)/(2x-3)
Let a = (3x + 4) / (2x-1), then x = (4 + a) / (2a-3)
So, f (a) = (11a-11) / (2a-3)
y = f(x) = (11x-11)/(2x-3)
Let (3x + 4) / (2x-1) = t, then x = (T + 4) / (2t-3)
So f (T) = x + 5 = (T + 4) / (2t-3) + 5 = (11t-11) / (2t-3)
That is, y = f (x) = 11 (x-1) / (2x-3)
Mathematics of senior one: given f (x) = 3x ^ 3 + 2x ^ 2, find the value of F (2) + F (- 2)
Sixteen
16。 Thank you for asking: how to calculate
Given that: (27 / 8) ^ (x-1) * (2 / 3) ^ (2x-3) = 4 / 9, then x=____
（27/8）^(x-1) × （2/3）^(2x-3)
=（3/2）＾（3X-3）×（2/3）＾（2X-3）
=（2/3）＾〔-（3X-3）〕×（2/3）＾（2X-3）
=（2/3）＾（-X）
4/9=（2/3）＾2
So - x = 2, x = - 2
It is known that: (27 / 8) ^ (x-1) * (2 / 3) ^ (2x-3) = 4 / 9, then x = - 2
It is known that A-B = 1, then the square of a-b-2b=
The square of a - the square of B - 2b = (a-b) (a + b) - 2b
=(a-b)-2b
=a-b
=1
Square of A-Square of b-2b
=(a+b)(a-b)-2b
=(a+b)×1-2b
=a+b-2b
=a-b
=1
Mathematical problems (factorization)
3x^2-2x+1/3
9(a+b)^2+30(a-b)(a+b)+25(a+b)^2
(x^2-2x)^2+2x(x-2)+1
2m^2 n^2+8mn+8+x(mn+2)^2
(a-b)^2(3a-2)+(2-3a)a^2
=1/3(3x-1)^2
First: (- 3x + 1) (- x + 1 / 3)
The second course: 2 (a + b) (32a + 2b)
Third: (2x (X-2) + 1) ^ 2
（1/3x-1/9)(9x-3)=（x-1/3)(3x-1）
4（a+b)(16a+b)
(x^2-2x+1)^2
(mn+2)^2(x+2)
b(3a-2)(b-2a)
Given that the square of a plus the square of B plus the square of C is equal to 2Ab plus 2BC plus 2Ac, how to prove that a = b = C
By substituting a = b = 1, C = 0 into the given conditions, we can get
a^2+b^2+c^2=1+1+0=0
2ab+2bc+2ac=2+0+0=0
Then a = B ≠ C
Therefore, the proposition is false and does not hold
a^2+b^2+c^2=2ab+2bc+2ac
a^2-2ab+b^2+c^2=2c(b+a)
(a-b)^2+c^2=-2c(a-b)
(a-b)^2+2c(a-b)+c^2=0
[(a-b)+c]^2=0
So A-B + C = 0
a=b-c
-b=c-a c=-c+a
c=-a+b
So a = b = C
a²+b²+c²=2ab+2bc+2ac
2a²+2b²+2c²-4ab-4bc-4ac=0
(a²-2ab+b²)+(b²-2bc+c²)+(a²-2ac+c²)=0
(a-b)²+(b-c)²+(a-c)²=0
a-b=0,b-c=0,a-c=0
a=b=c
（xy）²-3xy-10
2.x²-y²+y-¼
3. Given X & # 178; - 5x-1 = 0, find the value of X - (one part of x), X & # 178; + (one part of X & # 178;)
4. Fill in the blanks: 1x2x3x4 + 1 = (), # 178;;
               2x3x4x5+1=(  )²
               11x12x13x14+1=(  )²
               36x37x38x39+1=(      )²
The rule of the above formula is summed up in the words you have learned, and the language language you have learned is used to use the words you have learned to use the words you have learned to use the words you have learned to use the words you have learned to use the words you have learned to use the words you have learned to summarize the rule of the above formula, and the rule of the rule of the above formula, the rule of the above formula, which is & nbsp & nbsp; & nbsp & nbsp & nbsp & nbsp & nbsp; & nbsp & nbsp & nbsp; & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp; & & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp; & & nbsp & nbsp & & nbsp & & nbsp & nbsp & nbsp & nbsp & nbsp; & & & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp nbsp; & nbsp; & nbsp; & nbsp; & nbsp;                                         ）
And explain this rule with the knowledge of factorization
1. And it's also the result of 178; (3-3-xy-10 = (XY-5) (XY-5) (XY-5) (XY + 2) 2.x & \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\thevalue of X & # 178; -
(3a + 2b) square - (2B + a) square
Primary school students, right? You, 8A & # 178; + 8ab
9a^2+12ab+4b^2
4b^2+4AB+A^2
Expand and merge like items
=8a^2+8ab
Proof: the product of four consecutive integers plus 1 is the square of an integer
Let the four consecutive integers be n-1, N, N + 1, N + 2, then (n-1) n (n + 1) (n + 2) + 1, = [(n-1) (n + 2)] [n (n + 1)] + 1 = (N2 + n-2) (N2 + n) + 1 = (N2 + n) 2-2 (N2 + n) + 1 = (N2 + n-1) 2. Therefore, the product of four consecutive integers plus 1 is the square of an integer
The square of (- 3A + 2b)
It's a process
Complete square formula of Mathematics
How much is the square of (X-Y) - (x + y) (X-Y) =? To process, the first complete square formula, thank you