2 (3 + x) = 16.8-x to solve the equation 2 (3 + x) = 16.8-x to solve the equation

2 (3 + x) = 16.8-x to solve the equation 2 (3 + x) = 16.8-x to solve the equation

Simplification:
6+2x=16.8-x
2x+x=16.8-6
3x=10.8
x=3.6

Simplified evaluation (x + 2 / X & # 178; - 2x - X-1 / X & # 178; - 4x + 4) × X / x-4, where x = - 3

The original formula = [(x + 2) / X (X-2) - (x-1) / (X-2) &# 178;] × X / (x-4) = [(x + 2) (X-2) - x (x-1)] / X (X-2) &# 178; × X / (x-4) = (X & # 178; - 4-x & # 178; + x) / (X-2) &# 178; × 1 / (x-4) = 1 / (X-2) &# 178; when x = - 3, the original formula = 1 / (...)

The following are the univariate linear equations: A, x ^ 2 + 2x-3 = x ^ 2-x + 3 B, 1 / 3 x ^ 2 + 1 = 2x-5 C, X-1 / 3 = y-3x D, x + 1 = x + 1

The so-called one variable linear equation means that the equation contains only one unknown number, and the highest degree of the unknown number is once
The answer a is a quadratic equation about X, the answer B is the same, the answer c is a quadratic equation of two variables, the answer D is correct, it is a linear equation of one variable, although it is an identity, it conforms to the definition

F (x) = logax (x ＞ 0)
After using the formula of changing bottom, the method of quotient can't be used

f'(x)=(logax)'=(lnx/lna)'=1/(xlna)

-4A ^ 3 + 16A ^ 2 + 12a factorization process!

-4a³+16a²+12a
=(-4a)(a²-4a-3)
=(-4a)(a²-4a+4-7)
=(-4a)[(a-2)²-7]
=-4a(a-2+√7)(a-2-√7)

All nonnegative integers with inequality of x [(2x-1) absolute value] < 6 are

X must satisfy the following two equations
2X-1

Given that the function f (x) = 4x + m · 2x + 1 has and only has one zero point, the range of value of M is obtained and the zero point is obtained

Let 2x = t (T > 0), then T2 + MT + 1 = 0. When △ = 0, i.e. M2-4 = 0, M = - 2, t = 1. When m = 2, t = - 1 does not fit the problem, and it is rounded off. When △ > 0, i.e. m > 2 or m < - 2, the equation T2 + MT + 1 = 0 should have one positive and one negative root, i.e. t1t2 < 0, In conclusion, when m = - 2, ƒ (x) has a unique zero, which is x = 0

Given a point of intersection of quadratic function y = (k2-1) x2 + 2kx-4 and X axis, B (- 2,0), find K
I worked out that K has two values, one is - 1 and the other is 2. Which one do you want to give up~

When k = - 1
k²-1=0
At this point, the function is not a quadratic function, so the contradiction, rounding
Namely
k=2

For a three digit number, the sum of all digits is 6. Transpose the single digit number and the hundred digit number to form a new three digit number equal to 107 / 41 of the original number

Let ABC be the original number
a+b+c=6
The number of transposition of individual and hundred is CBA
cba=abc*107/41=abc*2.61 （1）
So C = 3A or C = 4A
Because C < 6, a = 1 when C = 3A
When a = 1, C = 3, B = 2 generations (1)
The verification is correct, so the original number is 123

The fourth power of X - the second power of 6x - 27 cross multiplication

1 -9
×
1 3
So the original formula = (X & # 178; - 9) (X & # 178; + 3)
=(x+3)(x-3)(x²+3)