Detailed process of solving equation: 0 = - 1 / 16 (x-4) ^ 2 + 3

Detailed process of solving equation: 0 = - 1 / 16 (x-4) ^ 2 + 3

1/16(X-4)^2=3
Multiply both sides by 16, (x-4) ^ 2 = 48,
The open root sign x-4 is equal to 48 under the positive and negative root sign,
So x is 48 plus 4 under the plus minus root

3 (1-x / 3) = 1-x / 5?
We need to use the equation

3(1-x/3)=1-x/5
3-x=1-x/5
2-x=-x/5
4/5x=2
x=5/2

Exercises with brackets for linear equation of one variable

3(x+2)=5x-3
3x+6=5x-3
5x-3x=6+3
2x=9
x=4.5
3x+3-2(x-1)=3
3x+3-2x+2=3
x=3-5
x=-2
(x-1)/2+(x+2)/3=6
3(x-1)+2(x+2)=36
3x-3+2x+4=36
5x=35
x=7

A problem on the derivation of implicit function
Xy = e ^ (x + y) finding dy / DX can be directly derived from both sides of X to get: dy / DX = (y-e ^ (x + y)) / (e ^ (x + y) - x)
But if I take the natural logarithm on both sides and convert it to: ln (XY) = x + y, then I take the derivative of X on both sides → (1 / XY) * (y + X * (dy / DX)) = 1 + dy / DX
The result is dy / DX = - 1. What's wrong with my second method?

Both methods are right
Do it directly
dy/dx=(y-e^(x+y))/(e^(x+y)-x)
Replace e ^ (x + y) with XY
That is dy / DX = [y-xy] / [xy-x]
Ln (XY) = x + y and then derive x on both sides
→ (1/xy)*(y+x*(dy/dx))=1+dy/dx
1/x+1/y*dy/dx=1+dy/dx
（y-1)/y*dy/dx=1/x-1=(1-x)/x
dy/dx=(y-xy)/(xy-x)
same

Factorization of X (2x-4y) (- 2x + 4)

x(2x-4y)(-2x+4)=4x(x-2y)(-x+2)

Given that the absolute value of X is equal to 4 and the absolute value of Y is equal to 5, then the value of 2x-y is []
To correct a mistake, there is another sentence: x > y

The absolute value of X is 4, the absolute value of Y is 5,
Then x = ± 4, y = ± 5
X＞Y
So x = ± 4, y = - 5
2x-y has two possibilities
13 ,-3

Given that f (x) is a quadratic function and satisfies the Group F (2x = 1) + F (x-1) = 5x ^ 2 + 3x + 6, find f (x)

This is where we are going to be the (x) and the (x) as (x) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\1, B = 1 / 3, C = 2F (x) = x &

Quadratic function y = X2 - (M + 1) x + 1, when x > 1, y increases with the increase of X, then the value range of M is (it is known that the symmetry axis of quadratic function y = AX2 + BX + C (a ≠ 0) is x = - B2A) ()
A. m≤1B. m≥1C. m≥-3D. m≤-3

When x ＞ 1, y increases with the increase of X. when x = 1, that is, x = - B2A = - − (M + 1) 2 × 1 = 1, the solution is: M = 1, only when the symmetry axis is on the left side of the line x = 1, can it have X ＞ 1, y increases with the increase of X. that is, when x = - − (M + 1) 2 × 1 ＜ 1, m meets the requirement. The solution is: m ≤ 1, so choose: a

For a three digit number, the sum of the numbers on the individual digit, the hundred digit number is equal to the number on the ten digit number, the number on the hundred digit number is 7 times larger than the sum of the numbers on the individual digit, the ten digit number and the hundred digit number, and the sum of the numbers on the individual digit, the ten digit number and the hundred digit number is 14

Let the number in the hundred be a, the number in the ten be B, and the number in the individual be C.A + C = B

A ^ 4-16 / 2A ^ 2B + 8b reduction

（a^2+4）(a+2)(a-2)/2b(a^2+4)
(a+2)(a-2)/2b