To make 3x ^ 3 + MX ^ 2 + NX + 42 divisible by x ^ 2-5x + 6, then m, n =?

To make 3x ^ 3 + MX ^ 2 + NX + 42 divisible by x ^ 2-5x + 6, then m, n =?

m=-8 n=-17
Obviously, the quotient should be 3x + 7, just multiply

5abc-{2a^b-[3abc-(4ab^-a^b)]},a=2/3,b=3/2,c=1

The original formula is 5abc - {2A & # 178; B - [3ABC - (4AB & # 178; - A & # 178; b)]}
=5abc-[2a²b-(3abc-4ab²+a²b)]
=5abc-(2a²b-3abc+4ab²-a²b)
=5abc-2a²b+3abc-4ab²+a²b
=(5abc+3abc)+(-2a²b+a²b)-4ab²
=8abc-a²b-4ab²
When a = 2 / 3, B = 3 / 2, C = 1
The original formula is 8 × 2 / 3 × 3 / 2 × 1 - (2 / 3) &# 178; × 3 / 2-4 × 2 / 3 × (3 / 2) &# 178;
=8×2/3×3/2×1-4/9×3/2-4×2/3×9/4
=8-2/3-6
=4/3

The problem of term shifting of linear equation with one variable
3x+20=4x-25
3x-4x=-25-20
-x=-45
x=45
2x+6=2.5x-7.5
2.5x-2x=6+7.5
0.5x=13.5
x=27
What's the difference between the two formulas?
I don't know why

It's not much different, but the principle is the same after changing the number

Understanding of implicit function derivation
As shown in the figure
① The derivative of X on both sides of the equation
②“y=y(x)"
③ How can we get the formula on the right by deriving x from the left side of the equation

e^y+xy-e=0;
Y is a function of X,
Take derivatives on both sides of the equation
Left: the result of e ^ y derivation is: (e ^ y) * y '
The result of XY derivation is y + X * y '
E is 0
So: (e ^ y) * y '+ y + X * y' = 0
Replacing y 'with dy / DX is the result

Factorization: (a + 1) ^ 2-4a
There should be steps

=a²+2a+1-4a
=a²-2a+1
=(a-1)²

Solving inequality 3 less than or equal to absolute value 3-2x less than 7

(1) When 3-2x ≥ 0, then 3 ≤ 3-2x < 7, 0 ≤ - 2x < 4, the solution is - 2 < x ≤ 0
(2) When 3-2x < 0, then 3 ≤ 2x-3 < 7,6 ≤ 2x < 10, the solution is 3 ≤ x < 5
To sum up: - 2 < x ≤ 0 or 3 ≤ x < 5

The number of zeros of function f (x) = / 2 ^ X-1 / - A is discussed

|2^x-1|=a
When a ＜ 0, X has no solution, that is, 0 zeros
When a = 0, 2 ^ X-1 = 0, x = 0, ｜ has one zero point
When 0 ＜ a ＜ 1, 2 ^ X-1 = - A or 2 ^ X-1 = a, x = log2 (1-A) or x = log2 (1 + a), there are two zeros
When a ≥ 1, 2 ^ X-1 ＞ 0, | 2 ^ X-1 | = 2 ^ X-1 = a, | x = log2 (1 + a), | has one zero
in summary:
When a < 0: 0 zeros
When a = 0: 1 zero point
When 0 < a < 1: 2 zeros
When a ≥ 1: 1 zero point

On the equation 2 ^ (2x) - M (2) ^ x + 4 = 0 (x) of X

Let u = 2 ^ x > 0
Then the original equation is reduced to u & # 178; - Mu + 4 = 0 and has positive solutions
So we need to judge the formula = M & # 178; - 16 > = 0 to get m > = 4 or M0, indicating that two of them have the same sign
So we need X1 + x2 = m > 0
So m > = 4

The sum of each digit of a three digit number is equal to 14, and the sum of a single digit number and a ten digit number is 2 greater than that of a hundred digit number. If the hundred digit number and a ten digit number are exchanged, the new number is 270 less than the original number, then the original three digit number is______ ．

Let x, y and Z be the numbers on the individual, ten and hundred digits, x + y + Z = 14x + y = Z + 2100z + 10Y + X − 270 = 100y + 10z + X. the solution is x = 5Y = 3Z = 6 ∧ the original three digit number is 635. So the answer to this question is: 635

If the three roots of the equation x to the third power = 3x-1 are x1, X2 and X3 respectively, where X1 〈 x2 〈 X3, then the interval where X2 is located__
1) Or (1.5,2) why

(0,1) Let f (x) = x ^ 3-3x + 1, then f (- 2) = - 2) ^ 3-3 (- 2) + 1 = - 10, f (0) > 0, f (1) = - 10
So there are zeros in the interval (- 2, - 1), (0,1) and (1,2)
f(1.5)=3.375-4.5+1