Given that a and B are the two roots of the equation x2-4x + M = 0, B and C are the two roots of the equation x2-8x + 5m = 0, then M=______ .

Given that a and B are the two roots of the equation x2-4x + M = 0, B and C are the two roots of the equation x2-8x + 5m = 0, then M=______ .

A + B = 4 ①, ab = m ②, B + C = 8 ③, BC = 5m ④, C-A = 4 ⑤ can be obtained from ① ③, BC = 5ab & nbsp; can be obtained from ② ④, when B = 0, M = 0; when B ≠ 0, ⑥ becomes C = 5A ⑦, the solution of simultaneous ⑤ and ⑦ is a = 1, C = 5, B = 3, ｜ M = 3. Therefore, M = 0 or 3
x+2y+3z＝12x+3y+z＝23x+y+2z＝3．
x+2y+3z＝1            ①2x+3y+z＝2              ②3x+y+2z＝3 &nbs...
Given that AB is the two roots of the equation x2-4x + M = 0 and BC is the two roots of the equation x2-8x + 5m = 0, then M=
Three
If x2 = Y3 = Z4 and 3x-2y + 5Z = - 20, then x + 3y-z=______ .
∵ x2 = Y3 = Z4, ∵ 6x = 4Y = 3Z, ∵ 3x-2y + 5Z = - 20, ∵ 6x-4y + 10z = - 40, ∵ z = - 4, ∵ x = - 2, y = - 3, ∵ x + 3y-z = - 2 + 3 × (- 3) - (- 4) = - 7
Given that the solution of equation x = 10-4x is the same as that of equation 8x + 5m = 11, then M=______ .
From (1) we get x = 2, the solution of ∵ equation x = 10-4x is the same as that of equation 8x + 5m = 11. Substituting x = 2 into (2) we get: 16 + 5m = 11 ∵ M = - 1
Given x + 2Y + 3x = 20, x + 3Y + 5Z = 31, find the value of X + y + Z
Don't do summer homework well again?
x+2y+3z=20
x+3y+5z=31
Subtract the one above from the one below
y+2z=11
Then the first equation is * 5 at the same time
5x+10y+15z=100
The second simultaneous * 3
3x+9y+15z=93
Subtract the second from the first
2x+y=7
Then the two equations are added together
2x+2y+2z=18
x+y+z=9
The equation (m-1) x ^ (M & # 178; - 5m + 6) + (M-4) x + 2 = 0 is a quadratic equation with one variable when m =,
X ^ (M & # 178; - 5m + 6) is the M & # 178; - 5m + 6 power of X. and + (M-4) x + 2 is the following formula, not the power of X (with process or explanation + points)
If the quadratic equation of one variable has two cases: first, the model ax ^ 2 + BX + C = 0 must satisfy two conditions at the same time: 1, M-1 ≠ 0; 2, m ^ 2-5M + 6 = 2, from 1, m ≠ 1, from 2, when m ^ 2-5M + 6 = 2, M = 1 or M = 4. Therefore, M = 4, the original equation is 3x ^ 2 + 2 = 0 ∵ 3x ^ 2 + 2 > 0, and the equation has no solution
(1)(4x-3y)^2 (2)(1.5a-2/3b)^2 (3)(25x^3+15^2-20x)÷(-5x) (4)(2/5y^3-7y^2+2/3y)÷2/3y
(5)(8a^3b-5a^2b^2)÷4ab (6)(2a)^3×b^4÷12a^3b^2 (7)(6/5a^3x^4-0.9ax^3)÷3/5ax^3
the sooner the better,
(1)(4x-3y)^2 =(4x)²-2×4x×3y+(3y)²=16x²-24xy+9y² (2)(1.5a-2/3b)^2=2.25a²-2ab+4b²/9 (3)(25x^3+15x^2-20x)÷(-5x) =25x^3÷(-5x) +15x^2÷(-5x) -20x ÷(-5x) =-5x²-3x+4 ...
(1)(4x-3y)^2
=16x^2-24xy+9y^2
(2)(1.5a-2/3b)^2
=2.25a^2-2ab+4/9b^2
(3)(25x^3+15^2-20x)÷(-5x)
=5(5x^3+45-4x)÷(-5x)
=(5x^3+45-4x)÷(-x)
=-5x^2 -5/x +4
(4) (2 / 5Y ^ 3-7y ^ 2 + 2... Expand)
(1)(4x-3y)^2
=16x^2-24xy+9y^2
(2)(1.5a-2/3b)^2
=2.25a^2-2ab+4/9b^2
(3)(25x^3+15^2-20x)÷(-5x)
=5(5x^3+45-4x)÷(-5x)
=(5x^3+45-4x)÷(-x)
=-5x^2 -5/x +4
(4)(2/5y^3-7y^2+2/3y)÷2/3y
=y(2/5y^2-7y+2/3)÷2/3y
=(2/5y^2-7y+2/3)*3/2
=3/5y^2-21/2y+1
(5)(8a^3b-5a^2b^2)÷4ab
=2a^2-5/4ab
(6)(2a)^3×b^4÷12a^3b^2
=8a^3×b^4÷12a^3b^2
=2/3b^2
(7)(6/5a^3x^4-0.9ax^3)÷3/5ax^3
=ax^3(6/5a^2x-9/10)÷3/5ax^3
=(6/5a^2x-9/10)*5/3
=2A ^ 2x-3 / 2
(1)(4x-3y)^2=16 x^2-12xy+9 y^2
(2)(1.5a-2/3b)^2 =9/4 a^2-ab+4/9 b^2
(3)(25x^3+15^2-20x)÷(-5x)=-5 x^3+4 x^2-45
(4)(2/5y^3-7y^2+2/3y)÷2/3y=3/5 y^2-21/2 y
It is known that the two roots of the equation (M + 1) x & # 178; - (m-1) X-2 = 0 about X are integers, so we can find the value of M
M can be - 2,0,1, because according to the formula x = [- B ± root sign (b ^ 2-4ac)] / 2a, that is, one of the roots of X is 1 and the other is - 2 / (M + 1). When m is - 2,0,1, X is an integer
How to solve the equations of two equations and three unknowns, such as 3x + 7Y + 11z = 315, 4x + 10Y + 11z = 420 to find the value of X, y, Z
4x+10y+11z-3x-7y-11z＝x+3y＝105
Then multiply both sides by 3 = 3x + 9y = 315
Then 3x + 9y-3x-7y = 0 = 2Y = 0
That is, y = 0
Substitute x + 3Y = 105 to get x = 105
Substituting the two results in Z = 0
The number of independent equations must be the same as the number of unknowns.
There are few equations, so we can't solve the specific value.
If the equations are independent, there is no solution. Or they are not independent.
You can take Z as a known value and find the solution of X, y with parameters.