The mathematician who is good at mathematics knows the univariate quadratic equation x & # 178; - (2k-1) x + K & # 178; - 2 = 0, two roots X1 and X2, and waits for 2 hours It is known that the univariate quadratic equation x & # 178; - (2k-1) x + K & # 178; - 2 = 0 has two roots X1 x2 and (x1 + x2) &# 178; - 3x1x2 = 12 for the value of K It's time for the specific process

The mathematician who is good at mathematics knows the univariate quadratic equation x & # 178; - (2k-1) x + K & # 178; - 2 = 0, two roots X1 and X2, and waits for 2 hours It is known that the univariate quadratic equation x & # 178; - (2k-1) x + K & # 178; - 2 = 0 has two roots X1 x2 and (x1 + x2) &# 178; - 3x1x2 = 12 for the value of K It's time for the specific process

x1+x2=2k-1
x1x2=k²-2
So (2k-1) &# 178; - 3 (K & # 178; - 2 = 12)
4k²-4k+1-3k²+6=12
k²-4k-5=(k+1)(k-5)=0
k=-1,k=5
If there is a solution, then (2k-1) &# 178; - 4 (K & # 178; - 2) > = 0
So k = - 1
Given X & # 178; - 5x = 14, find the value of (x-1) · (2x-1) - (x + 1)
x²-5x=14,
x²-5x-14=0,
(x-7)(x+2)=0
x=-2,7
If x = - 2
（x-1）·（2x-1）-（x+1）=16
If x = 7
（x-1）·（2x-1）-（x+1）=70
I don't know how to send letters~
It's too easy, "X-1 · (2x-1) - (x + 1)" is not clear enough.
Known X & # 178; – 5x = 14
x² – 5x-14=0
(x-7)(x+2)=0
X = 7 and x = - 2
Take 7 and - 2 into (x-1) · (2x-1) - (x + 1) to get 70 and 16, respectively
16 or 70
x^2-5x-14=0
(x-7)(x+2)=0
x1=7,x2=-2
Just bring them in separately
x²-5x=14
(x+2)(x-7)=0
X = - 2 or x = 7
1.x=-2
（x-1）·（2x-1）-（x+1）
=-3*(-5)-(-1)
=15+1=16
2.x=7
（x-1）·（2x-1）-（x+1）
=6*13-8
=78-8
=70
x²-5x=14
x²-5x-14=0
(x-7)(x+2)=0
X = 7 or x = - 2
（x-1）*（2x-1）-（x+1）
=2x²-3x+1-x-1
=2x²-4x
Substituting x = 7 or x = - 2 into the above formula, we get 2x & # 178; - 4x = 16 or 70
(x-1) * (2x-1) - (x + 1) = 16 or 70
Hope to help you
From the known (X-7) (x + 2) = 0, X is - 2uo7
When x = - 2, the solution is 16, and when x = 7shi, the solution is 70
Let x > 0, find y = 2x2 + 5x + 3x, when there is the minimum value, and explain the value of X at this time
If and only if 2x = 3x, x > 0, i.e. x = 62, the minimum value of 26 + 5 is obtained
It is said that no matter what real number x takes, the value of the algebraic formula x2 times + Y2 times - 10x + 8y + 45 is always positive
As long as the formula can be
It can be changed into:
(x-5)^2 (y 4)^2 4
The square of a number is greater than or equal to zero
The original formula is greater than 0
(X-5) square + (y + 4) square + 4
Two nonnegative numbers plus a positive number, it's always positive, right
Given that the real number XY satisfies x2 + y2 = 4, find the maximum value of T = 2x + y
The maximum value is twice the root 5, and the minimum value is minus twice the root 5
The square of (X-2) - the square of 9 (x + 1) = 0. How can this equation be solved by the square root method?
(x-2)^2 - 9(x+1)^2 = 0
(x-2)^2 - [3(x+1)]^2 = 0
According to the square difference formula: [(X-2) + 3 (x + 1)] [(X-2) - 3 (x + 1)] = 0
The results are as follows: (4x + 1) (- 2x - 3) = 0
So x = - 1 / 4 or - 3 / 2
The quadratic equation of one variable x 2-6x-5 = 0 is reduced to the form of (x + a) 2 = B
Solving the original equation can be reduced to x2-6x = 5, formula x2-6x + 9 = 5 + 9, that is (x-3) 2 = 14
Given the circle C: x ^ 2 + y ^ 2 = 5, make two tangent lines of the circle through the point Q (3, - 5), and find the equation of the straight line passing through the two tangent points
Let a (x1, Y1) and B (X2, Y2) be tangent points passing through a, then PA equation of tangent line passing through a is: X1X + y1y = 5, Pb equation of tangent line passing through B is: X2X + Y2Y = 5, P is substituted into two equations: 3x1-5y1 = 5, 3x2-5y2 = 5, a, B coordinates conform to the straight line 3x-5y-5 = 0, the equation of straight line of two tangent points is 3x-5y-5 = 0
Why "if the tangent point a (x1, Y1) is set, then the tangent PA equation passing through a is: X1X + y1y = 5"
Let the linear equation be y + 5 = K (x-3), then the absolute value of k = 3K + 5 divided by the square of K under the root sign + 1 can be obtained from the formula of distance from the point to the line, and the linear equation can be obtained by solving K
No matter what rational numbers x and y are, the value of x2 + y2-10x + 8y + 45 is ()
A. Positive number B. zero C. negative number D. non negative number
X2 + y2-10x + 8y + 45, = x2-10x + 25 + Y2 + 8y + 16 + 4, = (X-5) 2 + (y + 4) 2 + 4, ∵ (X-5) 2 ≥ 0, (y + 4) 2 ≥ 0, ∵ (X-5) 2 + (y + 4) 2 + 4 > 0
4X ^ 4Y ^ 4-1 / 2XY factorization
4x^4y^4-1/2xy
=1/2xy(8x³y³-1)
=1/2xy[(2xy)³-1]
=1/2xy(2xy-1)(4x²y²+2xy+1)