How to factorize y ^ 2-5y + 6 = 0 How to multiply a cross with instructions

How to factorize y ^ 2-5y + 6 = 0 How to multiply a cross with instructions

Cross multiplication: Y & # 178; = y times y, 6 = (- 2) x (- 3) or (- 1) x (- 6) Because - 2 + (- 3) = - 5, so draw a fork
So (Y-2) (Y-3) = 0
If and only if y = 2 or 3
Nanya middle school XXX

Solving the equations x + 2Z = 3 2x + y = 2 2Y + Z = 7 by substitution method

From 1: x = 3-2z
The formula is y = (7-z) / 2
Substituting 2: 6-4z + (7-z) / 2 = 2 --- > 8-8z + 7-z = 0, z = 5 / 3
x=3-2z=3-10/3=-1/3
y=(7-5/3)/2=8/3

Let f (x) = a ^ | x | + 2 / (a ^ x) (where constant a > 0 and a ≠ 1)
(1) When a = 10, the equation f (x) = m (where the constant M > 2 √ 2;
(2) If the minimum value of function f (x) on (- ∞, 2] is a constant independent of a, the value range of real number a is obtained

When x > 0, f (x) = a ^ x + 2 / (a ^ x) = a ^ x + 2 * a ^ (- x)
When x = 0, f (0) = 1 + 2 = 3
If x0, then f (x) = 10 ^ x + 2 * 10 ^ (- x) = M
Let 10 ^ x = k, because x > 0, so k > 1. Then K + 2 / k = M
That is, K ^ 2 - MK + 2 = 0
The solution is k = m / 2 + radical (m ^ 2 / 4 - 2) or K = m / 2 - radical (m ^ 2 / 4 - 2)
So x = LG (M / 2 + radical (m ^ 2 / 4 - 2)
Or x = LG (M / 2 - radical (m ^ 2 / 4 - 2)
Note that m ^ 2 / 4 - 2 is more than 0 because m > 2 * radical 2, so the above two roots are meaningful
If x0, f (x) = a ^ x + 2 * a ^ (- x), then f (x) is a continuous function in the domain
When x > 0, f (x) > = 2 * radical (a ^ x * 2 * a ^ (- x)) = 2 * radical 2
Take the equal sign if and only if a ^ x = 2 / A ^ X. in this case, a ^ (2x) = 2, so 2x = log (a, 2)
Because x

a²b²-ab-12=

(ab-4)(ab+3)

Ellipse A's square of X + B's square of y = 1, a, B are left and right vertices, P is different from a, B two points, straight line AP, BP slope product
The eccentricity of ellipse is - 1 / 2

Let P (XO, yo)
Kap*Kbp=【yo/（xo-a）】*【yo/（xo+a）】=-1/2
The result is: XO ^ 2 + 2yo ^ 2 = a ^ 2
P on the ellipse, XO ^ 2 / A ^ 2 + yo ^ 2 / b ^ 2 = 1
B ^ 2xo ^ 2 + A ^ 2yo ^ 2 = (AB) ^ 2
① If the coefficients of corresponding terms are equal, then
b^2=1,a^2=2,(ab)^2 =a^2
The solution is a = √ 2, B = 1, C = 1,
So e = √ 2 / 2

Solve the equation: 2 / 2 X-1 = x + 3

It's very simple. X-1 / 2 = x + 3
x－1＝2x＋6
-x＝7
x＝-7

It is known that A1 and A2 are two-dimensional column vectors, matrix A = (2A1 + A2, a1-a2), B = (A1, A2). If | a | = 6, then B =?, a =?,

A = (2a1+a2,a1-a2)
= (a1,a2) K
K =
2 1
1 -1
|K| = -2-1 = -3
So | a | = | B | K | = - 3 | B | = 6
So | B | = - 2

6 x = 6 solution equation

0.6x=6
x=6÷0.6
x=10

Let x 1 ~ U (0,1), x = max {x 1, x 2}, y = min {x 1, x 2}
Then the probability density of X and Y is?
The answer is 2x and 2 (1-y). Why,

FX(x)=P(X

Given that rational numbers x and M satisfy | x 4 | X-9 | = 13 - (m-2) & # 178;, find the maximum of | X-2 | X-8 |

The rational numbers x and m are known to be rational numbers x and m to satisfy the rational numbers x and m to satisfy the [x + 4 + + |x-9 + |x-9 | = 13 - (m-2) &; when x > 9, when x > 9, x + 4 + + |x-9 124 + |x-9-9 124\\124\\\\\\124\\124\\\\ + + + \124\themaximum value of | X-8 | is 20