1.(X+2)(X+3)+X^2-4= 2.X^2+XY-YZ-XZ= It's better to have a process. I just learned this

1.(X+2)(X+3)+X^2-4= 2.X^2+XY-YZ-XZ= It's better to have a process. I just learned this

1) Original formula = x ^ 2 + 5x + 6 + x ^ 2-4 = 2x ^ 2 + 5x + 2 = (2x + 1) (x + 2)
2) The original formula = (x ^ 2 + XY) - (YZ + XZ) = x (x + y) - Z (x + y) = (x-z) (x + y)

a^4-12a^3-28a^2
(x-y)^2+2(x-y)-15
(x^2+2x+3)(x+2x-2)-6
x^2+xy-2y^2-x+7y-6
1. It is known that x ^ 2 + KX + 12 can decompose factor in the range of real number, then integer K=_______
2. Decompose polynomial x ^ 2-x - √ (1 + √ 3) in the range of real number

1. Original formula = a ^ 2 (a ^ 2-12a-28)
=a^2(a-14)(a+3)
2. The original formula = (x-y-3) (X-Y + 5)
3. The original formula = (x ^ 2 + 2x + 3) ^ 2-5 (x ^ 2 + 2x + 3) - 6
=(x^2+2x+3-6)(x^2+2x+3+1)
=(x-1)(x+3)(x^2+2x+4)
4. The original formula = x ^ 2 + (Y-1) x - (2Y ^ 2-7y + 6)
=x^2+(y-1)x-(2y-3)(y-2)
=(x+2y-3)(x-y+2)
5. According to cross multiplication, k = 13, - 13,8, - 8,7. - 7
6. Using the root method, there is the original formula=
(x-(1/2)-(√(1+4√(√3+1)))/2)(x-(1/2)+(√(1+4√(√3+1)))/2)
LZ if there is anything you don't know, just ask me

(1) The second power of 4a-4a-1
(2) The power of 25A - the power of 4 (B + C)
(3) The second power of (x + y) + 6 (x + y) + 9

A & # 178; - 9b & # 178; 25X & # 178; - 361 + 6y + 9y & # 178; 49x & # 178; + 28x + 4 A & # 178; - 8A + 16 81m & # 178; - 144mn + 64n & # 178; help people to the end, I'm sorry to trouble you

If SiNx / (1 + COS) = 1 / 2, then TaNx / 2=

Let a = x / 2
sinx/(1+cosx)
=sin2a/(1+cos2a)
=2sinacosa/(1+2cos²a-1)
=2sinacosa/2cos²a
=sina/cosa
=tana=1/2
That is, TaNx / 2 = 1 / 2

75 × 3.2-2.5 × 0.72 by a simple method

0.75×3.2-2.5×0.72
=2.5*(0.3*3.2-0.72)
=2.5*0.24
=2.5*4*0.06
=10*0.06
=0.6
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Through the point m (2,1), make a straight line L intersection hyperbola x ^ 2-y ^ 2 / 2 = 1 at two points a and B, and M is the midpoint of AB, find the equation of straight line L

Let Y-1 = K (X-2) y = K (X-2) + 1 be substituted into hyperbolic equation to get x ^ 2 - (K (X-2) + 1) ^ 2 / 2 = 12x ^ 2 - (K (X-2) + 1) ^ 2-2 = 02x ^ 2 - (k ^ 2 (X-2) ^ 2 + 2K (X-2) + 1)) - 2 = 02x ^ 2 - (k ^ 2x ^ 2-4k ^ 2x + 4K ^ 2 + 2kx-4k + 1) - 2 = 02x ^ 2-k ^ 2x ^ 2 + 4K ^ 2-2kx + 4k-1-2 = 0 (2

Two parts x plus two parts 3x is five solutions

x/2+3x/2=5
4x/2=5
2x=5
x=2.5

It is known that the coordinates of the intersection point of the images of the two linear functions y = 2x-9 and y = - 3x + 6 (1) can be regarded as the solutions of the two binary linear equations?
Using the method of solving equations, the coordinates of the image intersection of the two linear functions are obtained

﹛y=2x-9
y=-3x+6
Coordinates of the intersection (3, - 3)

(8.4 × 2.5 + 9.7) △ 1.05 △ 1.5 + 8.4 △ 0.28)

=(8.4×10÷4+9.7)÷(0.7+30)
=(84÷4+9.7)÷30.7
=30.7÷30.7
=1

Find the minimum of the square of the distance from the curve X ^ 2-3xy-3y ^ 2 = 1 to the origin

The minimum value of the square of the distance from the curve X & sup2; - 3xy - 3Y & sup2; = 1 to the origin (0,0)
Using conditional derivative in Higher Mathematics,
f(x,y) = x² + y² + λ(x² - 3xy - 3y² - 1)
f'x(x,y) = 2x + λ(2x - 3y) = 0 ①
f'y(x,y) = 2y + λ(-3x - 6y) = 0 ②
x² - 3xy - 3y² = 1 ③
Two extreme points are obtained from (1) and (2): Y / x = - 1 / 3 or 3, substituting (3), x = ± √ (3 / 5) y = negative sign √ (1 / 15)
The minimum value of the square of distance X & sup2; + Y & sup2; = 3 / 5 + 1 / 15 = 2 / 3