It is known that the square of | a + B-2 | + ((AB-1)) = 0 Find a= b=

It is known that the square of | a + B-2 | + ((AB-1)) = 0 Find a= b=

Known: a + B = 4, ab = 2, find (1) a square B + AB square (2) a square + b square

Analysis
（1）a²b+ab²=ab(a+b)
2x4=8
(2) a²+b²=(a+b)²-2ab
=16-2x2
=12

If a + B = 7, ab = 5, find the sum of the squares of a + B, ab

a+b=7
Square on both sides
a²+b²+2ab=49
So a & # 178; + B & # 178;
=49-2ab
=49-2*5
=39
And ab = 5

Given that the radius of circle O is 1, PA is the tangent of circle O, a is the tangent point, PA = 1 chord, ab = root 2, find Pb

Vector NQ + vector QP + vector Mn - vector MP =? (process)

NQ+QP=NP
MN+NP=MP
MP-MP=0

As shown in the figure, in the polyhedral ABCDEF, the quadrilateral ABCD is a square, EF / / AB, EF ⊥ FB, ab = 2ef, ∠ BFC = 90 °, BF = FC, and the tangent value of dihedral angle d-bf-c is obtained
(2) Proving plane EDB ⊥ plane ABCD

Let AB = 2, then DC = 2, FC = √ 2 (⊿ BFC isosceles right angle) ∠ DCF = 90 & # 186; let Tan ∠ DFC = 2 / √ 2 = √ 2 (2) be FH ⊥ BC h ∈ BC o is the midpoint of BD. ohfe is a rectangle. HF ⊥ BC (three in one) ⊥ HF ⊥ a

Set a = {(x, y) x ^ 2 + MX + 1 = 0} set B = {(x, y) X-Y + 1 = 0 and 0 〈 x 〈 2} and a ∩ B ≠ empty set, find the value range of real number M
Detailed process

Set a = {(x, y) x ^ 2 + MX + 1 = 0} set B = {(x, y) X-Y + 1 = 0 and 0 〈 x 〈 2} and a ∩ B ≠ empty set,
That is, the equation x ^ 2 + MX + 1 = 0 has a solution in (0,2)
△>=0
If f (2) > 0, M > = 2 can be obtained
or
△>0
f(2)

The line y = KX + m intersects the ellipse 2x ^ 2 + y ^ 2 = 1 at two different points a and B, and intersects the Y axis at point P (0, m). If the vector AP = vector 3PB, find the value range of M
RT.

Let a (x1, Y1), B (X2, Y2), P (0, m)
Vector AP = vector 3PB
(-x1,m-y1)=3(x2,y2-m),x1=-3x2
Y = KX + m and ellipse 2x ^ 2 + y ^ 2 = 1
(k^2+2)x^2+2kmx+m^2-1=0
X1+x2=-2x2=-2km/(k^2+2),
x2=km/(k^2+2) (1)
X1*x2=( m^2-1)/(k^2+2)=-3x2^2,(2)
Substituting (1) into (2)
-3[km/(k^2+2)]^2=( m^2-1)/(k^2+2),
K ^ 2 = (2-2m ^ 2) / (4m ^ 2-1)
Delt=(2kmx)^2-4(k^2+2)(m^2-1)>0,
The results show that K ^ 2-2m ^ 2 + 2 > 0
Substituting K ^ 2 = (2-2m ^ 2) / (4m ^ 2-1), we get m ^ 2 (m ^ 2-1) / (4m ^ 2-1)

X2 + 3x-3 = 0, then the value of the algebraic formula x3-3x + 3

∵x²+3x-3=0
∴x²=3-3x
∴x³-3x+3
=x(3-x)+3x+3
=-x²+6x+3
=-(3-3x)+6x+3
=9x
Is that right?
It is necessary to solve the equation

It is known that in the parallelepiped abcd-a1b1c1d1, ABCD is rhombic and the angle a1ab = angle a1ad. It is proved that the plane aa1c1c is perpendicular to the plane ABCD

Because angle a1ab = angle a1ad, it is easy to prove that the projection h of a on the bottom surface is on the bisector AC of angle bad. Because ah is perpendicular to the bottom surface, the plane aa1c1c passing through ah is perpendicular to the plane ABCD