Solve the following equation with formula method: known root x + 2 + root X-1 + root X-2. Find the maximum and minimum value of Y Good words in the bonus!

Solve the following equation with formula method: known root x + 2 + root X-1 + root X-2. Find the maximum and minimum value of Y Good words in the bonus!

(x + 2), (x-1), (X-2) are increasing
So y is an increasing function
Domain of definition
x+2>=0,x-1>=0,x-2>=0
So it's x > = 2
So x = 2
The minimum value of Y is 2 + 1 + 0 = 3
There is no maximum

(2 + radical 3 + R) 2 + 1 = R2
Solve the equation, the 2 after brackets and the 2 after R are squares

R=-2
(2 + radical 2-2) ^ 2 + 1 = 4 = (- 2) ^ 2

Root sign 1 × 3 + 1 = root sign 4 = 2 root sign 2 × 4 + 1 = root sign 9 = 3 root sign 3 × 5 + 1 = root sign 16 = 4 root sign 4 × 6 + 1 = root sign 25 = 5 according to the above conditions, please find out the rule and express it with formula

√n（n+2）+1=n+1

The square of X, the square of Y - 2XY + 1
There are 9-12t + 4T square y square + y + quarter x square + XY + y square (x + y) square + 6 (x + y) + 9a square-2a (B + C) + (B + C) square 4xy square-4x square Y-Y cubic-a + 2A square-a cubic

This is the result of X \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\x + y + 3) & 178; a & 178; - 2A (B + C) + (B + C) & 178

Given nonzero vectors a, B, then a + B = 0 is a ‖ B? What is the condition

If a + B = 0, then a = - B, a / / B can be deduced
On the contrary, a / / B cannot deduce a + B = 0. Counterexample: B = 2A ≠ 0, then a and B are parallel, but a + B ≠ 0
So we should fill in unnecessary conditions

In △ ABC, the vertical bisectors of AB and AC intersect BC at points E and f respectively. If ∠ BAC = 115 °, then ∠ EAF=______ Degree

The vertical bisectors of AB and AC intersect BC at points E and f respectively, so: (1) ea = EB, then ∠ B = ∠ EAG, let ∠ B = ∠ EAG = x degree, (2) FA = FC, then ∠ C = ∠ FAH, let ∠ C = ∠ FAH = y, because ∠ BAC = 115 °, so x + y + ∠ EAF = 115 °, according to the triangle inner angle sum theorem, x + y + X + y + ∠ EAF = 180 °, the solution is ∠ EAF = 50 °

The sequence {an} is an arithmetic sequence, A1 = f (x + 1), A2 = 0, A3 = f (x-1), where f (x) = x2-4x + 2, then the general term formula an = ()
A. - 2n + 4B. - 2n-4c. 2n-4 or - 2n + 4D. 2n-4

∵ f (x) = x2-4x + 2, ∵ A1 = f (x + 1) = (x + 1) 2 − 4 (x + 1) + 2 = x2-2x-1, A3 = f (x − 1) = (x − 1) 2 − 4 (x − 1) + 2 = x2-6x + 7, and the sequence {an} is an arithmetic sequence, A2 = 0 ∵ a1 + a3 = 2A2 = 0, ∵ (x2-2x + 1) + (x2-6x + 7) = 2x2-4x6 = 0, the solution is: x = 1 or x = 3 & nbsp; & nbsp

30 points
In the known parallelogram ABCD, a (4,1,3), (2, - 5,1) and C (5,7, - 5) are used to find the coordinates of vertex D, the coordinates of diagonal intersection point O and iabi

Vector AB = (- 2, - 6, - 2), vector AC = (1,6, - 7), vector BC = (3,12, - 6), d = (x, y, z) when vector ad = vector BC, (x-4, Y-1, Z-3) = (3,12, - 6), D (7,13, - 3), midpoint o (9 / 2,4, - 1) when vector CD = vector AB, (X-5, Y-7, Z + 5) = (- 2, - 6, - 2), D (3,1, - 7), midpoint o (7 / 2

E. F is the point on the sides of AB and AC in the triangle ABC, connecting BF and CE to g, and the area of triangle beg, BGC and CGF is 8, 10 and 5, respectively
How to find the area of quadrilateral aegf?

Connect AG, let the area of triangle AEG be x, and the area of triangle AFG be y
(8 + x): y = 10:5 = 2:1
x :(y+5)=8:10=4:5
8+x=2y
5x=4y+20
x=12
y=10
Quadrilateral aegf area = x + y = 22

Help to solve a math problem of grade 4 (using ordinary formula instead of equation)
The title is as follows:
The mother is 32 years older than the son. Three years later, the mother's age is five times that of the son. How old is the son this year?
(do not use equation, use common formula)

[(32 + 3) - 5 × 3] / (5-1) = 5 years old