Evaluation: [log6 (3)] ^ 2 + log6 (18) / log2 (6) [log6(3)]^2+log6(18)/log2(6)

[log6(3)]^2+log6(18)/log2(6)
=[log6(3)]^2+log6(3^2 * 2)*[log6(2)]
=[log6(3)]^2+2log6(3)log6(2)+[log6(2)]^2
=[log6(3)+log6(2)]^2
=[log6(3*2)]^2
=1

Evaluate log2cos π 9 + log2cos2 π 9 + log2cos4 π 9

log2cosπ9+log2cos2π9+log2cos4π9=log2(cosπ9•cos2π9•cos49π)=log2(sin2π92sinπ9•sin4π92sin2π9•sin8π92sin4π9)=log218=-3．

Senior high school mathematics compulsory two yuan
What is the center coordinate of circle 2x2 (quadratic) + 2Y2 (quadratic) - 4x + 8y + 5 = 0?
A.（1,-2） B.（-1,2） C.（2,-4） D.（-2,4）
Please attach the detailed steps. Because I am not very good at math. Thank you

2x2 (quadratic) + 2Y2 (quadratic) - 4x + 8y + 5 = 0
2（x-1）^2+2(y+2)^2=5
(x-1)^2+(y+2)^2=5/2
So the center of the circle (1, - 2)
Choose a

Given that the root of Y-2 and the root of 2-3x are opposite to each other, find the value of X divided by y

Because: the root of Y-2 and the root of 2-3x are opposite to each other
So: √ (Y-2) + √ (2-3x) = 0
And because: √ (Y-2) ≥ 0, √ (2-3x) ≥ 0
So: Y-2 = 0, 2-3x = 0
y=2、x=2/3
X divided by y = 1 / 3

English translation
1.they do simpie jobs over and over again.
2.peopie would not like to do such jobs and would get bored.
There are already robots working in factories
There are already robots working in factories?

They always have to do monotonous work repeatedly
The second is that people always don't like to do such work and be disturbed

Mathematical problems, using formula method to solve quadratic equation of one variable
It is known that the univariate quadratic equation 4 / 1 x square - (m-2) x + m square = 0 (1) with respect to X has two unequal real roots, so we can find the range of M. (2) when the equation has real roots, we can find the maximum integer solution of M

(1) According to the meaning of the title
Δ＞0
∴ [-(m-2)]²－4×(¼)×m²＞0
The solution is: m < 1
(2) When the equation has real roots, Δ ≥ 0
∴[-(m-2)]²－4×(¼)×m²≥0
The solution is m ≤ 1
The largest integer solution of M is 1

What is the area formula of a rectangle?
What is a rectangle?

Rectangle is commonly known as rectangle
S distance = length * width

Prove inequality: when 0 ≤ x
When x > 0, x > in (1 + x)

Let f (x) = arctanx, G (x) = x, x > 0
f(0)=0,g(0)=0
f'(x)=1/(1+x²)＞0,g'(x)=1＞0
F '(x) - G' (x) = 1 / (1 + X & sup2;) - 1 = - X & sup2; / (1 + X & sup2;) ≤ 0, that is, f '(x) ≤ G' (x)
The function values of F (x) and G (x) at the left end point are the same. On [0, + ∞), f (x) and G (x) increase monotonically, and f '(x) ≤ G' (x), so there is arctanx ≤ x, which is equal only if x = 0
Let f (x) = x, G (x) = ln (1 + x), x > 0
f(0)=0,g(0)=0
f'(x)=1＞0,g'(x)=1/(1+x)＞0
F '(x) - G' (x) = 1-1 / (1 + x) = x / (1 + x) > 0, that is, f '(x) > G' (x)
The function values of F (x) and G (x) at the left end point are the same. On (0, + ∞), f (x) and G (x) increase monotonically and f '(x) > G' (x), so x > ln (1 + x)

Divide a piece of paper into several equal parts to form an approximate rectangle. The circumference of the rectangle is 8.28cm. What is the original area of the paper?
π is 3.14

Divide a piece of paper into several equal parts to form an approximate rectangle
The length of a rectangle is equal to half of the circumference of a circle = Πr
The width of a rectangle is equal to the radius r of a circle
(r+∏r)×2=8.28
r=1cm
The original area of this piece of paper is 3.14 * 1 & sup2; = 3.14cm & sup2;

1. If the side De of the inscribed square defg of the right triangle ABC coincides with the hypotenuse BC, then one side of the square is the middle term of the proportion of BD and EC
2. Isosceles right triangle ABC, BC as hypotenuse, D, e as AB, AC point, BD = 1 / 3AB, AE = 1 / 3aC, prove: angle ade = angle EBC
3. In the right triangle ABC, the angle BAC = 90 degrees, ad perpendicular to BC, PA = PD, BP extension line AC to g, GF perpendicular to BC to F, GF square = Ag * CG
four
In triangle ABC, De is parallel to BC, DF and EF intersect BC with G, h respectively, AF intersects de respectively, BC with Q and P respectively, proving: PG: Pb = pH: PC

Let AE = BD = 1, ad = EC = 2, EF = CF = root 2, be = 2 times root 2, then triangle ade is similar to FBE, then angle ade = angle EBC. 3. Lengthen FG, intersect Ba extension line to H. it can be proved that triangle AGH is similar to CGF, then Ag * CG = GH * GF, because PA = PD, and

Evaluation: [log6 (3)] ^ 2 + log6 (18) / log2 (6) [log6(3)]^2+log6(18)/log2(6)