1、 6x²-7x+1=0 2、 5x²-18=9x 3、 4x²-3x=52 4、 5x²=4-2x

1、 6x²-7x+1=0 2、 5x²-18=9x 3、 4x²-3x=52 4、 5x²=4-2x

1.(6x-1)(x-1)=0
X = 1 / 6 or 1
2.(5X-6)(X+3)=0
X = 6 / 5 or - 3
3.(4x+13)(x-4)=0
X = - 13 / 4 or 4
4. Use the root formula to get x = (- 1 + √ 21) / 5 or (- 1 - √ 21) / 5

One third (X-6) = one half to one fifth (x + 2) to solve the equation,

One third (X-6) = one half - one fifth (x + 2)
Multiply 30 to get 10 (X-6) = 15-6 (x + 2)
10x-60=15-6x-12
10x+6x=15-12+60
16x=63
x=63/16

（1） x²+x-6=0 （2）x²-√3x-1/4=0 （3）3x²-6-2=0
（4）4x²-6x=0 （5）x²+4x+8=4x+11 （6）x（2x-4x）=5-8x

(1) (x + 3) (X-2) = 0, ｜ x = - 3 or x = 2
(2) (3) calculate with the root formula
(4) 2X (2x-3) = 0, ｜ x = 0 or x = 3 / 2
(5) X & # 178; = 3. X = ± radical 3
(6) I don't have a book. Do you have the wrong question

27:2 / 3 108:96 simplify the process of comparison

2 of 27:3 = 27 * 3:2 = 81:2
108：96 =108/12:96/12=9:8

Given that α, β ∈ (π / 3,5 π / 6), if sin (α + π / 6) = 4 / 5, cos (β - 5 π / 6) = 5 / 13, then sin (α - β) =?

∵α,β∈（π/3,5π/6)
==>π/2

The area ratio of the inscribed circle to the circumscribed circle of an equilateral triangle is______ ．

The radius of the inscribed circle and circumscribed circle of an equilateral triangle is the edge center distance and radius of the equilateral triangle, and the ratio of the edge center distance and radius of the equilateral triangle is 1:2, so the ratio of the area is 1:4

Y = 3 ^ x + 3 ^ (- x) (x = / 0)

y=-x^2+2x
y=-(x-1)^2+1
When x = 1, y has a maximum value of 1
X & lt; 1 is a decreasing function, so in the range of 0 & lt; = x & lt; = 1, when x = 0, y is the minimum of 0
In the range of 3 & gt; = x & gt; 1, when x = 3, y is the smallest, which is - 3
So the range of Y is - 3 & lt; = y & lt; = 1

How to understand and use first order differential form invariance?
Solving dy with ysinx cos (X-Y) = 0
According to the invariance of the first-order differential form, the
D (ysinx) - D (COS (X-Y)) = 0. How do we get this?
I searched and still didn't understand
Let y = f (U), u = g (x). If u = g (x) is differentiable to x, and y = f (U) is differentiable to corresponding u, then y = f [g (x)] is differentiable to x, Dy = f [g (x)] 'DX = f' (U) g '(x) DX = f' (U) Du. We can know that dy = f '(U) Du remains unchanged no matter u is the differentiable function of independent variable or other independent variables. This is the form invariance of first order total differential
If x and y are not independent variables, we can still use the original differential formula when x and y are independent variables

It's explanation 1!
That is, compound function, differential, first for the outer function, in the inner function
Both the inner and outer layers are f '(U) Du = f' (x) DX
That is, the derivative is in the form of integral element D

For a rectangle, increase its length and width by 6 meters, increase its area by 1236 square meters, and find the original perimeter?

Method 1, using the primary school method: after the length and width of the rectangle are increased by 6 meters, two rectangles with the width of 6 meters are added, the length is the length and width of the original rectangle respectively; and a square with the side length of 6 meters. The total area of the three parts is 1236 square meters, so the sum of the length and width of the original rectangle is (1236-6 * 6) / 6 = 200 meters, so

Cube-b of the square of (1 + AB) - (a + b) (where a = - 56, B = 125)

(b3-b)/[(1+ab)2-(a+b)2]
=b(b2-1)/[a2(b2-1)-(b2-1)]
=b(b2-1)/[(a2-1)·(b2-1)]
=b/(a2-1)
=25/627
≈0.0399