-10. The second power of 3x - the third power of 5x, the power of 7x, the power of 4-9x. 5. Find the n power of X

-10. The second power of 3x - the third power of 5x, the power of 7x, the power of 4-9x. 5. Find the n power of X

Law: odd number is negative, even number is positive
The coefficients are all odd numbers, and the relationship with the number of terms is 2N-1
The power of X is equal to the number of terms
So the n power of X is (- 1) ^ n (2n-1) x ^ n

It is known that the 2x-y = 10 [(X-Y + y 2) - (X-Y) 2 + 2Y (X-Y)] / 4Y process

【（x²+y²）-（x-y)²+2y（x-y）】÷4y
=（x²+y²-x²+2xy-y²+2xy-2y²）÷4y
=（4xy-2y²）÷4y
=x-y/2
=1/2(2x-y)
=10/2
=5

Given that 2x-y = 10, how does the formula [(x ^ 2 + y ^ 2) - (X-Y) ^ 2 + 2Y (X-Y)] / 4Y (4xy-2y?) / 4Y becomes = 2Y (2x-y) △ 4Y

It is known that 2x-y = 10,
The formula [(x ^ 2 + y ^ 2) - (X-Y) ^ 2 + 2Y (X-Y)] / 4Y
=[x^2+y^2-x^2+2xy-y^2+2xy-2y^2]/4y
=[4xy-2y^2]/4y
=2y(2x-y)/4y
=2y*10/4y
=5

Given that 2x + 3y-4z = 0, 3x + 4Y + 5Z = 0, find the value of X + y + Z / (divided by) X-Y + Z. because I just started to learn ternary once, I have a lot of problems,

From formula 2-1:
X+Y=-9Z…… Type 3
1, x2-2
X+2Y=13Z…… Type 4
From formula 4-3:
When y = 22z is substituted into formula 3, we can get that:
X = - 31z, then there is:
X+Y+Z/X-Y+Z=-31Z+22Z+Z/-31Z-22Z+Z=8/-52=2/-13

If the value of 3x? - 4x + 1 is 0, then the value of 2x - three thirds of X? - 3 is

(3x + 4) (x-1) = 0, x = 1 or - 4 / 3. So the value is - 2, - 11 / 9

If the value of 2X-4 is 3, then the value of 4x? - 16x + 16 is——

4x²-16x+16
=（2x—4）²
∵2x—4=3
The original formula = (2X-4) 2 = 3? = 9

Given the quadratic power of X + X-1 = 0, find the value of the cubic power of x-2x + 2010

x²+x-1=0
x²=-x+1
So x 3 = x x x 2
=x(-x+1)
=-x²+x
=-(-x+1)+x
=2x-1
Original formula = (2x-1) -2x+2010
=2x-1-2x+2010
=2009

Given sin (x + π / 6) = 1 / 3, find sin (5 π / 6-x) + sin ^ 2 (π / 3-x)

sin(x+π/6)=1/3
sin(5π/6-x)=sin[π-(x+π/6)]=1/3
sin^2(π/3-x)=sin^2[π/2-(x+π/6)]=cos^2(x+π/6)=1-sin^2(x+π/6)=8/9
sin(5π/6-x)+sin^2(π/3-x)=1/3+8/9=11/9

Given sin (x + π / 6) = 1 / 4, find the value of sin (7 π / 6 + x) + cos (11 π / 6-x)

Sin (7 Pai / 6 x) = 1 / 4, cos (11 Pai / 6-x) = (15 under radical sign) / 4, the sum of these two formulas is equal to (1 radical 15) / 4

Then sin + 6 is the root of sin + 6?

cos(x-π/6)+sinx=4√3/5
√3/2cosx+3/2sinx=4√3/5
1/2cosx+√3/2sinx=4/5
sin(x+π/6)=4/5
sin(x+7π/6)=sin(x+π/6+π)=-sin(x+π/6)=-4/5