It is known that the 2m-3 power - 2 = m of X is the one variable linear equation of X. try to find the 2011 power value of formula (x-3)

It is known that the 2m-3 power - 2 = m of X is the one variable linear equation of X. try to find the 2011 power value of formula (x-3)

From the fact that the 2m-3 power-2 = m of X is the one variable linear equation of X, we can know that 2m-3 = 1, M = 2
Then X-2 = m, x = 4
The 2011 power of (x-3) = the 2011 power of 1
2011 = 1 of (x-3)
Hope to adopt, thank you!
High school mathematics linear equation formula
It is known that the line L, x = 1 + T cos @ y = t sin @ (t is a parameter) is very urgent
Very simple, use the formula (Sina) ^ 2 + (COSA) ^ 2 = 1
There are two cases
(1) If t = 0, the trajectory of the point is a straight line, and the equation is x = 1;
(2) If t ≠ 0, cos @ = (x-1) / T, sin @ = Y / T, so ((x-1) / T) ^ 2 + (Y / T) ^ 2 = 1
It is reduced to: (x-1) ^ 2 + (y) ^ 2 = T ^ 2, the locus of the point is the circle center (1,0), and the radius is t
x-1=tcos@,y=tsin@
Then (x-1) / y = (TCOS @) / (Tsin @) = cot@
y=tan@x-tan@
What's the problem??
Solving equation 2x to the second power - 8x + 3 = 0 by formula method second problem: 2 (x to the second power - 1) - 1 = 4x
∵ax^2+bx+c=0
Then: x = [- B ± √ (b ^ 2-4ac)] / (2a)
And 2x ^ 2-8x + 3 = 0
∴x=[8±√(-8)^2-4*2*3]/(2*2)
=(4±√10)/2
That is: x = (4 - √ 10) / 2, x = (4 + √ 10) / 2
2 (x ^ 2-1) = 4x
2x^2-4x-1=0
∴x=[4±√(-4)^2+4*2*1)/(2*2)
=(2±√6)/2
That is: x = (2 - √ 6) / 2, x = (2 + √ 6) / 2
High school mathematics problems: 1. All the methods to solve the linear equation 2. Chord length formula 3. The formula of the distance between the straight line and the straight line 4. The correlation of the curve equation
High school mathematics problems: 1. All the methods of solving linear equations
2. Chord length formula 3. Straight line and straight line distance formula 4. How to use the correlation point method and point difference method for solving curve equation
Thank you for your advice
(1) General formula: ax + by + C = 0 (where a and B are different at the same time) (2) oblique point formula: y-y0 = K (x-x0) (3) intercept formula: X / A + Y / b = 1 (4) oblique section formula: y = KX + B (k ≠ 0) (5) two point formula: (y-y0) / (y0-y1) = (x-x0) / (x0-x1) these are more common
Calculus theorem of higher numbers
What are the concepts of the following theorems in Higher Mathematics? Not only the Chinese translation of the names, but also the translation of the contents of the theorems
ROLLE,LAGRANGE,L'HOSPITAL,TAYLOR ,MACLAURIN,FERMAT,CAUCHY
Don't tell me to find it on Baidu, because Baidu and Google only use these theorems, but don't write them out The reason is to study abroad, the theorem here is not very easy to understand, so I hope to have Chinese help to understand
Rolle's theorem: if the function f (x) is continuous in the closed interval [a, b], differentiable in the open interval (a, b), and the function values at the end of the interval are equal, that is, f (a) = f (b), then there is at least one point ξ (a) in (a, b)
namely
Rolle's theorem, Lagrange's mean value theorem, Robida's law, Taylor's formula, Maclaurin's formula, Fermat's theorem, Cauchy's mean value theorem
Specific content of their own Baidu, it is easy to find, here is not detailed
The above theorems are very important basic theorems in calculus, be sure to master them!!!
Tangent equation formula from point to circle
The second chapter of analytic geometry, a compulsory course of mathematics in senior one
Passing a point and a garden
Finding the tangent of point and circle
Do you have a formula
yes.
Let the point be p (XO, yo) and the equation of the circle be (x-a) & sup2; + (y-b) & sup2; = R & sup2;, then the tangent equation of the point and the circle is: (xo-a) * (x-a) + (yo-b) * (y-b) = R & sup2
Here's the way
First, write the center coordinates (a, b) and radius R
Judge the relation between point and circle again
If the point is on a circle, there is a tangent
There are two points outside the circle
In the slope of a straight line passing through a point, write out the oblique formula of the point, and then turn it into a standard formula
Ax+By+C=0
Then find out the distance from the center of the circle to the straight line
|Aa+Bb+C|/√(A^2+B^2)
He should be equal to the radius
That is | AA + BB + C | / √ (a ^ 2 + B ^ 2) = R
12 / 4 of COS - 12 / 4 of sin = explanation process
The fourth power 12 of COS - the fourth power 12 of sin
=(cos²π/12+sin²π/12)(cos²π/12-sin²π/12)
=1·（cos²π/12-sin²π/12）
=cos2×π/12
=cosπ/6
=√3/2
Tangent equation formula on a circle
Let the equation of circle be (x-a) &# 178; + (y-b) &# 178; = R & # 178;
There is a point (x0, Y0) on the circle
Then the tangent through this point is (x-a) (x0-a) + (y-b) (y0-b) = R & # 178;
Note that it is a point outside the circle, not a point on the circle. I'm not asking about the method of finding. Note: it's not the tangent length formula. Let the equation of circle be (x + a) ^ 2 + (y + a) ^ 2 = R ^ 2
Derivation of high calculus formula
Integral derivation process of (1 + x2) under root sign
This is the second kind of commutative integral; let x = tant; DX = sec ^ 2tdt
Then: ∫ sqrt (1 + x ^ 2) DX = ∫ sec ^ 3tdt = ∫ sectd (tant)
=sect*tant-∫sect(sec^2t-1)dt
=sect*tant-∫sec^3tdt+∫sectdt
=sect*tant+ln|sect+tant|-∫sec^3tdt
∫ sec ^ 3tdt is the same as the left side of the equal sign. If you move the term to the left, you get 2 * ∫ sec ^ 3tdt
Divide 2 to get: ∫ sec ^ 3tdt = (1 / 2) * (sect * tant) + (1 / 2) * ln|sect + tant| + C
By changing t back to x, we get the following result:
∫sqrt(1+x^2)dx＝(1/2)*x*sqrt(1+x^2)+(1/2)*ln|sqrt(1+x^2)+x|+C
There is a formula
∫sqrt(a^2+x^2)dx＝(x/2)*sqrt(a^2+x^2)+(a^2/2)ln|sqrt(a^2+x^2)+x|+C
What is the tangent equation formula?
There are two points and a circle. How to find the equation of tangent?
What is the formula of tangent? What is it like?
Let the point be x (a, b) and the linear equation passing through point X be y-b = K (x-a), but the premise is that K exists. First, we discuss whether the line is tangent to the circle when K does not exist, then we establish the equation of the line and the circle, and get the quadratic equation. The discriminant of the quadratic equation is equal to 0. There is another way to solve K