Factorization ab-a-b-1 Factorization in the range of rational numbers

Factorization ab-a-b-1 Factorization in the range of rational numbers

The original topic cannot be decomposed into rational number range
If the title is ab-a-b + 1 = (A-1) (B-1)
If the title is ab + a-b-1 = (A-1) (B + 1)
If the title is AB-A + B-1 - (a + 1) (B-1)

Known: a = 10000, B = 9999, find the value of A2 + b2-2ab-6a + 6B + 9

∵ a = 10000, B = 9999, ∵ A-B = 10000-9999 = 1, then the original formula = (a-b) 2-6 (a-b) + 9 = 1-6 + 9 = 4

A is nxm matrix, B is mxn matrix, where n

It is known that R (AB) = R (E) = n
Because R (AB)

In the triangle ABC, the angle ACB is equal to 90 degrees, CD is perpendicular to AB, BF bisector angle ABC intersects CD with E, AC with F. find CE equal to CF

The problem is transformed into a proof: ∠ CFE = ∠ CEF ∠ CFE = ∠ cab + ∠ FBA (the outer angle of the triangle is equal to the sum of the two inner angles that don't want to be adjacent) ∠ CEF = ∠ DCB + ∠ CBF (as above) because BF is bisecting ∠ ABC, so ∠ CBF = ∠ FBA has ∠ cab + ∠ ACD = ∠ ACD + ∠ DCB = 90 ° so ∠ cab = ∠ DCB

The known function f (x) = a (x-1 / x) - 2lnx a belongs to R
Finding monotone interval of function f (x)

Given that the function f (x) = a (x-1 / x) - 2lnx a belongs to R, find the monotone interval of function f (x)
Analysis: ∵ function f (x) = a (x-1 / x) - 2lnx a belongs to R, and its definition field is x > 0
∴f’(x)=a(1+1/x^2)-2/x=[a(1+x^2)-2x]/x^2
Let a (1 + x ^ 2) - 2x > 0 = = > A = 2x / (1 + x ^ 2)
a(1+x^2)-2x=ax^2-2x+a
⊿=4-4a^2>=0==>-1

In △ abd, if OA · ob = ob · OC = OC · OA, then o is △ ABC's (a outer center, B vertical center)
If O is a point in the plane of △ ABC and satisfies (vector ob vector OC) · (vector ob + vector oc-2, vector OA) = 0, then the shape of △ ABC is ()

The first question is, if you make a difference between two, AC · ob = 0, then AC vertical ob, the others are the same
Second, BC · (AB + AC) = 0, which means AB + AC is perpendicular to BC, so there is ab = AC, isosceles triangle

As shown in the figure, in △ ABC, ab = AC, CD is the height on the edge of AB, and the proof is: ∠ BCD = 12 ∠ a

It is proved that a is AE ⊥ BC in E, Cd in F, and ∫ BAE + ⊥ B = 90 ° and ab = AC, ∫ BAE = 12 ⊥ BAC. And ∫ CD ⊥ AB, ∫ BCD + ⊥ B = 90 ° and ∫ BAE = ∫ BCD. ∫ BCD = 12 ∫ a

x. 2X, 4x are sequences? Where x is unknown!

It's a sequence. There's a certain rule that it's a sequence. It's an equal ratio sequence. A2 / A1 = A3 / A2 = 2. 2 is a constant, so it's a sequence

A (3,5,2), B (- 1,2,1) translate AB vector according to vector (2,3,6). What is the vector coordinate after translation?

AB=(-4,-3,-1)
No matter what kind of vector AB is translated, its coordinates will not change after translation
Because translation doesn't change direction or size
The vector coordinates after translation are still (- 4, - 3, - 1)

It is known that, as shown in the figure, △ ABC, point E is on the central line BD, ∠ DAE = ∠ abd

It is proved that: (1) ∵ - DAE = - abd, ∵ - ADE = - BDA, ∵ △ ade ∵ BDA. (2 points) ∵ - addd = dead, (2 points) & nbsp; that is, ad2 = de · dB. (1 point) (2) ∵ D is the midpoint on the AC side, ∵ ad = DC. ∵ addd = dead, ∵ dcbd = dedc, (2 points) ∵ - CDE = - BDC. (1 point)