When a is a value, the intercept of the two axes of the line (A-1) x + (3-A) y + a = 0 is equal

When a is a value, the intercept of the two axes of the line (A-1) x + (3-A) y + a = 0 is equal

（a-1）x+(3-a)y+a=0
Equal intercept: - A / (A-1) = - A / (3-A),
The solution is a = 2

Calculation of 17.5 + 6.35 + 42.5 + 3.65 * 6 + 60 by simple method

17.5+6.35+42.5+3.65*6+60
=17.5+42.5+60+6.35+3.65×6
=120+6.35+21.9
=148.25

Given the function y = 3x-6, if its image intersects the x-axis with point a, and its image intersects the y-axis with point B, then the area of the triangle AOB is?

If the image intersects with the X axis and the point a, then y = 0 and x = 2, then the distance from the origin to a = 2,
Intersection with y axis and point B, let x = 0, find y = - 6, then the distance from the origin to B = 6
Then the area of triangle AOB = 2 × 6 △ 2 = 6

The function f (x) = 13x3 + x2 − 2 is known. (I) let {an} be a sequence of positive numbers, and the sum of the first n terms is Sn, where a1 = 3. If the point (an, an + 12-2an + 1) (n ∈ n *) is on the image of function y = f ′ (x), prove that the point (n, Sn) is also on the image of function y = f ′ (x); (II) find the extremum of function f (x) in the interval (A-1, a)

(I) prove: because f (x) = 13x3 + x2 − 2, so f ′ (x) = x2 + 2x, from the point (an, an + 12-2an + 1) (n ∈ n +) on the image of function y = f ′ (x), and an ＞ 0 (n ∈ n +), so (an-1-an) (an + 1-an-2) = 0, so Sn = 3N + n (n − 1) 2 × 2 = N2 + 2n, and because f

The third power of (X-Y) × (X-Y) × (X-Y), the square of (X-Y), is written in the form of the nth power of (X-Y)

(x-y)×(x-y)^3×(x-y）^2
=(x-y)^6

If x ∈ [1,2], let x2 + 2x + a ≥ 0 "be true proposition, then the value range of real number a is
Why should we use the no proposition of the original proposition to transform it into the problem of finding the maximum? Another analysis is that if f (2) ≥ 0, then why is it not that if f (1) ≥ 0, then the minimum value can be established

F (x) = - X & # 178; + 2x + A is a parabola with an opening downward, and the axis of symmetry is x = 1;
Because the interval [1,2] is on the right side of the axis of symmetry,
Therefore, when x ∈ [1,2], f (1) is the maximum and f (2) is the minimum;
If x ∈ [1,2], f (x) ≥ 0 is constant, as long as the minimum f (2) ≥ 0;
The results show that f (1) = 8 + a ≥ 0, and the solution is a ≥ - 8;
That is: the value range of real number a is [- 8, + ∞)

-How much smaller is the sum of 4, - 9, + 7 than the sum of their absolute values?

-4-9 + 7 = - 6 | - 4 | + | - 9 | + | + 7 | = 20 20 - (- 6) = 26 small 26

The function f (x) = x ^ 3 + SiNx + 1 (x ∈ R), if f (a) = 2, what is the value of F (- a)

First, a is substituted into the original formula, the equation about a is derived, and then - A is brought in
F (a) = a ^ 3 + Sina + 1 = 2, so a ^ 3 + Sina = 1
f（-a）=（-a）^3+sin（-a）+1
=-（a^3）+（-sina）+1
=-（a^3+sina）+1
Because a ^ 3 + Sina = 1
So f (- a) = - 1 + 1 = 0

If AB = 20cm, MB = 8cm, OM = 10cm, find the radius of circle 0

If the perpendicular of AB through O is C, then
BC=1/2AB=10
∵ MB=8CM
∴MC=BC-BM=2
∵ OM=10
∴OC=√(10^-2^2 )
=√96
∴OB =√(BC^2+OC^2)
=√(10^2+√96^2)
=√196
=14
A: the radius of circle 0 is 14cm

Given that a ^ 2 + B ^ 2 = 2, (a + C) · (B + D) = 9, ABCD are non negative real numbers, find the min of C ^ 2 + D ^ 2

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