There are five story books and nine science and technology books mixed in the bookcase. Make sure you can take out at least one science and technology book and at least how many books

Science and technology books account for 9 / 14 of all the books. If the first five books are all storybooks, the science and technology book will take 5 + 1 = 6 (Times), so at least 6 books will be taken. But if the first book is science and technology book, it will only take 1 time, but to ensure the science and technology book, we must prepare for the worst, so 6 times

After the formula of equation x2-3x + P = 0, we get (x + m) 2 = 12. (1) find the value of constant P and m; (2) find the root of the equation

(1) ∵ x2-3x + P = 0, ∵ x2-3x = - P, x2-3x + (32) 2 = - P + (32) 2, (x-32) 2 = - P + 94, ∵ M = - 32, - P + 94 = 12, the solution is: P = 74, M = - 32; (2) ∵ x2-3x + P = 0, ∵ x-32) 2 = 12, x-32 = ± 22, that is, the solution of the equation is: X1 = 3 + 22, X2 = 3 − 22

As shown in the figure, the lines FB, ad and CE are parallel to each other, the area of △ ABC is 1, and the area of △ DEF is calculated

Because FB, ad and CE are parallel to each other
So s △ ADC = s △ ade
S△ADB＝S△ADF
Because s △ ABC = s △ ADC + s △ ADB = 1
So s △ ade + s △ ADF = 1
Because FB is parallel to CE
So s △ FBC = s △ FBE
So s △ fbc-s △ FBA = s △ fbe-s △ FBA
That is s △ FEA = s △ CBA = 1
S △ def = s △ ade + s △ ADF + s △ AEF = 2
The reason why this problem is confusing is that EF seems to be parallel to BC in the figure given in the problem. As long as you draw a new picture, it's easy to make it clear
By the way, are you two in class 10?

Let the equation of parabola C be x2 = 4Y, m (x0, Y0) be any point on the straight line L: y = - M (M > 0), and make two tangent Ma, MB and tangent points of parabola C through point M
When m changes, try to explore whether there is a point m on the line L, so that Ma ⊥ MB? If there are, there are several such stores. If not, explain the reason

Tangent: y-y0 = K (x-x0)
C：x²=4y
Lianlide: X & # 178; = 4K (x-x0) + 4y0
x²-4kx+4x0k-4y0=0
Tangent condition: Δ = 0
Δ=(4k)²-4(4x0k-4y0)=16k²-16x0k+16y0=0
k²-x0k+y0=0
Combined with Ma ⊥ MB
k1·k2=y0=-1
Then - 1 = - m, M = 1 > 0, and
Δ‘=x0²-4y0=x0²+4＞0
Therefore, when m = 1, any point on L satisfies the requirement of problem setting, that is, there are innumerable such points

As shown in the figure, in square ABCD, M is the midpoint of CD, e is a point on CD, and ∠ BAE = 2 ∠ dam

Prove: as shown in the figure, extend AB to F, make BF = CE, connect EF and BC, intersect at point n, in △ BFN and △ Cen, ∠ FBN = ∠ C = 90 °∠ BNF = ∠ cnebf = CE, ≌ BFN ≌ cen (AAS), ≌ BN = CN, EN = FN, and ∫ m is the midpoint of CD, ∫ ban = ∠ dam, ≌ BAE = 2 ≌ dam, ≌ ban =} Ba =}

If the quadratic equation 2x (kx-4) - x2 + 6 = 0 with respect to X has no real root, then the minimum integer value of K is______ ．

The equation is reduced to a general form: (2k-1) x2-8x + 6 = 0, ∵ the original equation is a quadratic equation of one variable and has no real root, ∵ 2k-1 ≠ 0 and △ 0, that is, △ = (- 8) 2-4 × (2k-1) × 6 = 88-48k ＜ 0, the solution is k ＞ 116. So the value range of K is k ＞ 116. Then the minimum integer value of K satisfying the condition is 2. So the answer is 2

Given that a (- 2,1) B (- 1,3) on the plane, the vector OC is the position vector corresponding to the vector AB, then the coordinates of point C

Because OC is the position vector corresponding to vector AB, that is to say, the vector OC is the unit vector parallel to vector ab
And because AB = (1,2), the vector OC is equal to ab divided by the module of ab. OC = (five fifths root sign five, two fifths root sign five)
So the coordinates of point C are (two fifths root, two fifths root, five)

If the remainder 2x-5 is obtained by dividing x ^ 2-5x + 6 by the polynomial f (x), then f (3)=__
The greatest common factor of polynomial x ^ 4 + 2x ^ 3-4x ^ 2-2x + 3 and x ^ 3 + 4x ^ 2 + X-6 is_
If we take the first polynomial f (x) and G (x) of 2x ^ 2-3x-2 as the initial polynomial, we can get the remainder formulas 2x + 3 and 4x-1 respectively. Then we can get the remainder formula by dividing 2x + 1 by F (x) - G (x)____

It should be noted that the meaning of "divide by" and "divide by" is completely different. It must be that the questioner is confused. Otherwise, the input is wrong. In fact, as long as you ask the pupils, you will know the difference between the two
For problems 1 and 3, the original problem is unsolved, or the result depends on the selection of F (x) and G (x)
Now the answer to the revised question is given
1. If x ^ 2-5x + 6 divides the polynomial f (x) into the remainder 2x-5, then f (3)=____ .
2. The greatest common factor of polynomials x ^ 4 + 2x ^ 3-4x ^ 2-2x + 3 and x ^ 3 + 4x ^ 2 + X-6 is___ .
3. If 2x ^ 2-3x-2 divides the polynomials f (x) and G (x) respectively to obtain the complements 2x + 3 and 4x-1, then the complements of 2x + 1 divided by F (x) - G (x) are____ .
1. Let Q (x) be the quotient of F (x) divided by x ^ 2-5x + 6, then
F(x)=(x^2-5x+6)*q(x)+(2x-5),
Where x = 3, f (3) = 0 * q (3) + 1 = 1
2. Division by rotation is similar to division by rotation of integers
The remainder formula of x ^ 4 + 2x ^ 3-4x ^ 2-2x + 3 divided by x ^ 3 + 4x ^ 2 + X-6 is 3x ^ 2 + 6x-9; the coefficient of the first term is changed to 1 to get x ^ 2 + 2x-3;
(it can be easily calculated by using the separation coefficient method, and it can also be calculated by using the long division method. To be simpler, we can use the short division comprehensive division method.)
The remainder of x ^ 3 + 4x ^ 2 + X-6 divided by x ^ 2 + 2x-3 is 0; (the quotient is (x + 2))
So the greatest common factor of x ^ 4 + 2x ^ 3-4x ^ 2-2x + 3 and x ^ 3 + 4x ^ 2 + X-6 is x ^ 2 + 2x-3
3. If we know, we can set
F(x)=(2x^2-3x-2)*q1(x)+(2x+3),
G(x)=(2x^2-3x-2)*q2(x)+(4x-1),
be
F(x)-G(x)=(2x^2-3x-2)*( q1(x)-q2(x) )+(-2x+4)
=(2x+1)*(x-2)*( q1(x)-q2(x) )+(-2x+4),
F (x) - G (x) = (2x + 1) * (X-2) * (Q1 (x) - Q2 (x)) + (- 2x + 4)
In the above formula, let x = - 1 / 2, we get:
F(-1/2)-G(-1/2)=0+5=5,
According to the remainder theorem, the remainder except f (x) - G (x) divided by (2x + 1) is f (- 1 / 2) - G (- 1 / 2) = 5

In the cube abcd-a'b'c'd ', m and N are the midpoint of AA' and BB 'respectively, and the cosine value of the angle formed by cm and d'n
The steps of the party's pursuit in Senior High School

Let Q be the midpoint of DD ',
In the cube abcd-a'b'c'd ', m and N are the midpoint of AA' and BB 'respectively
Then Bq / / d'n, QM / / BC
Then mbcq is a rectangle
The edge length in the cube abcd-a'b'c'd'is 2,
Then MQ = BC = 2, MB = CQ = radical 5
CM=QN=3
Let cm and QN intersect at point o,
Then om = OQ = 1.5
Cosine value of angle formed by cm and QN = (1.5 ^ 2 + 1.5 ^ 2-2 ^ 2) / (2 * 1.5 * 1.5) = 1 / 9
The cosine of the angle between cm and d'n = 1 / 9

The square of a minus 2Ab plus the square of B minus the square of C

Solution
Factorization
A square - 2Ab + b square - C square
=(a-b) square-c square
=[(a-b)-c][(a-b)+c]
=(a-b-c)(a-b+c)

There are five story books and nine science and technology books mixed in the bookcase. Make sure you can take out at least one science and technology book and at least how many books