The value of polynomial f (x) = 7x7 + 6x6 + 5x5 + 4x4 + 3x3 + 2x2 + X is obtained by Qin Jiushao's algorithm when x = 2

The value of polynomial f (x) = 7x7 + 6x6 + 5x5 + 4x4 + 3x3 + 2x2 + X is obtained by Qin Jiushao's algorithm when x = 2

f（x）=（（7x+6）+5）x+4）x+3）x+2）x+1）x
V0=7，
V1=7×2+6=20，
V2=20×2+5=45，
V3=45×2+4=94，
V4=94×2+3=191，
V5=191×1+2=384，
V6=384×2+1=769，
V7=769×2=1538，
∴f（2）=1538
That is, when x = 2, the function value is 1538

The Qin Jiushao algorithm is used to calculate the function f (x) = 2x4 + 3x3 + 5x-4. When x = 2, the multiplication is performed () times A. 4 B. 5 C. 6 D. 7

When using Qin Jiushao algorithm to calculate the value of polynomials,
The degree of multiplication is the same as the exponent of the highest degree term of the unknown number of polynomials,
 a total of four multiplication operations were performed,
Therefore, a

Using Qin Jiushao algorithm to calculate the function f (x) = 2x4 + 3x3 + 5x-4, when x = 2, the function value is______ .

Qin Jiushao's algorithm is as follows: F (x) = 2x4 + 3x3 + 5x-4 = x (2x3 + 3x2 + 5) - 4 = x [x (2x2 + 3x) + 5] - 4 = x {x [x (2x + 3)] + 5} - 4
When x = 2, f (x) = 2 × {2 × [2 × (2 × 2 + 3)] + 5} - 4 = 62
So the answer is: 62

The value of function f (x) + 2x ^ 4 + 3x ^ 3 + 5x-4 at x = 3 is calculated by Qin Jiushao algorithm

Let t = x square = 9, then the original formula = 2T square + 3Tx + 5x-4, and bring t = 9, x = 3 to get the original formula = 173

( X-2 x) In the expansion of 6, the constant term is______ .

The general term of the expansion is tr + 1 = (- 2) rc6rx3-r
Let 3-R = 0 and R = 3
So the constant term of the expansion is (- 2) 3c63 = - 160
So the answer is: - 160

-The constant term of the Square-1 / 2x-1 of X is

If t = 2x-1, then x = (T + 1) / 2
Then the original formula = - (T ^ 2 + 2T + 5) / 4T
The constant term is - 1 / 2

The general form, the coefficient of quadratic term, the coefficient of first order and the term of constant are taken as the square of 1 / 3 (x + 1) = 2x - 1 / 5 2 (x-1) square-7 = (x + 2) (X-2) into the general form? 4x square + 3x-5 = - 2x + 7 is the coefficient of the first term?

(x + 1) 2 / 3 = 1 / (2x-5) (x? + 2x + 1) (2x-5) = 3 general form: 2x? - x? - 8x-8 = 0, coefficient of quadratic term: - 1 coefficient of primary term: - 8 constant term: - 82 (x-1) 2? - 7 = (x + 2) (X-2) 2x? - 4x + 2-7 = x? - 4 general form: x? - 4x-1 = 04x? + 3x

The square of the polynomial - x + 2x-3 is_______ Times_______ Term formula_______ The constant term is________

The square of polynomial - x + 2x - 3 is a quadratic trinomial, and the constant term is - 3

If the maximum value of the function f (x) = root 3 * 2x + 2cos square + m in the interval [0, π / 2] is 6, the value of constant M and the minimum value of this function when x belongs to R are obtained, and the corresponding set of values of X is obtained Thank you ~ ~ that root and square can't play,

If the maximum value of the function f (x) = root 3 * 2x + 2cos square + m in the interval [0, π / 2] is 6, the value of constant M and the minimum value of this function when x belongs to R are obtained, and the corresponding set of values of X is obtained
Analysis: ∵ function f (x) = 2 √ 3x + 2 (cosx) ^ 2 + M
F‘(x)=2√3-4cosxsinx=2√3-2sin2x>0
The function f (x) increases monotonically in the domain of definition
The maximum value of ∵ f (x) in the interval [0, π / 2] is 6
∴f(π/2)=√3π+m＝6==>m=6-√3π
∴f(x)=2√3x+2(cosx)^2+6-√3π
The function has no minimum value when x belongs to R

If the value of (ax squared-2x + 6) + [- 3x squared + (a + b) x + 1] is a constant, then a-b=______

a=3 b=-1 a-b=4