Question

a) A(x) = x¹⁷ - 12x¹⁶ + 12x¹⁵ - 12x¹⁶ +...+ 12x - 1 tại x = 11

b) Cho x - y = 5 *Tính B = 6x - 6y + 10 - 3ax + 3ay + 15a

c) Cho x - 2y = 6 *Tính C = \(\frac{x-y}{x+6}\) x−yx+6 (x ≠y và x≠-6)

d) Cho x,y,z ≠0 và x - y - z = 0 . Tính GT của BT sau:

D = (1 - \(\frac{z}{x}\) ) . ( 1 - \(\frac{y}{x}\)) . (1 + \(\frac{y}{z}\) )

Answer 1

a) Với x = 11 <=> 12 = x+1

\(A\left(x\right)=x^{17}-\left(x+1\right)x^{16}+\left(x+1\right)x^{15}-...+12x-1\)

\(A\left(x\right)=12x-11=12.11-1=120\)

b) \(B=6x-6y+10-3ax+3ay+15a\)

\(B=6\left(x-y\right)+10-3a\left(x-y\right)+15a\)

\(B=6.5+10-3.a.5+15a\)

\(B=40\)

c)\(C=\frac{x-y}{x+6}=\frac{x-y}{x+x-2y}=\frac{x-y}{2\left(x-y\right)}=\frac{1}{2}\left(x-2y=6\right)\)

\(C=\frac{2x+6}{3x-2y}+\frac{2y-6}{4y-x}\)

\(C=\frac{2x+1-2y}{3x-2y}+\frac{2y-x+2y}{4y-x}\)

\(C=1+1=2\)

d) ta có : x-y-x = 0

\(\Rightarrow\left\{{}\begin{matrix}x-z=y\\x-y=z\\x=y+z\end{matrix}\right.\).Thay vào B, ta có :

\(B=\frac{x-z}{x}.\frac{y-x}{y}.\frac{z+y}{z}\)

\(B=\frac{y}{x}.\frac{\left(-z\right)}{y}.\frac{x}{z}\)

B= -1