a, Theo bài ra,ta có:
f(x) = ax+b
⇒ $\left \{ {{f(2)=2a+b} \atop {f(-1)=-a+b}} \right.$
Mà f(2)=3, f(-1)=9 (Bài cho)
⇒ $\left \{ {{2a+b=3} \atop {-a+b=9}} \right.$
⇒ $\left \{ {{2a+b=3} \atop {2b-2a=18}} \right.$
⇒ 2a+b+2b-2a=3+18
⇒3b=21
⇒b=7
Mà -a+b=9 (theo trên)
⇒7-a=9
⇒ a=-2
Vậy a=-2, b=7
b, Theo bài ra, ta có:
N=$\frac{9}{\sqrt{x}-5}$ ( x$\neq$ 25, x∈N)
Để N có giá trị nguyên với x$\neq$ 25, x∈N thì $\frac{9}{\sqrt{x}-5}$ có giá trị nguyên
Hay 9 chia hết cho $\sqrt{x}-5$
⇒ $\sqrt{x}-5$ ∈{1;3;9;-1;-3;-9}
Vì √x≥0∀x nên $\sqrt{x}-5$ ≥-5∀x
Do đó: $\sqrt{x}-5$ ∈{1;3;9;-1;-3}
⇒ √x ∈{6;8;14;4;2}
⇒ x ∈{36;64;196;16;4} ( Thỏa mã điều kiện)
Vậy x ∈{36;64;196;16;4} thì N có giá trị nguyên
c, Theo bài ra, ta có:
P(x)= 100$x^{100}$+99$x^{99}$+98$x^{98}$+...+2x²+x+1
⇒ P(1)= 100+99+98+...+2+1+1
P(1)= (100+1).100÷2 +1
P(1)= 5050+1
P(1)= 5051
Vậy P(1)= 5051
d, Theo bài ra, ta có:
f(x)= $x^{21}$-2020$x^{20}$+2020$x^{19}$-2020$x^{18}$+...+2020x³-2020x²+2020x
⇒ f(2019)= $2019^{21}$-2020.$2019^{20}$+2020.$2019^{19}$-2020.$2019^{18}$+...+
+2020.2019³-2020.2019²+2020.2019
⇒ f(2019)= $2019^{21}$-2019.$2019^{20}$-$2019^{20}$+2019.$2019^{19}$+$2019^{19}$-
-2019.$2019^{18}$+$2019^{18}$+...+2019.2019³+2019³-2019.2019²-2019²+2019.2019+2019
⇒ f(2019)= $2019^{21}$-$2019^{21}$-$2019^{20}$+$2019^{20}$+$2019^{19}$-$2019^{19}$+
+$2019^{18}$+...+$2019^{4}$+2019³-2019³-2019²+2019²+2019
⇒ f(2019)=2019
Vậy f(2019)=2019
