a: Ta có: x=31

nên x-1=30

Ta có: \(A=x^3-30x^2-31x+1\)

\(=x^3-x^2\left(x-1\right)-x^2+1\)

\(=x^3-x^3+x^2-x^2+1\)

=1

c: Ta có: x=16

nên x+1=17

Ta có: \(C=x^4-17x^3+17x^2-17x+20\)

\(=x^4-x^3\left(x+1\right)+x^2\left(x+1\right)-x\left(x+1\right)+20\)

\(=x^4-x^4-x^3+x^3+x^2-x^2-x+20\)

\(=20-x=4\)

d: Ta có: x=12

nên x+1=13

Ta có: \(D=x^{10}-13x^9+13x^8-13x^7+...+13x^2-13x+10\)

\(=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-x^7\left(x+1\right)+...+x^2\left(x+1\right)-x\left(x+1\right)+10\)

\(=10-x\)

=-2

d: Ta có: x=12

nên x+1=13

Ta có: \(D=x^{10}-13x^9+13x^8-13x^7+...+13x^2-13x+10\)

\(=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-x^7\left(x+1\right)+...+x^2\left(x+1\right)-x\left(x+1\right)+10\)

\(=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^3+x^2-x^2-x+1+9\)

\(=-x+10=-2\)

Ta có

P = x 10   –   13 x 9   +   13 x 8   –   13 x 7   +   …   -   13 x   +   10

=   x 10   –   12 x 9   –   x 9   +   12 x 8   +   x 8   –   12 x 7   –   x 7   +   12 x 6   +   …   + x 2   –   12 x   –   x   +   10     =   x 9 ( x   –   12 )   –   x 8 ( x   –   12 )   +   x 7 ( x   –   12 )   -   …   +   x ( x   –   12 )   –   x   +   10

Thay x = 12 vào P ta được

P   =   12 9 . ( 12   –   12 )   –   12 8 ( 12   –   12 )   +   12 7 ( 12   –   12 )             -   …   +   12 ( 12   –   12 )   –   12   +   10

= 0 + … + 0 – 2 = -2

Vậy P = -2

Đáp án cần chọn là: A

S(x) = x^9(x - 12) -x^8(x - 12) + x^7(x - 12) + . . . +x(x-12) - (x - 12) - 2

Suy ra:    S(x) = -2 

x = 79 => 80 = x + 1 thay vòa Px ta có 

P(x) = x^7 - ( x + 1) x^6 + .... + (x + 1) x + 15

         = x^7 - x^7- x^6 + ... + x^2 + x  +15

          = x + 15

          = 79 + 15 

           = 94

Ý B tương tự

\(Q_{\left(x\right)}=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)

\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+..+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)

\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)

\(=1\)

\(a.P(x)=x^7-80x^6+80x^5-80x^4+....+80x+15\)

\(=x^7-79x^6-x^6+79x^5+x^5-79x^4-....-x^2+79x+x+15\)

\(=x^6(x-79)-x^5(x-79)+x^4(x-79)-....-x(x-79)+x+15\)

\(=(x-79)(x^6-x^5+x^4-....-x)+x+15\)

Thay x = 79 vào biểu thức trên , ta có

\(P(79)=(79-79)(79^6-79^5+79^4-...-79)+79+15\)

\(=0+79+15\)

\(=94\)

Vậy \(P(x)=94\)khi x = 79

\(b.Q(x)=x^{14}-10x^{13}+10x^{12}-.....+10x^2-10x+10\)

\(=x^{14}-9x^{13}-x^{13}+9x^{12}+.....-x^3+9x^2+x^2-9x-x+10\)

\(=x^{13}(x-9)-x^{12}(x-9)+.....-x^2(x-9)+x(x-9)-x+10\)

\(=(x-9)(x^{13}-x^{12}+.....-x^2+x)-x+10\)

Thay x = 9 vào biểu thức trên , ta có

\(Q(9)=(9-9)(9^{13}-9^{12}+.....-9^2+9)-9+10\)

\(=0-9+10\)

\(=1\)

Vậy \(Q(x)=1\)khi x = 9

\(c.R(x)=x^4-17x^3+17x^2-17x+20\)

\(=x^4-16x^3-x^3+16x^2+x^2-16x-x+20\)

\(=x^3(x-16)-x^2(x-16)+x(x-16)-x+20\)

\(=(x-16)(x^3-x^2+x)-x+20\)

Thay x = 16 vào biểu thức trên , ta có

\(R(16)=(16-16)(16^3-16^2+16)-16+20\)

\(=0-16+20\)

\(=4\)

Vậy \(R(x)=4\)khi x = 16

\(d.S(x)=x^{10}-13x^9+13x^8-13x^7+.....+13x^2-13x+10\)

\(=x^{10}-12x^9-x^9+12x^8+.....+x^2-12x-x+10\)

\(=x^9(x-12)-x^8(x-12)+....+x(x-12)-x+10\)

\(=(x-12)(x^9-x^8+....+x)-x+10\)

Thay x = 12 vào biểu thức trên , ta có

\(S(12)=(12-12)(12^9-12^8+....+12)-12+10\)

\(=0-12+10\)

\(=-2\)

Vậy \(S(x)=-2\)khi x = 12

Hình như đây là toán lớp 7 có trong phần trắc nghiệm của thi HSG huyện

Chúc bạn học tốt , nhớ kết bạn với mình

a, x = 79 => x + 1 = 80

Ta có:\(P\left(x\right)=x^7-80x^6+80x^5-80x^4+...+80x+15\)

\(=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+...+\left(x+1\right)x+15\)

\(=x^7-x^7-x^6+x^6+x^5-x^5-x^4+...+x^2+x+15\)

\(=x+15=79+15=94\)

Còn lại tương tự

\(Q_{\left(x\right)}=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)

\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+..+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)

\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)

\(=1\)

Lời giải:

a) Với \(x=79\)

\(P(x)=x^7-80x^6+80x^5-80x^4+...+80x+15\)

\(=(x^7-79x^6)-(x^6-79x^5)+(x^5-79x^4)-....-(x^2-79x)+x+15\)

\(=x^6(x-79)-x^5(x-79)+x^4(x-79)-...-x(x-79)+x+15\)

\(=(x^6-x^5+x^4-...-x)(x-79)+x+15\)

\(=(x^6-x^5+x^4-...-x)(79-79)+79+15=79+15=94\)

b) Hoàn toàn tương tự phần a.

\(Q(x)=(x^{14}-9x^{13})-(x^{13}-9x^{12})+(x^{12}-9x^{11})-...+(x^2-9x)-x+10\)

\(=x^{13}(x-9)-x^{12}(x-9)+x^{11}(x-9)-...+x(x-9)-x+10\)

\(=(x-9)(x^{13}-x^{12}+x^{11}-...+x)-x+10\)

\(=(9-9)(x^{13}-x^{12}+...+x)-9+10=0-9+10=1\)

c)

\(R(x)=(x^4-16x^3)-(x^3-16x^2)+(x^2-16x)-x+20\)

\(=x^3(x-16)-x^2(x-16)+x(x-16)-x+20\)

\(=(x-16)(x^3-x^2+x)-x+20\)

Với $x=16$ thì $Q(x)=(16-16)(x^3-x^2+x)-16+20=0-16+20=4$

d)

\(S(x)=(x^{10}-12x^9)-(x^9-12x^8)+(x^8-12x^7)-....+x(x-12)-x+10\)

\(=x^9(x-12)-x^8(x-12)+x^7(x-12)-...+x(x-12)-x+10\)

\(=(x-12)(x^9-x^8+x^7-..+x)-x+10\)

\(=(12-12)(x^9-x^8+x^7-...+x)-12+10=-12+10=-2\)

a)

$A=(1^2-2^2)+(3^2-4^2)+....+(2003^2-2004^2)+2005^2$

$=(1-2)(1+2)+(3-4)(3+4)+....+(2003-2004)(2003+2004)+2005^2$

$=-(1+2)-(3+4)-...-(2003+2004)+2005^2$

$=-(1+2+3+...+2004)+2005^2=-\frac{2004.2005}{2}+2005^2$

$=2005^2-1002.2005=2005(2005-1002)=2011015$

b)

$B=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$

$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$

$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$

$=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$

$=(2^{16}-1)(2^{16}+1)(2^{32}+1)-2^{64}$

$=(2^{32}-1)(2^{32}+1)-2^{64}$

$=2^{64}-1-2^{64}=-1$

c) Do $x=16$ nên $x-16=0$

$R(x)=x^4-17x^3+17x^2-17x+20$

$=(x^4-16x^3)-(x^3-16x^2)+x^2-16x-x+20$

$=x^3(x-16)-x^2(x-16)+x(x-16)-x+20$

$=x^3.0-x^2.0+x.0-x+20=-x+20=-16+20=4$

d) Do $x=12$ nên $x-12=0$. Khi đó:

$S(x)=(x^{10}-12x^9)-(x^9-12x^8)+(x^8-12x^7)-....+(x^2-12x)-x+10$

$=x^9(x-12)-x^8(x-12)+x^7(x-12)-....+x(x-12)-x+10$

$=(x-12)(x^9-x^8+x^7-....+x)-x+10$

$=0-x+10=-x+10=-12+10=-2$