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a) Ta có: \(P\left(x\right)=x^7-80x^6+80x^5-80x^4+...+80x+15\)
\(=x^7-x^6\left(x+1\right)+x^5\left(x+1\right)-...+x\left(x+1\right)+15\)
\(=x^7-x^7-x^6+x^6+x^5-...+x^2+x+15\)
\(=x+15\)
Thay x=79 vào biểu thức \(P\left(x\right)=x+15\), ta được:
\(P\left(79\right)=79+15=94\)
\(Q_{\left(x\right)}=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+..+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)
\(=1\)
\(a.P(x)=x^7-80x^6+80x^5-80x^4+....+80x+15\)
\(=x^7-79x^6-x^6+79x^5+x^5-79x^4-....-x^2+79x+x+15\)
\(=x^6(x-79)-x^5(x-79)+x^4(x-79)-....-x(x-79)+x+15\)
\(=(x-79)(x^6-x^5+x^4-....-x)+x+15\)
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, \(P\left(x\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+...+15\)
\(=x^7-x^7-x^6+x^6+x^5-x^5-x^4+...+15=15\)
b) Thay x+1=10 ta được:
Q(x) = \(x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+...+\left(x+1\right)x^2-\left(x+1\right)x+\left(x+1\right)\) \(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1=1\)
d) Thay x+1=13, ta được:
S(x) = \(x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...+\left(x+1\right)x^2-\left(x+1\right)x+10\)
\(=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^3+x^2-x^2-x+10=-12+10=-2\)
x=9
\(9^{14}-10.9^{13}+10.9^{12}-10.9^{11}+..+10.9^2-10.9+10\)
\(9^{14}-\left(9+1\right).9^{13}+\left(9+1\right).9^{12}+..+\left(9+1\right).9^2-\left(9+1\right)9+10\)
\(9^{14}-9^{14}-9^{13}+9^{13}+9^{12}-..+9^3+9^2-9^2-9+10=1\)
Vậy......
Bài 2:
a) Vì x = 79 => x + 1 = 80
\(P\left(x\right)=x^7-80x^6+80x^5-80x^4+.....+80x+15\)
\(\Rightarrow P\left(x\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+.....+\left(x+1\right)x+15\)
\(=x^7-x^7-x^6+x^6+x^5-x^5-x^4+....+x^2+x+15\)
\(=x+15\)
Thay x = 79 vào đa thức ta được:
79 + 15 = 94
b) Vì x = 9 => x + 1 = 10
\(Q\left(x\right)=x^{14}-10x^{13}+10x^{12}-10x^{11}+.....+10x^2-10x+10\)
\(\Rightarrow Q\left(x\right)=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+....+\left(x+1\right)x^2-\left(x+1\right)x+10\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+....+x^3+x^2-x^2-x+10\)
\(=-x+10\)
\(=-9+10=1\)
P/s: Ko chắc nhé!
Bài 1:
a/ \(\left(2x-1\right)\left(x^2-x+1\right)-2x^3+3x^2=2\)
\(\Rightarrow2x\left(x^2-x+1\right)-1\left(x^2-x+1\right)-2x^3+3x^2=2\)
\(\Rightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2=2\)
\(\Rightarrow3x-1=2\)
\(\Rightarrow3x=2+1=3\)
\(\Rightarrow x=3:3=1\)
b/ \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Rightarrow x\left(x^2+2x+4\right)+1\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Rightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Rightarrow6x+20=0\)
\(\Rightarrow6x=0-20=-20\)
\(\Rightarrow x=-\frac{20}{6}=-\frac{10}{3}\)
c/ \(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Rightarrow\left[x\left(x+2\right)+1\left(x+2\right)\right]\left(x+5\right)-x^3-8x^2=27\)
\(\Rightarrow\left(x^2+2x+x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Rightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Rightarrow x^2\left(x+5\right)+3x\left(x+5\right)+2\left(x+5\right)-x^3-8x^2=27\)
\(\Rightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2=27\)
\(\Rightarrow17x+10=27\)
\(\Rightarrow17x=27-10=17\)
\(\Rightarrow x=17:17=1\)
a, tách 80=79+1
sau đó phân phối ra
b, tách 10=9+1
phân phối tiếp
phần a ) là \(P\left(x\right)=x^7-80x^6-80x^5-80x^4\)\(+...+80x+5\)nha ình chép thiếu