Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a.\) Vì \(x=14\) \(\Rightarrow\) \(x+1=15;\) \(x+2=16;\) \(2x+1=29;\) và \(x-1=13\)
Khi đó, biểu thức trên trở thành:
\(x^5-15x^4+16x^3-29x^2+13x=x^5-\left(x+1\right)x^4+\left(x+2\right)x^3-\left(2x+1\right)x^2+\left(x-1\right)x\)
\(=x^5-x^5-x^4+x^4+2x^3-2x^3-x^2+x^2-x\)
\(x^5-15x^4+16x^3-29x^2+13x=-x=-14\)
\(b.\) Làm tương tự
- Charlotte-
Mấy bài kia phá tung tóe rồi rút gọn hết sức xong thay x vào, làm câu c thôi nhé:
c) \(C=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
riêng câu này ta thay x = 9 vào luôn, vậy ta có:
\(C=9^{14}-10\cdot9^{13}+10\cdot9^{12}-10\cdot9^{11}+...+10\cdot9^2-10\cdot9+10\)
\(=9^{14}-\left(9+1\right)\cdot9^{13}+\left(9+1\right)\cdot9^{12}-\left(9+1\right)\cdot9^{11}+...+\left(9+1\right)\cdot9^2-\left(9+1\right)\cdot9+10\)
\(=9^{14}-9^{14}-9^{13}+9^{13}+9^{12}-9^{12}-9^{11}+...+9^3+9^2-9^2-9+10\)
\(=-9+10\)
\(=1\)
a) Ta có : \(x=31\Rightarrow30=x-1\)
Thay vào biểu thức ta được:
\(A=x^3-\left(x-1\right).x^2-x^2+1=x^3-x^3+x^2-x^2+1=1\)
b) Ta có: \(x=9\Rightarrow x+1=10\)
Thay vào biểu thức ta được
\(B=x^{14}-\left(x+1\right).x^{13}+\left(x+1\right).x^{12}-\left(x+1\right).x^{11}+.....+x^2.\left(x+1\right)=\left(x+1\right).x+\left(x+1\right)\)
\(\Leftrightarrow B=x^{14}-x^{14}-x^{13}+x^{13}+....+x^3+x^2=x^2+2x+1\)
\(\Leftrightarrow B=x^2-x^2-2x-1=-2.9-1=-19\)
\(x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+..+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)
\(=1\)
a, 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)
\(S(x)=(x^{10}-12x^9)-(x^9-12x^8)+(x^8-12x^7)-....+x(x-12)-x+10\)
\(=x^9(x-12)-x^8(x-12)+x^7(x-12)-...+x(x-12)-x+10\)
\(=(x-12)(x^9-x^8+x^7-..+x)-x+10\)
\(=(12-12)(x^9-x^8+x^7-...+x)-12+10=-12+10=-2\)
Nếu \(x=25\)
\(\Rightarrow\left\{{}\begin{matrix}26=x+1\\27=x+2\\47=2x-3\\77=3x+2;50=2x;24=x-1\end{matrix}\right.\) ( * )
Thay ( * ) vào C , ta được :
\(C=x^7-\left(x+1\right)x^6+\left(x+2\right)x^5-\left(2x-3\right)x^4-\left(3x+2\right)x^3+2x.x^2+x-\left(x-1\right)\)
\(=x^7-x^7-x^6+x^6+2x^5-2x^5+3x^4-3x^4-2x^3+2x^3+x-x+1\)
\(=1\)
Vậy \(C=1\) tại \(x=25\)
\(C=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+..+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)
\(=1\)