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Ta có : x = 7 ⇒ x + 1 = 8
Thay x + 1 = 8 vào A , ta được :
A = x15 - ( x + 1)x14 + ( x + 1)x13 - ( x + 1)x12 +....- ( x + 1)x2 + ( x + 1)x - 5
A = x15 - x15 - x14 + x14 + x13 - x13 - x12 +....- x3 - x2 + x2 + x - 5
A = x - 5 = 7 - 5 = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ \(x=7\Rightarrow x+1=8\) thay vào B ta được :
\(B=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...-\left(x+1\right)x^2+\left(x+1\right)x-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+......-x^3-x^2+x^2+x-5\)
\(=x-5=7-5=2\)
Vậy B = 2
\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)
\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)
\(=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x=7
nên x+1=8
\(A=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-5\)
\(=x-5=7-5=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=x^{15}-8x^{14}+8x^{13}-...-8x^2+8x-5\)
Vì \(x=7\) \(\Rightarrow\) \(x+1=8\) \(\left(\text{*}\right)\)
Thay \(\left(\text{*}\right)\) vào \(A\), ta được:
\(A=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-...-\left(x+1\right)x^2+\left(x+1\right)x-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-...-x^3-x^2+x^2+x-5\)
\(A=x-5\)
Tại \(x=7\) thì khi đó, \(A=7-5=2\)
Vậy, giá trị cua biểu thức \(x^{15}-8x^{14}+8x^{13}-...-8x^2+8x-5\) là \(2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)
\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)
\(=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x=7=>x+1=8
B=x15-8x14+8x13-8x12+....-8x2+8x-5
=x15-(x+1)x14+(x+1)x13-(x+1)x12+...-(x+1)x2+(x+1)x-5
=x15-x15-x14+x14+x13-x13+x12+...-x3-x2+x2+x-5
=x-5
=7-5
=2
Vậy B=2
\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)
\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)
\(=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có B = 715 - 8.714 + 8.713 - 8.712 + ... - 8.72 + 8.7 – 5
= 715 - 8.(714 - 713 + 712 - .... + 72 - 7) - 5
Đặt C = 714 - 713 + 712 - .... + 72 - 7
=> 7C = 715 - 714 + 713 - .... + 73 - 72
Lấy 7C cộng C theo vế ta có :
7C + C = ( 715 - 714 + 713 - .... + 73 - 72) + (714 - 713 + 712 - .... + 72 - 7)
8C = 715 - 7
=> C = \(\left(7^{15}-7\right).\frac{1}{8}\)
Khi đó B = \(7^{15}-8.\left(7^{15}-7\right).\frac{1}{8}-5=7^{15}-7^{15}+7-5=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(x=7\)\(\Rightarrow x+1=8\)
\(B=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-........-\left(x+1\right)x^2+\left(x+1\right)x-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-......-x^3-x^2+x^2+x-5\)
\(=x-5=7-5=2\)
Với x = 7 ta có 8 = x + 1
Thay 8 = x + 1 vào biểu thức B ta có \(B=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...-\left(x+1\right)x^2+\left(x+1\right)x-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...-x^3-x^2+x^2+x-5\)
\(=x-5\)
Thay x = 7 vào biểu thức B đã thu gọn ta được B = 7 - 5 = 2
Vậy B = 2
x=7
=>x+1=8
=> A= x^15 - 8x^14 + 8x^13 - 8x^12 +....- 8x^2 + 8x - 5
=x15-(x+1)x14+(x+1)x13-(x+1)x12+...-(x+1)x2+(x+1)x-5
=x15-x15-x14+x14+x13-x13-x12+...-x3-x2+x2+x-5
=x-5
=>A=7-5=2
Vậy A=2 khi x=7
\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)
\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)
\(=2\)