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x=7 nen x+1=8
\(A=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-...+x^3\left(x+1\right)-x^2\left(x+1\right)+x\left(x+1\right)-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+...+x^4+x^3-x^3-x^2+x^2+x-5\)
=x-5
=2
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
\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)
\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)
\(=2\)
a: \(=\dfrac{\left(2\cdot547+1\right)\cdot3}{547\cdot211}-\dfrac{546}{547\cdot211}-\dfrac{4}{547\cdot211}\)
\(=\dfrac{2735}{547\cdot211}=\dfrac{5}{211}\)
b: x=7 nên x+1=8
\(x^{15}-8x^{14}+8x^{13}-8x^{12}+...-8x^2+8x-5\)
\(=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-x^{12}\left(x+1\right)+...-x^2\left(x+1\right)+x\left(x+1\right)-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}-...-x^3-x^2+x^2+x-5\)
=x-5=7-5=2
Ta có : \(x=7\Rightarrow\left\{{}\begin{matrix}8=x+1\\5=x-2\end{matrix}\right.\)
\(\Rightarrow B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...-8x^2+8x-5\\ \\ =x^{15}-\left(x+1\right)x^{14}+...+\left(x+1\right)x-\left(x-2\right)\\ \\=x^{15}-x^{15}-x^{14}+...+x^2+x-x+2\\ \\=2\)
\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)
\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)
\(=2\)
Gợi ý:
Đặt:
\(\frac{1}{117}=a\)
\(\frac{1}{119}=b\)
Đến đây bạn thế a, b vào A rồi thu gọn, sau đó tính
\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)
\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)
\(=2\)
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
\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)
\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)
\(=2\)
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 = \(7^{15}-8.\left(7^{15}-7\right).\frac{1}{8}-5=7^{15}-7^{15}+7-5=2\)