Bạn thay 12 = x + 1 vào B rồi khai triển ra nó sẽ tự triệt tiêu hết 

Tại x=11

\(\Rightarrow f\left(x\right)=x^{17}-\left(x+1\right)x^{16}+\left(x+1\right)x^{15}-...+\left(x+1\right)x-1\)

\(f\left(x\right)=x^{17}-x^{17}-x^{16}+x^{16}+x^{15}-...+x^2+x-1\)

\(f\left(x\right)=x-1\)

\(f\left(x\right)=10\)

\(x=11\Leftrightarrow12=x+1\)

Mà \(f\left(x\right)=x^{17}-12x^{16}+12x^{15}-12x^{14}+........+12x-1\)

\(\Leftrightarrow f\left(x\right)=x^{17}-\left(x+1\right)x^{16}+\left(x+1\right)x^{15}-.......+\left(x+1\right)x-1\)

\(\Leftrightarrow f\left(x\right)=x^{17}-x^{17}-x^{16}+x^{16}+x^{15}-.....+x^2+x-1\)

\(\Leftrightarrow f\left(x\right)=x-1\)

Mà \(x=11\)

\(\Leftrightarrow f\left(11\right)=11-1=10\)

Vậy \(f\left(11\right)=10\)

Bài 1. 

a. \(3^4.5^4-\left(15^2+1\right)\left(15^2-1\right)=15^4-\left(15^4-1\right)=1\)

b. \(x=11\Rightarrow x+1=12\)

Từ đây, ta có: \(x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+111=-x+111=-11+111=100\)

Bài 2. 

\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

Câu 1:

Với \(x=11\Rightarrow12=x+1\) ta có: \(x^{17}-12x^{16}+12x^{15}-....+12x-1\)

\(=x^{17}-\left(x+1\right)x^{16}+\left(x+1\right)x^{15}-\left(x+1\right)x^{14}+...+\left(x+1\right)x-1\)

\(=x^{17}-x^{17}-x^{16}+x^{16}+x^{15}-x^{15}-x^{14}+...-x^3-x^2+x^2+x+1\)

\(=x+1\)

\(=12\)

Câu 2:

Do \(VT>0\Rightarrow VP>0\Rightarrow x>0\Rightarrow\) tất cả các biểu thức dưới dấu trị tuyệt đối đều dương, phương trình trở thành:

\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)

\(\Leftrightarrow100x+\frac{1+2+3+...+100}{101}=101x\)

\(\Rightarrow x=\frac{100.101}{2.101}=50\)

Câu 3:

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

\(A=\left(n+3\right)\left(n-1\right)\left(n+1\right)\)

Do n lẻ \(\Rightarrow n=2k+1\)

\(\Rightarrow A=\left(2k+4\right).2k.\left(2k+2\right)=8k.\left(k+1\right)\left(k+2\right)\)

Do \(k\left(k+1\right)\left(k+2\right)\) là tích 3 số nguyên liên tiếp nên chia hết cho 6

\(\Rightarrow A⋮\left(8.6\right)\Rightarrow A⋮48\)

@Đinh Tuấn Việt

kết bn đi

\(B=x^{17}-12.x^{16}+12.x^{15}-12.x^{14}+...-12.x^2+12x-1\)

\(=11^{17}-\left(11+1\right)11^{16}+\left(11+1\right)11^{15}-\left(11+1\right)11^{14}+...-\left(11+1\right)11^2+\left(11+1\right)11-1\)

\(=11^{17}-11^{17}-11^{16}+11^{16}+11^{15}-11^{15}-11^{14}+...-11^3-11^2+11^2+11-1\)

\(=11-1=10\)

Vậy B = 10

B= x^17 - 12x^16 +12x^15 - 12x^14 +...-12x^2+12x-1

  = x^17-11x^16-x^16+11x^15 +x^15-11x^14-x^14+...-x^2+11x+x-1

  = (x-11)( x^16-x^15+x^14-...-x +1) 

  = 0 ( vì x=11)