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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
\(\Rightarrow A⋮\left(8.6\right)\Rightarrow A⋮48\)
\(F\left(x\right)=2x^3-7x^2+12x+a\)
\(G\left(x\right)=x+2\)
\(F\left(x\right):G\left(x\right)=2x^2-11x+34\) dư \(a-68\)
Để \(F\left(x\right)⋮G\left(x\right)\Rightarrow a-68=0\Rightarrow a=68\)
a) Với x = 11 <=> 12 = x+1
\(A\left(x\right)=x^{17}-\left(x+1\right)x^{16}+\left(x+1\right)x^{15}-...+12x-1\)
\(A\left(x\right)=12x-11=12.11-1=120\)
b) \(B=6x-6y+10-3ax+3ay+15a\)
\(B=6\left(x-y\right)+10-3a\left(x-y\right)+15a\)
\(B=6.5+10-3.a.5+15a\)
\(B=40\)
c)\(C=\frac{x-y}{x+6}=\frac{x-y}{x+x-2y}=\frac{x-y}{2\left(x-y\right)}=\frac{1}{2}\left(x-2y=6\right)\)
\(C=\frac{2x+6}{3x-2y}+\frac{2y-6}{4y-x}\)
\(C=\frac{2x+1-2y}{3x-2y}+\frac{2y-x+2y}{4y-x}\)
\(C=1+1=2\)
d) ta có : x-y-x = 0
\(\Rightarrow\left\{{}\begin{matrix}x-z=y\\x-y=z\\x=y+z\end{matrix}\right.\).Thay vào B, ta có :
\(B=\frac{x-z}{x}.\frac{y-x}{y}.\frac{z+y}{z}\)
\(B=\frac{y}{x}.\frac{\left(-z\right)}{y}.\frac{x}{z}\)
B= -1
a)Đang suy nghĩ...
b)\(M\left(x\right)=\left(x^2-3x\right)+\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
a) \(12x^{11}-15x^7-6x^5+2018\)
\(=3x^5.\left(4x^6-5x^2-2\right)+2018\)
\(=3x^5.0+2018\)
\(=2018\)
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 \(f\left(11\right)=10\)