Bằng 18005168122313401

có fải là (x^7-14x^5+49x^3-36x+1) mod 210

đúng ko bạn????

\(4x^3-36x=0\)

\(x.\left[\left(2x\right)^2-6^2\right]=0\)

\(x.\left(2x-6\right)\left(2x+6\right)=0\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=0\\2x-6=0\end{cases}}\)hoặc \(2x+6=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)hoặc \(x=-3\)

KL:...............................................

tích mình với

ai tích mình

mình tích lại

thanks

\(\left(3x-5\right)^2-\left(x+1\right)^2=0\)

\(\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\)

\(\left(2x-6\right)\left(4x-4\right)=0\)

\(2.\left(x-3\right).4.\left(x-1\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

Vậy \(x=3\)hoặc \(x=1\)

c) tương tự

2) 

\(A=\left(5n-2\right)^2-\left(2n-5\right)^2\)

\(A=\left(5n-2-2n+5\right)\left(5n-2+2n-5\right)\)

\(A=\left(3n+3\right)\left(7n-7\right)\)

\(A=3.\left(n+1\right).7.\left(n-1\right)\)

\(A=21.\left(n+1\right)\left(n-1\right)\)

Ta có: \(21⋮21\)

\(\Rightarrow\)\(21.\left(n+1\right)\left(n-1\right)⋮21\)

\(\Rightarrow A⋮21\)

  đpcm

Câu b tương tự

\(\Leftrightarrow x\left(x^6-14x^4+49x^2-36\right)=0\)

\(\Leftrightarrow x\left(x^2-1\right)\left(x^2-4\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=1\\x^2=4\\x^2=9\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm1\\x=\pm2\\x=\pm3\end{matrix}\right.\)

(5n-2)^2-(2n-5)^2=(5n-2-2n+5)(5n-2+2n-5)=(3n+3)(7n-7)=3(n+1)7(n-1)=21(n^2-1) chia het cho 21