vào đây nha https://coccoc.com/search/math#query=+3%5E%E2%88%921%C2%B73%5En%2B6%C2%B73%5En%E2%88%921%3D7%C2%B736++

\(3^{-1}\cdot3^n+6\cdot3^{n-1}=7\cdot3^6\)

\(3^{n-1}+6\cdot3^{n-1}=7\cdot3^6\)

\(3^{n-1}\left(1+6\right)=7\cdot3^6\)

\(3^{n-1}\cdot7=7\cdot3^6\)

\(\Rightarrow3^{n-1}=3^6\)

\(\Rightarrow n-1=6\)

\(n=6+1=7\)

a/ \(127^2+146\cdot127+73^2=127^2+2\cdot73\cdot127+73^2=\left(127+73\right)^2=200^2=40000\)

b/ \(100^2-99^2+98^2-97^2+...+2^2-1^2=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)

\(=100+99+98+97+...+2+1=\frac{\left(100\right)\left(100+1\right)}{2}=5050\)

c/ \(34^2+66^2+68\cdot66=34^2+66^2+2\cdot34\cdot66=\left(34+66\right)^2=100^2=10000\)

=x³-2x²+6x²-12x+5x-10

=x²(x-2)+6x(x-2)+5(x-2)

=(x-2)(x²+6x+5)

=(x-2)(x²+x+5x+5)

=(x-2)[x(x+1)+5(x+1)]

=(x-2)(x+1)(x+5)

\(x^8+x^4+1\)

\(=x^8+2x^4-x^4+1\)

\(=\left(x^8+2x^4+1\right)-x^4\)

\(=\left(x^4+1\right)^2-\left(x^2\right)^2\)

\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\)

\(\frac{2n^2-n+2}{2n+1}=\left(n-1\right)+\frac{3}{2n+1}\)

Để \(\left(2n^2-n+2\right)\)chia hết \(\left(2n+1\right)\)thì \(3\)chia hết \(2n+1\)

\(\Rightarrow2n+1\)là ước của 3.

mà -1 ; 1; -3 ; 3 là ước của 2

\(\cdot2n+1=-1\Rightarrow n=-1\)(nhận)

\(\cdot2n+1=1\Rightarrow n=0\)(nhận)

\(\cdot2n+1=-3\Rightarrow n=-2\)(nhận)

\(\cdot2n+1=3\Rightarrow n=1\)(nhận)

Vậy \(n=-2;-1;0;1\)thi \(2n^2-n+2\)chai hết cho 2n +1.

   x⁴ - 25x² + 114
= ( x² )² - 2x . 25/2 + 625/4 - 625/4 + 114
= ( x² + 25/2 )² - ( 13/2 )²
= ( x² + 25/2 - 13/2 )( x² + 25/2 + 13/2 )
= ( x² + 6 )( x² + 19 )
sai sót j sửa giúp mình nha ^_^