Ăn đầu buồi

ko có câu trả lời

\(S=\dfrac{1}{5^2}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{2022}}\)

=>\(25\cdot S=1+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2020}}\)

=>\(25S-S=1+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2020}}-\dfrac{1}{5^2}-\dfrac{1}{5^4}-...-\dfrac{1}{5^{2022}}\)

=>\(24S=1-\dfrac{1}{5^{2022}}\)

=>\(S=\dfrac{1}{24}-\dfrac{1}{24\cdot5^{2022}}< \dfrac{1}{24}\)

\(S=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+4+2\right)+1\)

\(=\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)+1\)

\(=\left(x^2+5x+4+1\right)^2\)

\(=\left(x^2+5x+5\right)^2\) là SCP (đpcm)

\(S=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

Đặt \(t=x^2+5x+5\Rightarrow\) pt trở thành \(\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\)

\(=\left(x^2+5x+5\right)^2\)

Vì \(x\in Z\Rightarrow x^2+5x+5\in Z\Rightarrow\left(x^2+5x+5\right)^2\) là số chính phương

Ta có S=1+3+3^2+...+3^2011 chia hết cho 4

            =(1+3)+(3^2+3^3)+...+(3^2010+3^2011)

             =1.(1+3)+3^2.(1+3)+...+3^2010.(1+3)

             =1.4+3^2 .4+...+3^2010 .4

              =4.(1+3^2+...+3^2010) chia hết cho 4

           Vậy: S chia hết cho 4

S = 1 + 3 + 3² +..+3⁹⁹

=> S = (1 + 3) + ... + (3^98 + 3^99)

=> S = (1 + 3) + ... + 3^98.(1 + 3)

=> S = 4 + ... + 3^98.4

=> S = 4.(1 +... + 3^98) chia hết cho 4 (Đpcm)