\(1...1\) (2n chữ số)

\(=1+10+10^2+...+10^{2n-1}\)

\(=\dfrac{10^{2n}-1}{9}\)

 \(4...4\) (n chữ số) 

\(=4.\left(1+10+10^2+...+10^{n-1}\right)\)

\(=4.\dfrac{10^n-1}{9}\)

\(=\dfrac{4.10^n-4}{9}\)

Viết lại: \(D=\dfrac{10^{2n}-1}{9}+\dfrac{4.10^n-4}{9}+1=\dfrac{\left(10^n\right)^2-1+4.10^n-4+9}{9}=\dfrac{\left(10^n\right)^2+4.10^n+4}{9}=\left(\dfrac{10^n+2}{3}\right)^2\)

Vì \(\left\{{}\begin{matrix}10^nmod3=1\\2mod3=2\end{matrix}\right.\Rightarrow\left(10^n+2\right)⋮3\)

\(\Rightarrow\dfrac{10^n+2}{3}\) là số tự nhiên

Vậy D là số chính phương.

(x+y)² + (x-y)² = x² +2xy +y² + x² - 2xy + y² = 2x² + 2y²

\(\left(x+y\right)^2+\left(x-y\right)^2\)

\(=x^2+2.x.y+y^2+x^2-2x.y+y^2\)

\(=x^2+2xy+y^2+x^2-2xy+y^2\)

\(=2x^2+y^2+y^2\)

\(=2x^2+2y^2\)

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

\(x^2+2x-8\)

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

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

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

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

\(\Rightarrow\left(2x+1\right)^4-\left(2x+1\right)^2=0\)

\(\left(2x+1\right)^2\left[\left(2x+1\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x+1\right)^2=0\\\left(2x+1\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\\left(2x+1\right)^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\2x+1=1\\2x+1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=0\\x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[\left(2x+1\right)^2\right]^2=\left(2x+1\right)^2\)

\(\Leftrightarrow\left(2x+1\right)^2=\left(2x+1\right)\)

\(\Leftrightarrow2x+1=1\left(x\ne-\dfrac{1}{2}\right)\)

\(\Leftrightarrow x=0\)

\(2x\left(x-5\right)-x\left(x-10\right)+1=26\)

\(2x^2-10x-x^2+10x-25=0\)

\(x^2-25=0\)

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

\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

\(7\left(2x-5\right)-5\left(7x-2\right)+2\left(5x+7\right)=\left(x-2\right)-\left(x+4\right)\)

\(14x-35-35x+10+10x+14=-6\)

\(-11x-11=-6\)

\(x=-\dfrac{5}{11}\)

\(10x-5-32+12x=7\)

\(22x=44\)

\(x=2\)

Ai giúp đc k 

\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)

\(a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)

\(a^2+b^2+c^2=ab+bc+ca\)

\(\dfrac{1}{2}\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)=0\)

\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)

\(< =>a=b=c\)

a) \(x^4+2x^3+x^2=\left(x^2+x\right)^2\)

b) \(x^3-x+3x^2y+3xy^2+y^3-y=\left(x-y\right)^3-\left(x+y\right)\)

c) \(5x^2-10xy+5y^2-20z^2=\left(\sqrt{5x}-\sqrt{5y}\right)^2-20z^2\)

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