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(x+y)2 + (x-y)2 = x2 +2xy +y2 + x2 - 2xy + y2 = 2x2 + 2y2
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\(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right)\\ =\left[\left(x^2+1\right)^2-x^2\right]\left(x^4-x^2+1\right)\left(x^8-x^4+1\right)\\ =\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right)\\ =\left[\left(x^4+1\right)^2-x^4\right]\left(x^8-x^4+1\right)\\ =\left(x^8+x^4+1\right)\left(x^8-x^4+1\right)\\ =\left(x^8+1\right)^2-x^8\\ =x^{16}+x^8+1\)
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\(x^2+2x-8\)
\(=x^2-2x+4x-8\)
\(=x\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x+4\right)\)
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\(\left(2x+1\right)^4=\left(2x+1\right)^2\)
\(\Rightarrow\left(2x+1\right)^4-\left(2x+1\right)^2=0\)
\(\left(2x+1\right)^2\left[\left(2x+1\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x+1\right)^2=0\\\left(2x+1\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\\left(2x+1\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\2x+1=1\\2x+1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=0\\x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[\left(2x+1\right)^2\right]^2=\left(2x+1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)^2=\left(2x+1\right)\)
\(\Leftrightarrow2x+1=1\left(x\ne-\dfrac{1}{2}\right)\)
\(\Leftrightarrow x=0\)
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\(2x\left(x-5\right)-x\left(x-10\right)+1=26\)
\(2x^2-10x-x^2+10x-25=0\)
\(x^2-25=0\)
\(\left(x-5\right)\left(x+5\right)=0\)
\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
\(7\left(2x-5\right)-5\left(7x-2\right)+2\left(5x+7\right)=\left(x-2\right)-\left(x+4\right)\)
\(14x-35-35x+10+10x+14=-6\)
\(-11x-11=-6\)
\(x=-\dfrac{5}{11}\)
\(10x-5-32+12x=7\)
\(22x=44\)
\(x=2\)
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\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)
\(a^2+b^2+c^2=ab+bc+ca\)
\(\dfrac{1}{2}\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)=0\)
\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)
\(< =>a=b=c\)
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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) \(x^2+4x+3=x^2+3x+x+3=x\left(x+3\right)+\left(x+3\right)=\left(x+1\right)\left(x+3\right)\)
\(1...1\) (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.