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8 tháng 1 2019

\(\frac{1}{\left(x+29\right)^2}+\frac{1}{\left(x+30\right)^2}=\frac{\left(x+30\right)^2}{\left(x+29\right)^2\left(x+30\right)^2}+\frac{\left(x+29\right)^2}{\left(x+29\right)^2\left(x+30\right)^2}\)

\(=\frac{x^2+60x+900+x^2+58x+841}{\left(x+29\right)^2\left(x+30\right)^2}\)

\(=\frac{2x^2+118x+1741}{\left(x+29\right)^2\left(x+30\right)^2}\)

11 tháng 7 2019

\(\frac{\left(x^{95}+x^{94}\right)+.....+\left(x+1\right)}{\left(x^{31}+x^{30}\right)+.....+\left(x+1\right)}=\frac{x^{94}\left(x+1\right)+......+\left(x+1\right)}{x^{30}\left(x+1\right)+.....+\left(x+1\right)}=\frac{x^{94}+x^{92}+....+x^2+1}{x^{30}+x^{28}+....+x^2+1}=\frac{\left(x^2+1\right)x^{92}+x^{88}\left(x^2+1\right).....+\left(x^2+1\right)}{\left(x^2+1\right)x^{28}+\left(x^2+1\right)x^{24}+....+\left(x^2+1\right)}=\frac{x^{92}+x^{88}+......+x^4+1}{x^{28}+x^{24}+.....+x^4+1}=\frac{x^{88}\left(x^4+1\right)+x^{80}\left(x^4+1\right)+....+\left(x^4+1\right)}{x^{24}\left(x^4+1\right)+x^{16}\left(x^4+1\right)+.....+\left(x^4+1\right)}=\frac{x^{88}+x^{80}+....+1}{x^{24}+x^{16}+...+1}\)

\(=\frac{x^{80}\left(x^8+1\right)+x^{64}\left(x^8+1\right)+.....+\left(x^8+1\right)}{x^{16}\left(x^8+1\right)+\left(x^8+1\right)}=\frac{x^{80}+x^{64}+.....+1}{x^{16}+1}=\frac{x^{64}\left(x^{16}+1\right)+.....+x^{16}+1}{x^{16}+1}=x^{64}+x^{32}+1\)

17 tháng 2 2020

a)\(\frac{201-x}{99}+1+\frac{203-x}{97}+1+\frac{205-x}{95}+1=0\)

\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)

\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)

\(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\ne0\Rightarrow300-x=0\Rightarrow x=300\)

b)\(\frac{2-x}{2002}+1=\frac{1-x}{2003}+2-\frac{x}{2004}\)

\(\Leftrightarrow\frac{2004-x}{2002}=\frac{1-x}{2003}+1+1-\frac{x}{2004}\)

\(\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}+\frac{2004-x}{2004}\)

\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)

\(\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\ne0\Rightarrow2004-x=0\Rightarrow x=2004\)

c)\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}-2=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}-2\)

\(\Leftrightarrow\frac{x^2-10x-2000}{1971}+\frac{x^2-10x-2000}{1973}=\frac{x^2-10x-2000}{29}+\frac{x^2-10x-2000}{27}\)

\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27}\right)=0\)

\(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27}\ne0\)

\(\Rightarrow x^2-10x-2000=0\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)