4n+1-3.4n
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a)\(\left(x^2+4-4x\right)\left(x^2+4+4x\right)\)
b)\(x\left(y+1\right)+\left(y+1\right)=\left(y+1\right)\left(x+1\right)\)
c)\(\left(x+y\right)^2-2\left(x+y\right)=\left(x+y\right)\left(x+y-2\right)\)


\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{10}{20}-\frac{1}{20}\)
\(=\frac{9}{20}\)



PT <=> \(\frac{4}{5}x^2\left(\frac{x}{3}-\frac{1}{2}\right)-\frac{4}{3}x^2\left(\frac{1}{5}x-\frac{2}{3}\right)-\frac{22}{45}x^2-\left(\frac{1}{5}x-\frac{2}{3}\right)=0\)
<=> \(x^2\left(\frac{4x}{15}-\frac{2}{5}-\frac{4x}{15}+\frac{8}{9}-\frac{22}{45}\right)-\left(\frac{1}{5}x-\frac{2}{3}\right)=0\)
<=> \(x^2.0-\frac{1}{5}x+\frac{2}{3}=0\)
<=> \(\frac{1}{5}x=\frac{2}{3}\Rightarrow x=\frac{2}{3}:\frac{1}{5}=\frac{10}{3}\)
Vậy....


\(4^{n+1}-3.4^n=4^n.4-3.4^n=4^n.\left(4-3\right)=4^n\)