4) \(4-x^2+2xy-y^2\)

\(=4-\left(x-y\right)^2\)

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

5) \(9x^2-1+9xy+3y\)

\(=\left(3x-1\right)\left(3x+1\right)+3y\left(3x+1\right)\)

\(=\left(3x+1\right)\left(3x+3y-1\right)\)

Ta có:

\(\dfrac{1}{2}< 1;\dfrac{3}{4}< 1;\dfrac{5}{6}< 1;...;\dfrac{999}{1000}< 1\)

\(\Rightarrow\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{999}{1000}< 1\)

\(\Rightarrow p< 1\)

5) \(4b^2-2b+a-a^2\)

\(=\left(2b-a\right)\left(2b+a\right)-\left(2b-a\right)\)

\(=\left(2b-a\right)\left(2b+a-1\right)\)

6) \(b^2-9c^2+4+4b\)

\(=\left(b+2\right)^2-9c^2\)

\(=\left(b+3c+2\right)\left(b-3c+2\right)\)

hàm số \(y=ax+b\) song song \(y=2x-2023\)

\(\Rightarrow\left\{{}\begin{matrix}a=2\\b\ne-2023\end{matrix}\right.\)

hàm số \(y=ax+b\) đi qua điểm \(A\left(-2;1\right)\)

\(\Rightarrow1=-2.2+b\)\(\Rightarrow b=5\)

\(\dfrac{x}{3}=\dfrac{1}{x}\left(đk:x\ne0\right)\)

\(\Leftrightarrow x^2=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)

\(\left(\left|x\right|-1,5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x\right|=1,5\\x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1,5\\x=-1,5\\x=-1\end{matrix}\right.\)

\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+..+\dfrac{1}{97.99}+\dfrac{1}{98.100}-\dfrac{49}{99}\)

\(=\dfrac{1}{2}\left[\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{99.100}\right)\right]-\dfrac{49}{99}\)

\(=\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..+\dfrac{1}{98}-\dfrac{1}{100}\right]-\dfrac{49}{99}\)

\(=\dfrac{1}{2}\left[1-\dfrac{1}{99}+\dfrac{1}{2}-\dfrac{1}{100}\right]-\dfrac{49}{99}\)

\(=\dfrac{1}{2}\left[\dfrac{98}{99}+\dfrac{49}{100}\right]-\dfrac{49}{99}=\dfrac{14651}{19800}-\dfrac{49}{99}=\dfrac{49}{200}\)

\(\dfrac{1}{5+2\sqrt{3}}+\dfrac{1}{5-2\sqrt{3}}\)

\(=\dfrac{5-2\sqrt{3}+5+2\sqrt{3}}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}\)

\(=\dfrac{10}{25-12}=\dfrac{10}{13}\)

\(4.5=2.10\Rightarrow\left\{{}\begin{matrix}\dfrac{4}{2}=\dfrac{10}{5}\\\dfrac{4}{10}=\dfrac{2}{5}\end{matrix}\right.\)

\(6.63=9.42\Rightarrow\left\{{}\begin{matrix}\dfrac{6}{9}=\dfrac{42}{63}\\\dfrac{6}{42}=\dfrac{9}{63}\end{matrix}\right.\)

\(14.15=10.21\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{14}{10}=\dfrac{21}{15}\\\dfrac{14}{21}=\dfrac{10}{15}\end{matrix}\right.\)

\(ab=cd\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{d}{b}\\\dfrac{a}{d}=\dfrac{c}{b}\end{matrix}\right.\)