\(A=27x^3+27x^2+9x+1=\left(3x\right)^3+3.\left(3x\right)^2.1+3.\left(3x\right).1^2+1^3\)

\(=\left(3x+1\right)^3=10^3=1000\)

em cảm ơn 

\(2x^2-6xy+15y-5x\)

\(=2x\left(x-3y\right)-5\left(x-3y\right)\)

\(=\left(2x-5\right)\left(x-3y\right)\)

\(2x\left(x-17\right)+\left(17-x\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-17\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=17\end{cases}}\)

\(9x^2-18x=0\)

\(\Leftrightarrow9x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(B=\left(5x-4y\right)^2-\left(6x+4y\right)\left(5x-4y\right)+\left(3x+2y\right)^2\)

\(B=\left(5x-4y\right)\left(5x-4y-6x-4y\right)+\left(3x+2y\right)^2\)

\(B=\left(5x-4y\right)\left(-x-8y\right)+\left(3x+2y\right)^2\)

\(B=-5x^2-40xy+4xy+32y^2+9x^2+12xy+4y^2\)

\(B=4x^2-24xy+36y^2\)

\(B=x^2-6xy+6y^2\)

Bài chưa đc ktra lại đâu . Có gì sai sót thì bỏ qua

a) \(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2n\left(2n+2\right)}\)

\(=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2n\left(2n+2\right)}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2n}-\frac{1}{2n+2}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2n+2}\right)\)

b) \(A=\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}\)

\(< \frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2n\left(2n+2\right)}\)

\(< \frac{1}{4}\)

\(abc=a+b+c\Leftrightarrow\frac{abc}{abc}=\frac{a+b+c}{abc}\)

\(\Leftrightarrow1=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=Q\)

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

\(\Rightarrow P=3^2-2Q=9-2=7\)

\(\left(5x-4y\right)^2-\left(6x+4y\right).\left(5x-4y\right)+\left(3x+2y\right)^2\)

\(=\left(5x-4y\right)^2-2.\left(5x-4y\right).\left(3x+2y\right)+\left(3x+2y\right)^2\)

\(=\left(5x-4y-3x-2y\right)^2\)

\(=\left(2x-6y\right)^2\)

\(=4x^2-24xy+36y^2\)