Trả lời:

\(27a^2b^2-18ab+3\)

\(=3\left(9a^2b^2-6ab+1\right)\)

\(=3\left[\left(3ab\right)^2-2.3ab.1+1^2\right]\)

\(=3\left(3ab-1\right)^2\)

a: \(27a^2b^2+18ab+3\)

\(=3\left(9a^2b^2+6ab+1\right)\)

\(=3\left\lbrack\left(3ab\right)^2+2\cdot3ab\cdot1+1^2\right\rbrack\)

\(=3\left(3ab+1\right)^2\)

b: \(5x^2-y+5xy-x\)

=5x(x+y)-(x+y)

=(x+y)(5x-1)

c: \(2x^3y^2-8x^3-12x^2y-6xy^2-y^3+x^2y^3\)

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

\(=\left(2x+y\right)\left\lbrack x^2y^2-\left(2x+y\right)^2\right\rbrack\)

=(2x+y)(xy-2x-y)(xy+2x+y)

\(a,\Leftrightarrow x^2-x+2021x-2021=0\\ \Leftrightarrow\left(x-1\right)\left(x+2021\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2021\end{matrix}\right.\\ b,\Leftrightarrow-5x^2+15x+x-3=0\\ \Leftrightarrow\left(x-3\right)\left(1-5x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

b: \(-5x^2+16x-3=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

\(x^3+3x^2y+3xy^2+y^3-x-y=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)

Ta có : \(x^3+3x^2y+3xy^2+y^3-x-y.\)

\(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)

\(=x^2\left(x-6\right)-24\left(x-6\right)=\left(x^2-24\right)\left(x-6\right)\)

\(x^3-6x^2-24x+144=\left(x^3-6x^2\right)-\left(24x-144\right)=x^2\left(x-6\right)-24\left(x-6\right)=\left(x-6\right)\left(x^2-24\right)\)

bằng phương pháp dùng hằng đẳng thức ak??

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x² - 7xy + y²

= (x² - 2xy + y²) - 5xy

= (x - y)² - 5xy

còn lại bn xem thu gọn được ko thì làm nhé!!

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a) 27a2b2 – 18ab + 3 = 3(9a2b2 – 6ab + 1) = 3(3ab – 1)2

\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(=\left(a+b+c\right)\left(ab+bc\right)+\left(a+b+c\right)ac-abc\)

\(=\left(ab+b^2+bc\right)\left(a+c\right)+\left(a+c\right)ac+abc-abc\)

\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)

\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)