\(\Leftrightarrow5\left(x^4+2x^2+1\right)+2\left(y^6+2y^3+1\right)=13\)

\(\Leftrightarrow5\left(x^2+1\right)^2+2\left(y^3+1\right)^2=13\)

\(\Leftrightarrow\left(x^2+1\right)^2=\dfrac{13-2\left(y^3+1\right)^2}{5}\le\dfrac{13}{5}< 4\)

\(\Rightarrow x^2+1< 2\Rightarrow x^2< 1\)

\(\Leftrightarrow x=0\)

\(\Rightarrow y^6+2y^3-3=0\Rightarrow\left[{}\begin{matrix}y^3=1\Rightarrow y=1\\y^3=-3\left(ktm\right)\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(0;1\right)\)

bn học Δ chx nhỉ

\(6x^2-9xy+10x-15y\\ =3x\left(2x-3y\right)+5\left(2x-3y\right)\\ =\left(3x+5\right)\left(2x-3y\right)\\ 27x^3+36x^2y+12xy^2\\ =3x\left(9x^2+12xy+4y^2\right)\\ =3x\left(3x+2y\right)^2\)

a) \(6x^2-9xy+10x-15y=\left(6x^2-9xy\right)+\left(10x-15y\right)\)

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

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

b) \(27x^3+36x^2y+12xy^2=3x\left(9x^2+12xy+4y^2\right)\)

                                        \(=3x\left[\left(3x\right)^2+2\cdot3x\cdot2y+\left(2y\right)^2\right]\)

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

                                   \(\)

Ta có:\(10x^2-12xy+5y^2+11y-36x+38\)

=\(9x^2-12xy+4y^2+x^2-36x+324+y^2+2.\frac{11}{2}y+\frac{11^2}{4}-\frac{1265}{4}\)

=\(\left(3x-2y\right)^2+\left(x-18\right)^2+\left(y+\frac{11}{2}\right)^2-\frac{1265}{4}\)

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