Gọi số cần lập là \(\overline{a_1a_2a_3a_4a_5a_6}\)

\(\Rightarrow a_1\ge3\)

\(\Rightarrow a_1\) có 4 cách chọn (3;4;5;6)

5 chữ số còn lại có \(A_5^5=120\) cách

\(\Rightarrow4.120=480\) số thỏa mãn

\(a_1,=\left(x^3+x^2-2x^2-2x+3x+3\right):\left(x+1\right)\\ =\left(x+1\right)\left(x^2-2x+3\right):\left(x+1\right)\\ =x^2-2x+3\\ a_2,=\left(2x^2+3y^2\right)^2:\left(2x^2+3y^2\right)=2x^2+3y^2\\ b_1,=\left(x^3-7x^2+x^2-7x-2x+14\right):\left(x-7\right)\\ =\left(x-7\right)\left(x^2+x-2\right):\left(x-7\right)\\ =x^2+x-2\\ b_2,=\left(8ab-7m^2n\right)\left(8ab+7m^2n\right):\left(8ab+7m^2n\right)=8ab-7m^2n\\ c,=\left(3x-2y^2\right)\left(9x^2+6xy^2+4y^4\right):\left(3x-2y^2\right)\\ =9x^2+6xy^2+4y^4\\ d,=\left(3x+2y^2\right)\left(9x^2-6xy^2+4y^4\right):\left(9x^2-6xy^2+4y^4\right)\\ =3x+2y^2\)

Tách ra với để dọc lại đi.

b: Ta có: \(N=a^3+b^3+3ab\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)

\(=1-3ab+3ab\)

=1

\(2,\\ 1,=20\sqrt{3}+20\sqrt{3}+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=40\sqrt{3}+\sqrt{3}=41\sqrt{3}\\ 2,A=\dfrac{2\sqrt{x}-9-x+9+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{2\sqrt{x}-x+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ c,A< 1\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-3}< 0\Leftrightarrow\sqrt{x}-3< 0\left(4>0\right)\\ \Leftrightarrow x< 9\Leftrightarrow0\le x< 9\)

\(3,\\ 1,A=\sqrt{2}-1-\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}=\sqrt{2}-1-\sqrt{2}=-1\\ 2,\\ a,P=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{4}\left(x\ge0;x\ne4\right)\\ P=\dfrac{4\left(\sqrt{x}+2\right)}{4\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\\ b,P< 1\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-1< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-2}< 0\Leftrightarrow\sqrt{x}-2< 0\left(4>0\right)\\ \Leftrightarrow x< 4\Leftrightarrow0\le x< 4\)