HOC24
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Môn học
Chủ đề / Chương
Bài học
\(a) x^4-1\\=(x^2)^2-1\\=(x^2-1)(x^2+1)\\=(x-1)(x+1)(x^2-1)\\b) 4x^4+81= (2x^2)^2+36x^2+81-36x^2\\=(2x)^2+2.2x^2.9+9^2-36x^2\\=(2x^2+9)^2-(6x)^2\\=(2x^2+9-6x)(2x^2+9+6x)\)
\(a) x^2y+xy^3-xy-y^3\\=(x^2y+xy^3)-(xy+y^3)\\=xy(x+y^2)-y(x+y^2)\\=(x+y^2)(xy-y)\\=y(x+y^2)(x-1)\\b)2x^2+5x+8(xem lại đề)\\c)x^2-10x+21\\=x^2-3x-7x+21\\=x(x-3)-7(x-3)\\=(x-3)(x-7)\)
Dzịt lên chức nhanh thiệt ha!
Mấy hôm nay ôn thi sấp mặt :((. Bùn :((
Anh chép sai đề rồi!
A= \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+9}{9-x}\left(x\ge0;x\ne1;x\ne4\right)\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{-3\sqrt{x}—9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{-3}{\sqrt{X}-3}\\ \)