Question

Phân tích đa thức thành nhân tử :

a. \(\sqrt{ax}+\sqrt{by}-\sqrt{ay}-\sqrt{bx}\)

b. \(\sqrt{a^2-b^2}-\sqrt{a^3+b^3}\)

c. \(x-3\sqrt{x}-18\)

d. \(x\sqrt{x}+4x-12\sqrt{x}-27\)

Answer 1

a/ \(=\sqrt{a}\left(\sqrt{x}-\sqrt{y}\right)-\sqrt{b}\left(\sqrt{x}-\sqrt{y}\right)=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{a}-\sqrt{b}\right)\)

b/ \(=\sqrt{\left(a-b\right)\left(a+b\right)}-\sqrt{\left(a+b\right)\left(a^2-ab+b^2\right)}\)

\(=\sqrt{a+b}\left(\sqrt{\left(a-b\right)\left(a^2-ab+b^2\right)}\right)\)

c/ \(=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{81}{4}=\left(\sqrt{x}-\frac{3}{2}-\frac{9}{2}\right)\left(\sqrt{x}-\frac{3}{2}+\frac{9}{2}\right)=\left(\sqrt{x}-6\right)\left(\sqrt{x}+3\right)\)

Answer 2

\(a.\sqrt{ax}+\sqrt{by}-\sqrt{ay}-\sqrt{bx}\\ =\left(\sqrt{ax}-\sqrt{ay}\right)-\left(\sqrt{bx}-\sqrt{by}\right)\\ =\sqrt{a}\left(\sqrt{x}-\sqrt{y}\right)-\sqrt{b}\left(\sqrt{x}-\sqrt{y}\right)\\ =\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{x}-\sqrt{y}\right)\)

\(b.\sqrt{a^2-b^2}-\sqrt{a^3+b^3}\\ =\sqrt{\left(a+b\right)\left(a-b\right)}-\sqrt{\left(a+b\right)\left(a^2-ab+b^2\right)}\\ =\sqrt{a+b}\left(\sqrt{a-b}-\sqrt{a^2-ab+b^2}\right)\)

\(c.x-3\sqrt{x}-18=x-6\sqrt{x}+3\sqrt{x}-18\\ =\sqrt{x}\left(\sqrt{x}-6\right)+3\left(\sqrt{x}-6\right)\\ =\left(\sqrt{x}+3\right)\left(\sqrt{x}-6\right)\)

\(d.x\sqrt{x}+4x-12\sqrt{x}-27=\left(\sqrt{x^3}-27\right)+\left(4x-12\sqrt{x}\right)\\ =\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)+4\sqrt{x}\left(\sqrt{x}-3\right)\\ =\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9+4\sqrt{x}\right)\\ =\left(\sqrt{x}-3\right)\left(x+7\sqrt{x}+9\right)\)

(có gì sai mong mọi người góp ý)