`ĐKXĐ: x^2 -6xy+9y^2 >= 0`

`<=>x^2 -2.3.x.y+(2y)^2 >= 0`

`<=>(x-2y)^2 >=0`(luôn đúng `forall x ,y`)

Vậy biểu thức được xác định ` forall x,y in R`

\(a,3x+5\sqrt{x}-2\\ =3x+6\sqrt{x}-\sqrt{x}-2\\ =3\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)\\ =\left(3\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\\ b,16-x^2\\ =4^2-x^2\\ =\left(4-x\right)\left(4+x\right)\\ c,x^2+\sqrt{x}\\ =\sqrt{x^4}+\sqrt{x}\\ =\sqrt{x}\left(\sqrt{x^3}+1^3\right)\\ =\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)

\(\dfrac{2}{3}-\dfrac{1}{3}\left(\dfrac{4}{9}-\dfrac{5}{6}\right):\dfrac{7}{12}\\ =\dfrac{2}{3}-\dfrac{1}{3}.\left(-\dfrac{7}{18}\right):\dfrac{7}{12}\\ =\dfrac{2}{3}+\dfrac{7}{54}.\dfrac{12}{7}\\ =\dfrac{2}{3}+\dfrac{2}{9}\\ =\dfrac{6}{9}+\dfrac{2}{9}\\ =\dfrac{8}{9}\)

\(a,x-6\sqrt{x}+9\\ =\sqrt{x^2}-2.3\sqrt{x}+3^2\\ =\left(\sqrt{x}-3\right)^2\\ b,x-\sqrt{x}+\dfrac{1}{4}\\ =\sqrt{x^2}-2.\dfrac{1}{2}\sqrt{x}+\left(\dfrac{1}{2}\right)^2\\ =\left(\sqrt{x}-\dfrac{1}{2}\right)^2\\ c,9x-6\sqrt{x}+1\\ =\left(3\sqrt{x}\right)^2-2.3\sqrt{x}.1+1^2\\ =\left(3\sqrt{x}-1\right)^2\\ d,x+5\sqrt{x}-14\\ =x+7\sqrt{x}-2\sqrt{x}-14\\ =\sqrt{x}\left(\sqrt{x}+7\right)-2\left(\sqrt{x}+7\right)\\ =\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)\)

\(\sqrt{29^2-20^2}\\ =\sqrt{\left(29-20\right)\left(29+20\right)}\\ =\sqrt{9.49}\\ =\sqrt{9}.\sqrt{49}\\ =3.7\\ =21\)

\(\dfrac{-2}{3}+1,2.1\dfrac{1}{3}-\dfrac{2}{15}\\ =\dfrac{-2}{3}+\dfrac{6}{5}.\dfrac{4}{3}-\dfrac{2}{15}\\ =\dfrac{-2}{3}+\dfrac{8}{5}-\dfrac{2}{15}\\ =\dfrac{-10}{15}+\dfrac{24}{15}-\dfrac{2}{15}\\ =\dfrac{12}{15}\\ =\dfrac{4}{5}\)

ukm

\(a,x\sqrt{x}-2x+\sqrt{x}\\ =\sqrt{x}\left(x-2\sqrt{x}+1\right)\\ =\sqrt{x}\left(\sqrt{x}-1\right)^2\\ b,3x-9\sqrt{x}\\ =3\sqrt{x}\left(\sqrt{x}-3\right)\\ c,25x-9\\ =\left(5\sqrt{x}\right)^2-3^2\\ =\left(5\sqrt{x}-3\right)\left(5\sqrt{x}+3\right)\\ =d,x\sqrt{x}-8\\ =\sqrt{x^3}-2^3\\ =\left(\sqrt{x}-2\right)\left(x-2\sqrt{x}+4\right)\)