Question

Giải phương trình:

a. \(\sqrt{2}.x-\sqrt{50}=0;\)            b. \(\sqrt{3}.x+\sqrt{3}=\sqrt{12}+\sqrt{27};\)

c. \(\sqrt{3}.x^2-\sqrt{12}=0;\)           d. \(\dfrac{x^2}{\sqrt{5}}-\sqrt{20}=0.\)

Answer 1

a, \(\sqrt{2}x-\sqrt{50}=0\Leftrightarrow\sqrt{2}x-5\sqrt{2}=0\Leftrightarrow\sqrt{2}\left(x-5\right)=0\Leftrightarrow x=5\)

b, \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\Leftrightarrow\sqrt{3}\left(x+1\right)=5\sqrt{3}\Leftrightarrow x+1=5\Leftrightarrow x=4\)

c, \(\sqrt{3}x^2-\sqrt{12}=0\Leftrightarrow\sqrt{3}\left(x^2-2\right)=0\Leftrightarrow x^2-2=0\Leftrightarrow x=\pm\sqrt{2}\)

d, \(\dfrac{x^2}{\sqrt{5}}-\sqrt{20}=0\Leftrightarrow\dfrac{1}{\sqrt{5}}\left(x^2-10\right)=0\Leftrightarrow x^2-10=0\Leftrightarrow x=\pm\sqrt{10}\)

Answer 2

a) √2.x - √50 = 0 √2.x = √50 x =

x = = √25 = 5.

b) ĐS: x = 4.

c) √3. - √12 = 0 √3. = √12 = =

= √4 = 2 x = √2 hoặc x = -√2.

d) ĐS: x = √10 hoặc x = -√10.

Answer 3

a) \(\sqrt{2}.x-\sqrt{50}=0\Leftrightarrow\sqrt{2}\left(x-\sqrt{25}\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)

b) \(\sqrt{3}.x+\sqrt{3}=\sqrt{12}+\sqrt{27}\Leftrightarrow\sqrt{3}\left(x+1\right)=\sqrt{3}\left(\sqrt{4}+\sqrt{9}\right)\Leftrightarrow x+1=2+3\Leftrightarrow x=4\)

c) \(\sqrt{3}.x^2-\sqrt{12}=0\Leftrightarrow\sqrt{3}\left(x^2-\sqrt{4}\right)=0\Leftrightarrow x^2-2=0\Leftrightarrow x=\pm\sqrt{2}\)

d) \(\dfrac{x^2}{\sqrt{5}}-\sqrt{20}=0\Leftrightarrow\dfrac{x^2}{\sqrt{5}}=\sqrt{20}\Leftrightarrow x^2=10\Leftrightarrow x=\pm\sqrt{10}\)