Question

[Alien]

Tìm x, biết:

1/ √3 × x - 3 = √27

2/ √2 × x - √28 = √32

3/ √6 × x - 2√6 = √54

4/ √3 × x - √2 × x = √3 + √2

Answer 1

[YangSu]

\(1,\sqrt{3}x-3=\sqrt{27}\)

\(\Leftrightarrow\sqrt{3}x-3=3\sqrt{3}\)

\(\Leftrightarrow\sqrt{3}\left(x-\sqrt{3}\right)=3\sqrt{3}\)

\(\Leftrightarrow x-\sqrt{3}=3\)

\(\Leftrightarrow x=3+\sqrt{3}\)

\(2,\sqrt{2}x-\sqrt{28}=\sqrt{32}\)

\(\Leftrightarrow\sqrt{2}x-2\sqrt{7}=4\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}x=4\sqrt{2}+2\sqrt{7}\)

\(\Leftrightarrow x=\dfrac{\sqrt{2^2}\left(2\sqrt{2}+\sqrt{7}\right)}{\sqrt{2}}\)

\(\Leftrightarrow x=\sqrt{2}\left(2\sqrt{2}+\sqrt{7}\right)\)

\(\Leftrightarrow x=4+\sqrt{14}\)

\(3,\sqrt{6}x-2\sqrt{6}=\sqrt{54}\)

\(\Leftrightarrow\sqrt{6}\left(x-2\right)=3\sqrt{6}\)

\(\Leftrightarrow x-2=3\)

\(\Leftrightarrow x=5\)

\(4,\sqrt{3}x-\sqrt{2}x=\sqrt{3}+\sqrt{2}\)

\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)x=\sqrt{3}+\sqrt{2}\)

\(\Leftrightarrow x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)