Học giải bài 33 trang 19 SGK Toán lớp 9 tập 1

Đề bài
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) \(\dfrac{x^2}{\sqrt 5 } - \sqrt {20} = 0\)
Giải
a)
\(\sqrt{2}.x - \sqrt{50} = 0\)
\(\Leftrightarrow \sqrt{2}x=\sqrt{50}\)
\(\Leftrightarrow x=\dfrac{\sqrt{50}}{\sqrt{2}}\)
\(\Leftrightarrow x =\sqrt{\dfrac{50}{2}}\)
\(\Leftrightarrow x= \sqrt{25}\)
\(\Leftrightarrow x= \sqrt{5^2}\)
\(\Leftrightarrow x=5\).
Vậy \(x=5\).
b)
\(\sqrt{3}.x + \sqrt{3} = \sqrt{12} + \sqrt{27}\)
\( \Leftrightarrow \sqrt{3}.x = \sqrt{12} + \sqrt{27} - \sqrt{3}\)
\(\Leftrightarrow \sqrt{3}.x=\sqrt{4.3}+\sqrt{9.3}- \sqrt{3}\)
\(\Leftrightarrow \sqrt{3}.x=\sqrt{4}. \sqrt{3}+\sqrt{9}. \sqrt{3}- \sqrt{3}\)
\(\Leftrightarrow \sqrt{3}.x=\sqrt{2^2}. \sqrt{3}+\sqrt{3^3}. \sqrt{3}- \sqrt{3}\)
\(\Leftrightarrow \sqrt{3}.x=2 \sqrt{3}+3\sqrt{3}- \sqrt{3}\)
\(\Leftrightarrow \sqrt{3}.x=(2+3-1).\sqrt{3}\)
\(\Leftrightarrow \sqrt{3}.x=4\sqrt{3}\)
\(\Leftrightarrow x=4\).
Vậy \(x=4\).
c)
\(\sqrt{3}x^2-\sqrt{12}=0\)
\(\Leftrightarrow \sqrt{3}x^2=\sqrt{12}\)
\(\Leftrightarrow \sqrt{3}x^2=\sqrt{4.3}\)
\(\Leftrightarrow \sqrt{3}x^2=\sqrt{4}.\sqrt 3\)
\(\Leftrightarrow x^2=\sqrt{4}\)
\(\Leftrightarrow x^2=\sqrt{2^2}\)
\(\Leftrightarrow x^2=2\)
\(\Leftrightarrow \sqrt{x^2}=\sqrt{2}\)
\(\Leftrightarrow |x|= 2\)
\(\Leftrightarrow x= \pm 2\).
Vậy \(x= \pm 2\).
d)
\(\dfrac{x^{2}}{\sqrt{5}}- \sqrt{20} = 0\)
\(\Leftrightarrow \dfrac{x^2}{\sqrt{5}}=\sqrt{20}\)
\(\Leftrightarrow x^2=\sqrt{20}.\sqrt{5}\)
\(\Leftrightarrow x^2=\sqrt{20.5}\)
\(\Leftrightarrow x^2=\sqrt{100}\)
\(\Leftrightarrow x^2=\sqrt{10^2}\)
\(\Leftrightarrow x^2=10\)
\(\Leftrightarrow \sqrt{x^2}=\sqrt {10}\)
\(\Leftrightarrow |x|=\sqrt{10}\)
\(\Leftrightarrow x=\pm \sqrt{10}\).
Vậy \(x= \pm \sqrt{10}\).