Bài 33 trang 19 SGK Toán 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) \(\dfrac{x^2}{\sqrt 5 } - \sqrt {20}  = 0\)

Hướng dẫn giải

+ \(\sqrt{x^2}=|x|\)

+ \(|x|=x\)   nếu \(x \ge 0\).

   \(|x|=-x\)  nếu \( x<0\).

+\(\dfrac{\sqrt a}{\sqrt b}=\sqrt{\dfrac{a}{b}}\).

Lời giải chi tiết

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)

 \(\sqrt{3}x^2-\sqrt{12}=0\)

 \(\Leftrightarrow \sqrt{3}x^2=\sqrt{12}\)

\(\Leftrightarrow \sqrt{3}x^2=\sqrt{4.3}\)

\(\Leftrightarrow \sqrt{3}x^2=\sqrt{4}.\sqrt 3\)

\(\Leftrightarrow x^2=\sqrt{4}\)

\(\Leftrightarrow x^2=\sqrt{2^2}\)

\(\Leftrightarrow x^2=2\)

\(\Leftrightarrow \sqrt{x^2}=\sqrt{2}\)

\(\Leftrightarrow |x|= \sqrt 2\)

\(\Leftrightarrow x= \pm \sqrt 2\).

Vậy \(x= \pm\sqrt 2\).

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 \(x= \pm \sqrt{10}\).