Bài 28 trang 18 SGK Toán 9 tập 1

Đề bài

Tính:

a) \( \sqrt{\dfrac{289}{225}}\);                                 b) \( \sqrt{2\dfrac{14}{25}}\);

c) \( \sqrt{\dfrac{0,25}{9}}\) ;                               d) \( \sqrt{\dfrac{8,1}{1,6}}\).

Lời giải chi tiết

a) Ta có:

\(\sqrt{\dfrac{289}{225}}=\dfrac{\sqrt{289}}{\sqrt{225}}=\dfrac{\sqrt {17^2}}{\sqrt{15^2}}=\dfrac{17}{15}\).

b) Ta có:

\(\sqrt{2\dfrac{14}{25}}=\sqrt{\dfrac{2.25+14}{25}}=\sqrt{\dfrac{50+14}{25}}\)

\(=\sqrt{\dfrac{64}{25}}=\dfrac{\sqrt{64}}{\sqrt{25}}=\dfrac{\sqrt{8^2}}{\sqrt{5^2}}=\dfrac{8}{5}\).

c) Ta có:

\(\sqrt{\dfrac{0,25}{9}}=\dfrac{\sqrt{0,25}}{\sqrt{9}}=\dfrac{\sqrt{0,5^2}}{\sqrt{3^2}}=\dfrac{0,5}{3}\)

\(=0,5.\dfrac{1}{3}=\dfrac{1}{2}.\dfrac{1}{3}=\dfrac{1}{6}\).

d) Ta có:

\(\sqrt{\dfrac{8,1}{1,6}}=\sqrt{\dfrac{81.0,1}{16.0,1}}=\sqrt{\dfrac{81}{16}}=\dfrac{\sqrt{81}}{\sqrt{16}}=\dfrac{\sqrt{9^2}}{\sqrt{4^2}}=\dfrac{9}{4}\).

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