Bài 11 trang 11 SGK Toán 9 tập 1

Tính:

a) \(\sqrt{16}.\sqrt{25} + \sqrt{196}:\sqrt{49}\);

b) \(36:\sqrt{2.3^2.18}-\sqrt{169}\);

c) \(\sqrt{\sqrt{81}}\);

d) \( \sqrt{3^{2}+4^{2}}\).

Lời giải

a) Ta có: \(\sqrt{16}.\sqrt{25} + \sqrt{196}:\sqrt{49}\)

\(=\sqrt{4^2}.\sqrt{5^2}+\sqrt{14^2}:\sqrt{7^2}\)

\(=\left| 4 \right| . \left| 5 \right| + \left| {14} \right| : \left| 7 \right|\)

\(=4.5+14:7 \)

\(=20+2=22 \).

b) Ta có:

 \(36:\sqrt{2.3^2.18}-\sqrt{169} = 36: \sqrt{(2.3^2).18}-\sqrt{13^2} \)

                                         \(=36:\sqrt{(2.9).18} - \left| 13 \right| \)

                                         \(=36:\sqrt{18.18}-13\)  

                                         \(=36:\sqrt{18^2}-13 \)

                                         \(=36: \left|18 \right| -13\)

                                         \(=36:18-13 \)

                                         \(=2-13=-11\).

 

c) Ta có: \(\sqrt{81}=\sqrt{9^2}=\left| 9 \right| = 9\).

 \( \Rightarrow \sqrt{\sqrt{81}}\)=\(\sqrt{9}= \sqrt{3^2}=\left| 3 \right| =3\).

d) Ta có: \(\sqrt{3^{2}+4^{2}}=\sqrt{16+9}=\sqrt{25}=\sqrt{5^2}=\left|5 \right| =5\).


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