Học giải bài 11 trang 11 SGK Toán lớp 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) \( \sqrt{3^{2}+4^{2}}\).
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: \(\sqrt{3^{2}+4^{2}}=\sqrt{16+9}=\sqrt{25}=\sqrt{5^2}=\left|5 \right| =5\).