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a) \(\sqrt{16}\cdot\sqrt{25}+\sqrt{196}:\sqrt{49}\)
\(=\sqrt{16\cdot25}+\sqrt{196:49}\)
\(=20+2=22\)
b) \(36:\sqrt{2\cdot3^2\cdot18}-\sqrt{169}\)
\(=36:\sqrt{324}-\sqrt{169}\)
\(=36:18-13=2-13=-11\)
c) \(\sqrt{\sqrt{81}}\)
\(=\sqrt{9}=3\)
d) \(\sqrt{3^2+4^2}\)
\(=\sqrt{9+16}=\sqrt{25}=5\)
a) \(\sqrt{16}.\sqrt{25}+\sqrt{196}\div\sqrt{49}\)
\(=4.5+14:7\)
\(=20+2=22\)
b) \(36:\sqrt{2.3^2.18}-\sqrt{169}\)
\(=36:18-13=-11\)
c) \(\sqrt{\sqrt{81}}=\sqrt{9}=3\)
d) \(\sqrt{3^2+4^2}=\sqrt{25}=5\)
1. Rút gọn biểu thức:
a) \(\sqrt{\dfrac{81}{25}.\dfrac{49}{16}.\dfrac{9}{196}}=\sqrt{\dfrac{81}{25}}.\sqrt{\dfrac{49}{16}}.\sqrt{\dfrac{9}{4.49}}=\dfrac{9}{5}.\dfrac{7}{4}.\dfrac{3}{2.7}=\dfrac{9.3}{5.4.2}=\dfrac{27}{40}\)
b) \(\sqrt{72}-5\sqrt{2}-\sqrt{49.3}+\sqrt{48}+\sqrt{12}=\)
\(=\sqrt{9.4.2}-5\sqrt{2}-\sqrt{49.3}+\sqrt{16.3}+\sqrt{4.3}\)
\(=3.2\sqrt{2}-5\sqrt{2}-7\sqrt{3}+4\sqrt{3}+2\sqrt{3}\)
\(=6\sqrt{2}-5\sqrt{2}-7\sqrt{3}+4\sqrt{3}+2\sqrt{3}\)
\(=\sqrt{2}-\sqrt{3}\)
c) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}=\) \(=\left|2-\sqrt{3}\right|+\left|2+\sqrt{3}\right|=2-\sqrt{3}+2+\sqrt{3}=4\)
d) \(\sqrt{5}+\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}=\)
\(=\sqrt{5}+\sqrt{4.5}-\sqrt{9.5}+3\sqrt{9.2}+\sqrt{9.4.2}\)
\(=\sqrt{5}+2\sqrt{5}-3\sqrt{5}+3.3\sqrt{2}+3.2\sqrt{2}\)
\(=\sqrt{5}+2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
\(=15\sqrt{2}\)
a. Ta thấy \(\left(a\sqrt{5}\right)^2=\left(a\sqrt{3}\right)^2+\left(a\sqrt{2}\right)^2\Rightarrow AB^2=BC^2+AC^2\)
\(\Rightarrow\Delta ABC\)vuông tại C
b. \(\sin B=\frac{AC}{AB}=\frac{\sqrt{2}}{\sqrt{5}}=\frac{\sqrt{10}}{5};\cos B=\frac{CB}{AB}=\frac{\sqrt{3}}{\sqrt{5}}=\frac{\sqrt{15}}{5}\)
\(\tan B=\frac{AC}{AB}=\frac{\sqrt{6}}{3};\cot B=\frac{\sqrt{6}}{2}\)
\(\sin A=\cos B=\frac{\sqrt{15}}{5};\cos A=\sin B=\frac{\sqrt{10}}{5}\)
\(\tan A=\cot B=\frac{\sqrt{6}}{2};\cot A=\tan B=\frac{\sqrt{6}}{3}\)
a)\(\sqrt{25}+2\sqrt{49}=5+2\cdot7=5+14=19\)
b) \(\sqrt{16}\cdot\sqrt{25}+\sqrt{169}:\sqrt{49}=4\cdot5+13:7=20+\dfrac{13}{7}\) = \(\dfrac{153}{7}\)
c) \(\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{7}=3-\sqrt{7}+\sqrt{7}=3\)
d) \(2\sqrt{3}-\sqrt{75}+2\sqrt{12}=2\sqrt{3}-5\sqrt{3}+4\sqrt{3}\) \(=\sqrt{3}\)
1
a,\(\sqrt{\dfrac{36}{121}}=\sqrt{\dfrac{6^2}{11^2}}=\dfrac{6}{11}\)
\(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}=\sqrt{\dfrac{81}{100}}=\sqrt{\dfrac{9^2}{10^2}}=\dfrac{9}{10}\)
Bài làm:
Bài 1:
a)\(\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}\)
= 4.5 + 14 : 7
= 20 + 2
= 22
b)\(36:\sqrt{2.3^2.18}-\sqrt{169}\)
= 36 : 18 - 14
= 2 - 14
= - 12
c)\(\sqrt{\sqrt{81}}\) = \(\sqrt{9}\) = 3
d)\(\sqrt{3^2+4^2}\)
= \(\sqrt{9+16}\)
= \(\sqrt{25}\)
= 5
Làm sai rồi
Mình biết làm rồi
Chỉ là ra đề cho người khác thôi
Sai chỗ nào vậy Thái Viết Nam
\(\sqrt{169}=13\) chứ ko phải 14