\(=4\cdot5+13:7=20+\dfrac{13}{7}=\dfrac{153}{7}\)
tính
a) \(\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}\)
b) 36 : \(\sqrt{2.3^2.18}-\sqrt{169}\)
c) \(\sqrt{\sqrt{81}}\)
d) \(\sqrt{3^2+4^2}\)
Thực hiện phép tính:
a. \(\sqrt{25}+2\sqrt{49}\)
b. \(\sqrt{16}.\sqrt{25}+\sqrt{169}:\sqrt{49}\)
c. \(\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{7}\)
d. \(2\sqrt{3}-\sqrt{75}+2\sqrt{12}\)
Tính
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
b) \(\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}\)
c) \(-0.3.\sqrt{\left(-0.3\right)^2}\)
d) \(36:\sqrt{2.3^2.18}-\sqrt{169}:13\)
Bạn nào giải từng bước giúp mình với, mình cảm ơn nhiều
\(2.\sqrt{16-\sqrt{81}+}\sqrt{25}.\sqrt{49}\)
\(\sqrt{16}\cdot\sqrt{25}+\frac{\sqrt{196}}{\sqrt{49}}\)
Căn thức bậc 2 và hằng đẳng thức \(\sqrt{A^2}=\left|A\right|\)
1, \(\sqrt{\left(2\sqrt{3}-3\right)^2}\)=
2. \(\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}\)
3. 36 :\(\sqrt{18^2}-\sqrt{169}\)
4. \(\sqrt{\sqrt{81}}\)
5 . \(\sqrt{3^2}+4^2\)
Tính :
\(a)\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}\)
\(b)\sqrt{\sqrt{81}}\)
2\(2\sqrt{\frac{16}{3}}-5\sqrt{\frac{32}{25}}+14\sqrt{\frac{18}{49}}\)
e,\(\sqrt{\dfrac{9}{169}}\)
f,\(\sqrt{1\dfrac{9}{16}}\)
g,\(\dfrac{\sqrt{2300}}{\sqrt{23}}\)
h,\(\dfrac{\sqrt{12,5}}{\sqrt{0,5}}\)
