\(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}=4\sqrt{b}+2.2\sqrt{10b}-3.3\sqrt{10b}=4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}=4\sqrt{b}-5\sqrt{10b}\)

4) \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\)

\(=\sqrt{4^2\cdot b}+2\sqrt{2^2\cdot10b}-3\sqrt{3^2\cdot10b}\)

\(=4\sqrt{b}+2\cdot2\sqrt{10b}-3\cdot3\sqrt{10b}\)

\(=4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)

\(=4\sqrt{b}+\left(4\sqrt{10b}-9\sqrt{10b}\right)\)

\(=4\sqrt{b}-5\sqrt{10b}\)

`a, sqrt(16b) + 2 sqrt(40b) - 3 sqrt(90b)`

`= 4sqrtb + 2sqrt(8.5b) - 3 sqrt(9.10b)`

`= 4 sqrt b + 4sqrt(10b) - 9 sqrt(10b)`

`= 4sqrtb-5sqrt(10b)`.

2) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)

\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}\)

\(=\left(7-6+1\right)\sqrt{2}\)

\(=2\sqrt{2}\)

3) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}\)

\(=3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\)

\(=\left(3-4+7\right)\sqrt{a}\)

\(=6\sqrt{a}\)

4) \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\)

\(=4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)

\(=4\sqrt{b}-5\sqrt{10b}\)

Gấp nha 

\(=7\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}=7\sqrt{b}-5\sqrt{10b}\)

ĐS: a)

b)

c)

d)

a) \(\sqrt{75}+\sqrt{48}-\sqrt{300}\) = \(5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\) = \(-\sqrt{3}\)

b) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\) = \(7\sqrt{2}-6\sqrt{2}+\sqrt{2}\) = \(2\sqrt{2}\)

c) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}\) = \(3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\) = \(6\sqrt{a}\)

d) \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\) = \(4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)

= \(4\sqrt{b}-5\sqrt{10b}\)

\(A=\sqrt{4^2b}+2\sqrt{2^2\cdot10b}-3\sqrt{3^2\cdot10b}=4\sqrt{b}+4\sqrt{10}\cdot\sqrt{b}-9\sqrt{10}\cdot\sqrt{b}\)

\(=4\sqrt{b}-5\sqrt{10}\sqrt{b}=\left(4-5\sqrt{10}\right)\sqrt{b}\)

Rut gon A = √16b+2√40b−3√90bva`b≥0

 A=√42b+2√22·10b−3√32·10b=4√b+4√10·√b−9√10·√b

=4√b−5√10√b=(4−5√10)√b

câu a ở phần mẫu của cụm đầu tiên cái \(\left(\sqrt{a+\sqrt{b}}\right)^2\rightarrow\left(\sqrt{a}+\sqrt{b}\right)^2\) giúp em với ạ ( em cảm ơn )

a

\(=\dfrac{a-2\sqrt{ab}+b+4\sqrt{ab}}{a+2\sqrt{ab}+b-4\sqrt{ab}}.\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}.\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2.\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^3}{\left(\sqrt{a}-\sqrt{b}\right)^3}\)