C=\(\sqrt{12}\)+\(\sqrt{27}\)-\(\sqrt{48}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


a: =2015+6-5=2016
b: =10căn 2+5căn 2-6căn 2=9căn 2
c: =3căn 3-4căn 3-5căn 3=-6căn 3
d: =2căn 3+3căn 3-4căn 3=căn 3
\(A=2015+6-5==2015+1=2016\)
\(B=5\sqrt{2^3}+\sqrt{5^2.2}-2\sqrt{3^2.2}\\ =10\sqrt{2}+5\sqrt{2}-6\sqrt{2}\\ =\left(10+5-6\right)\sqrt{2}=9\sqrt{2}\)
\(C=\sqrt{3^3}-2\sqrt{2^2.3}-\sqrt{5^2.3}\\ =3\sqrt{3}-4\sqrt{3}-5\sqrt{3}\\ =\left(3-4-5\right)\sqrt{3}=-6\sqrt{3}\)
\(D=\sqrt{2^2.3}+\sqrt{3^3}-\sqrt{4^2.3}\\ =2\sqrt{3}+3\sqrt{3}-4\sqrt{3}\\ =\left(2+3-4\right)\sqrt{3}=\sqrt{3}\)

a: \(A=6\sqrt{3}+10\sqrt{3}-12\sqrt{3}=4\sqrt{3}\)
b: \(B=7\sqrt{3}+5\sqrt{3}-12\sqrt{3}=0\)
c: \(=12\sqrt{2}-6+3\left(9-4\sqrt{2}\right)=12\sqrt{2}-6+27-12\sqrt{2}=21\)
d: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)

\(1,=2\sqrt{3}-3\sqrt{3}+4\sqrt{3}=3\sqrt{3}\\ 2,=\left(2\sqrt{6}+2\sqrt{5}-4\sqrt{5}\right):5=\dfrac{2\sqrt{6}}{5}-\dfrac{2\sqrt{5}}{5}\\ 3,=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-\dfrac{9\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\\ 4,Sửa:\dfrac{1}{\sqrt{5}-\sqrt{3}}-\dfrac{1}{\sqrt{5}+\sqrt{3}}\\ =\dfrac{\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)
1) \(=2\sqrt{3}-3\sqrt{3}+4\sqrt{3}=3\sqrt{3}\)
2) \(=\left(2\sqrt{6}+2\sqrt{5}-4\sqrt{5}\right)=\dfrac{2\sqrt{6}}{5}+\dfrac{2\sqrt{5}}{5}-\dfrac{4\sqrt{5}}{5}\)
3) \(=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)
4) \(=\dfrac{\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{5-3}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)

a) Ta có: \(A=\sqrt{12}+2\sqrt{27}-3\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}-12\sqrt{3}\)
\(=-4\sqrt{3}\)
b) Ta có: \(C=\sqrt{20a}+4\sqrt{45a}-2\sqrt{125a}\)
\(=2\sqrt{5a}+12\sqrt{5a}-10\sqrt{5a}\)
\(=4\sqrt{5a}\)

\(A=\sqrt{12}+2\sqrt{27}+3\sqrt{45}-9\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}+9\sqrt{5}-36\sqrt{3}\)
\(=9\sqrt{5}-28\sqrt{3}\)
\(B=\left(\sqrt{48}-2\sqrt{75}+\sqrt{108}-\sqrt{147}\right):\sqrt{3}\)
\(=4-2\cdot5+6-7\)
\(=4-10+6-7\)
=-7
A=\(\sqrt{12}\)+2\(\sqrt{27}\)+3\(\sqrt{45}\) -9\(\sqrt{48}\)
=\(\sqrt{4.3}\) +2\(\sqrt{9.3}\)+3\(\sqrt{9.5}\) -9\(\sqrt{16.3}\)
=2\(\sqrt{3}\) +6\(\sqrt{3}\)+9\(\sqrt{5}\) -36\(\sqrt{3}\)
=\(\sqrt{3}\)(2+6-36) + 9\(\sqrt{5}\)
=9\(\sqrt{5}\)- 28\(\sqrt{3}\)

\(\sqrt{12}+4\sqrt{27}-3\sqrt{48}\)
\(=\sqrt{2^2\cdot3}+4\sqrt{3^2\cdot3}-3\sqrt{4^2\cdot3}\)
\(=2\sqrt{3}+4\cdot3\sqrt{3}-3\cdot4\sqrt{3}\)
\(=2\sqrt{3}+12\sqrt{3}-12\sqrt{3}\)
\(=2\sqrt{3}\)
\(=2\sqrt{3}+4\cdot3\sqrt{3}-3\cdot4\sqrt{3}\)
\(=2\sqrt{3}+12\sqrt{3}-12\sqrt{3}\)
=2căn 3

Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(=\dfrac{4\sqrt{2}-3\sqrt{2}}{3\sqrt{2}-4\sqrt{2}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=-1-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-6-\sqrt{6}}{6}\)

\(=\dfrac{2\sqrt{8}-2\sqrt{3}}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=\dfrac{-2}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}=\dfrac{-3}{\sqrt{6}}=\dfrac{-3\cdot\sqrt{6}}{6}=\dfrac{-\sqrt{6}}{2}\)

\(=2\sqrt{80\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{5\sqrt{3}}-4\sqrt{45\sqrt{3}}\)
\(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-12\sqrt{5\sqrt{3}}\)
=0
\(C=\sqrt{12}+\sqrt{27}-\sqrt{48}=2\sqrt{3}+3\sqrt{3}-4\sqrt{3}=\sqrt{3}\)
C=1.1732050808