a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)

b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)

c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)

d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)

\(I=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)

\(=\left(2\sqrt{3}-5\sqrt{3}.\sqrt{3^2}+2\sqrt{2^2}.\sqrt{3}\right):\sqrt{3}\)

\(=\left(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}\right):\sqrt{3}\)

\(=-5\sqrt{3}.\dfrac{1}{\sqrt{3}}\)

\(=-5\)

\(K=\sqrt{125}-4\sqrt{45}+3\sqrt{20}-\sqrt{80}\)

\(=\sqrt{5^2.5}-4\sqrt{3^2.5}+3\sqrt{2^2.5}-\sqrt{4^2.5}\)

\(=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}\)

\(=\sqrt{5}.\left(5-12+6-4\right)\)

\(=-5\sqrt{5}\)

\(L=2\sqrt{9}+\sqrt{25}-5\sqrt{4}\)

\(=2\sqrt{3^2}+\sqrt{5^2}-5\sqrt{2^2}\)

\(=2.3+5-5.2\)

\(=1\)

\(N=2\sqrt{32}-5\sqrt{27}-4\sqrt{8}+3\sqrt{75}\)

\(=2.4\sqrt{2}-5.3\sqrt{3}-4.2\sqrt{2}+3.5\sqrt{3}\)

\(=8\sqrt{2}-8\sqrt{2}-15\sqrt{3}+15\sqrt{3}\)

\(=0\)

\(O=2\sqrt{3.5^2}-3\sqrt{3.2^2}+\sqrt{3.3^2}\)

\(=2.5\sqrt{3}-3.2\sqrt{3}+3\sqrt{3}\)

\(=10\sqrt{3}-6\sqrt{3}+3\sqrt{3}\)

\(=7\sqrt{3}\)

\(L=\dfrac{2\sqrt{3}-15\sqrt{3}+8\sqrt{3}}{\sqrt{3}}=2-15+8=-5\)

\(K=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}\)

L=2*3+5-5*2=5-4=1

N=8căn 2-8căn2-15căn3+15căn 3=0

O=10căn 3-6căn3+3căn3=7căn 3

Ối giời! Bấm máy tính đi bn! Người ta sinh ra cái máy tính là để làm mấy việc này mà. :D

\(1.\sqrt{12}-\sqrt{27}+\sqrt{3}\\ =2\sqrt{3}-3\sqrt{3}+\sqrt{3}\\ =0\)

\(2.\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\\ =6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}\\ =0\)

\(3\sqrt{112}-7\sqrt{216}+4\sqrt{54}-2\sqrt{252}-3\sqrt{96}\\ =12\sqrt{7}-42\sqrt{6}+12\sqrt{6}-12\sqrt{7}-12\sqrt{6}\\ =-42\sqrt{6}\)

\(4.\sqrt{18}-3\sqrt{80}-2\sqrt{50}+2\sqrt{45}\\ =3\sqrt{2}-12\sqrt{5}-10\sqrt{2}+6\sqrt{5}\\ =-7\sqrt{2}-6\sqrt{5}\)

\(5.6\sqrt{12}-\sqrt{20}-2\sqrt{27}+\sqrt{125}\\ =12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}\\ =6\sqrt{3}+3\sqrt{5}\)

\(6.2\sqrt{18}-3\sqrt{80}-5\sqrt{147}+5\sqrt{245}-3\sqrt{98}\\ =6\sqrt{2}-12\sqrt{5}-35\sqrt{3}+35\sqrt{5}-21\sqrt{2}\\ =-15\sqrt{2}-35\sqrt{3}+23\sqrt{5}\)

a) \(A=2\sqrt{8}-3\sqrt{32}+\sqrt{50}\)

\(A=2\sqrt{4.2}-3\sqrt{16.2}+\sqrt{25.2}\)

\(A=2.2\sqrt{2}-3.4\sqrt{2}+5\sqrt{2}\)

\(A=4\sqrt{2}-12\sqrt{2}+5\sqrt{2}\)

\(A=\left(4-12+5\right)\sqrt{2}\)

\(A=-3\sqrt{2}\)

b) \(B=\sqrt{12}+4\sqrt{27}-3\sqrt{48}\)

\(B=\sqrt{4.3}+4\sqrt{9.3}-3\sqrt{16.3}\)

\(B=2\sqrt{3}+4.3\sqrt{3}-3.4\sqrt{3}\)

\(B=2\sqrt{3}\)

c) \(C=\sqrt{20a}+4\sqrt{45a}-2\sqrt{125a}\left(a\ge0\right)\)

\(C=\sqrt{4.5a}+4\sqrt{9.5a}-2\sqrt{25.5a}\)

\(C=2\sqrt{5a}+4.3\sqrt{5a}-2.5\sqrt{5a}\)

\(C=2\sqrt{5a}+12\sqrt{5a}-10\sqrt{5a}\)

\(C=\left(2+12-10\right)\sqrt{5a}\)

\(C=4\sqrt{5a}\)

a) ta có \(2\sqrt{8}=2\sqrt{4.2}=4\sqrt{2},3\sqrt{32}=3\sqrt{16.2}=12\sqrt{2},\sqrt{50}=\sqrt{25.2}=5\sqrt{2}\)                               \(\Rightarrow A=4\sqrt{2}-12\sqrt{2}+5\sqrt{2}=-3\sqrt{2}\)                                                                                              b) ta có \(\sqrt{12}=\sqrt{4.3}=2\sqrt{3},4\sqrt{27}=4\sqrt{9.3}=12\sqrt{3},3\sqrt{48}=3\sqrt{16.3}=12\sqrt{3}\Rightarrow B=2\sqrt{3}+12\sqrt{3}-12\sqrt{3}=26\sqrt{3}\)c) ta có \(\sqrt{20a}=\sqrt{4.5a}=2\sqrt{5a},4\sqrt{45a}=4\sqrt{9.5a}=12\sqrt{5a},2\sqrt{125a}=2\sqrt{25.5a}=10\sqrt{5a}\Rightarrow C=2\sqrt{5a}+12\sqrt{5a}-10\sqrt{5a}=4\sqrt{5a}\)   

a, -3 căn 2 

b, 2 căn 3

c, 4 căn (5a)

\(a.\sqrt{72}-5\sqrt{2}+3\sqrt{12}\\ =6\sqrt{2}-5\sqrt{2}+6\sqrt{3}\\ =\sqrt{2}+6\sqrt{3}\\ b.6\sqrt{\dfrac{1}{2}}-\dfrac{2}{\sqrt{2}}-5\sqrt{2}\\ =3\sqrt{2}-\sqrt{2}-5\sqrt{2}\\ =-3\sqrt{2}\\ c.\dfrac{\sqrt{8}-2}{\sqrt{2}-1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{3}{\sqrt{3}}\\ =2+1+\sqrt{3}-\sqrt{3}\\ =3\\ d.\sqrt[3]{64}+\sqrt[3]{27}-2\sqrt[3]{-8}\\ =4+3+4\\ =11\)

\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)

\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)