\(A=\left(\sqrt{5}+3\right)\left(5-\sqrt{15}\right)=5\sqrt{5}-5\sqrt{3}+15-3\sqrt{15}\)

Bạn ghi nhầm đề thì phải, ngoặc đầu là \(\sqrt{5}+\sqrt{3}\) mới rút gọn được theo HĐT số 3

\(B=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)

\(=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\)

\(C=1-\left(3\sqrt{5}-2\sqrt{5}-\sqrt{3}\right)\left(2\sqrt{5}-3\sqrt{5}-\sqrt{3}\right)\)

\(=1-\left(\sqrt{5}-\sqrt{3}\right)\left(-\sqrt{5}-\sqrt{3}\right)=1+\left(5-3\right)=3\)

\(D=\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{2}{3}}\right).\sqrt{6}=\frac{\left(3-2\right)}{\sqrt{6}}.\sqrt{6}=1\)

a) \(2\sqrt{20}-\sqrt{50}+3\sqrt{80}-\sqrt{320}=2\sqrt{2^2.5}-\sqrt{5^2.2}+3\sqrt{4^2.5}-\sqrt{8^2.5}\\ =4\sqrt{5}-5\sqrt{2}+12\sqrt{5}-8\sqrt{5}=8\sqrt{5}-5\sqrt{2}\)

b) \(\sqrt{32}-\sqrt{50}+\sqrt{18}=\sqrt{4^2.2}-\sqrt{5^2.2}+\sqrt{3^2.2}=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)

c) \(3\sqrt{3}+4\sqrt{2}-5\sqrt{27}=3\sqrt{3}+4\sqrt{2}-5\sqrt{3^2.3}=3\sqrt{3}+4\sqrt{2}-15\sqrt{3}=4\sqrt{2}-12\sqrt{3}\)

d) \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)-\sqrt{3}\left(\sqrt{\sqrt{3}+1}-1\right)}{\left(\sqrt{\sqrt{3}+1}-1\right)\left(\sqrt{\sqrt{3}+1}+1\right)}\\ =\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1\right)}{\left(\sqrt{3+1}\right)^2-1^2}\\ =\dfrac{2\sqrt{3}}{\sqrt{3}}=2\)

e)\(\left(2+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=2^2-\left(\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)^2=4-\left(\dfrac{9+6\sqrt{3}+3}{3+2\sqrt{3}+1}\right)\\ =4-\left(\dfrac{6\left(2+\sqrt{3}\right)}{2\left(2+\sqrt{3}\right)}\right)=4-3=1\)

b) \(\sqrt{32}-\sqrt{50}+\sqrt{18}=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=\left(4-5+3\right)\sqrt{2}=2\sqrt{2}\)

Tính hả bạn?

\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\\ =\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\\ =\left(-\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+\sqrt{2}\right)\\ =\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)\\ =\left(3\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2=9.3-2=25\)

\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)

\(=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)

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

\(=9.3-2=27-2=25\)

(\(\sqrt{32}-\sqrt{50}+\sqrt{27}\))(\(\sqrt{27}+\sqrt{50}-\sqrt{32}\))

= \(\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\)\(\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)

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

= \(\left(3\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)

= 27 - 2 = 25

\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)

\(=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)

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

\(=27-2\)

\(=25\)

\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)

\(\Leftrightarrow\left(4\sqrt{2}+3\sqrt{3}-5\sqrt{2}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)

\(\Leftrightarrow\left(4\sqrt{2}+3\sqrt{2}-5\sqrt{2}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)

\(\Leftrightarrow\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)

\(\Leftrightarrow\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+2\right)\)

\(\Leftrightarrow\left(3\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)

\(\Rightarrow25\)

Vậy: BT = 25

P/s: từ dòng thứ 2 trở xuống bạn tự phân ... Vấn đề là ở bạn thôi :)))

\(=\left(\sqrt{27}\right)^2-\left(\sqrt{50}-\sqrt{32}\right)^2\)

\(=27-2=25\)

\(C=\sqrt{15-6\sqrt{6}}+\sqrt{33+12\sqrt{6}}=\sqrt{9-2.3\sqrt{6}+6}+\sqrt{24+2.3.2\sqrt{6}+9}=3-\sqrt{6}+2\sqrt{6}+3=6+\sqrt{6}\) \(D=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\) \(F=\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\)

\(C=\sqrt{15-6\sqrt{6}}+\sqrt{33+12\sqrt{6}}=\sqrt{\left(\sqrt{9}-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}+\sqrt{9}\right)^2}=3-\sqrt{6}+2\sqrt{6}+3=6+\sqrt{6}\)

\(D=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

\(\Rightarrow\sqrt{2}D=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}-1-\sqrt{3}-1=-2\)

\(\Rightarrow D=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)

\(F=\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=\left(3\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2=27-2=25\)

\(1.A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)

\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)

\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)

a) Ta có: \(4\sqrt{28}+3\sqrt{63}-3\sqrt{112}-2\sqrt{175}\)

\(=8\sqrt{7}+9\sqrt{7}-12\sqrt{7}-10\sqrt{7}\)

\(=-5\sqrt{7}\)

b) Ta có: \(\sqrt{5}\left(\sqrt{5}-3\sqrt{20}+2\sqrt{80}\right)\)

\(=\sqrt{5}\left(\sqrt{5}-6\sqrt{5}+8\sqrt{5}\right)\)

\(=\sqrt{5}\cdot3\sqrt{5}=15\)

c) Ta có: \(\left(\sqrt{\dfrac{16}{3}}-\sqrt{\dfrac{25}{3}}\right)\cdot\sqrt{3}\)

\(=\dfrac{-1}{\sqrt{3}}\cdot\sqrt{3}\)

=-1

e) Ta có: \(\left(\sqrt{\dfrac{32}{3}}-\sqrt{54}+\sqrt{\dfrac{50}{3}}\right)\cdot\sqrt{6}\)

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

\(=\dfrac{9\sqrt{12}}{\sqrt{3}}-18\)

\(=0\)

f) Ta có:  \(\left(\sqrt{6}-2\right)\left(\sqrt{3}+\sqrt{2}\right)\)

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

\(=\sqrt{2}\)

1.\(D=\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\frac{10\sqrt{3}}{3}\)\(=\frac{-17\sqrt{3}}{3}\)

2.\(A=27-\left(\sqrt{32}-\sqrt{50}\right)^2=25\)

\(B=1-\left(\left(-\sqrt{3}\right)^2-\left(\sqrt{20}-\sqrt{45}\right)^2\right)\)\(=1-\left(-2\right)=3\)