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

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

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

\(=27-2\)

\(=25\)

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

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

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

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

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

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

\(\Rightarrow25\)

Vậy: BT = 25

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

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

\(=27-2=25\)

\(1,4\sqrt{5}-5\sqrt{2}+12\sqrt{5}-8\sqrt{5}=8\sqrt{5}-5\sqrt{2}\)

\(2,\left(\sqrt{27}+\sqrt{32}-\sqrt{50}\right)\left(\sqrt{27}-\sqrt{32}+\sqrt{50}\right)\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

= ( 3 √3-√ 2) *( 3 √ 3 +√ 2)

=(3 √ 3)^2-2

=25

bạn có thể kt lại bằng máy tính

\(A=10-\left(\sqrt{32}-\sqrt{8}-\sqrt{27}\right)\left(\sqrt{8}-\sqrt{32}-\sqrt{27}\right)\)

\(A=10-\left[-\sqrt{27}+\left(\sqrt{32}-\sqrt{8}\right)\right]\left[-\sqrt{27}-\left(\sqrt{32}-\sqrt{8}\right)\right]\)

\(A=10-\left[\left(-\sqrt{27}\right)^2-\left(\sqrt{32}-\sqrt{8}\right)^2\right]\)

\(A=10-\left(27-8\right)\)

\(A=-9\)

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

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

a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)

b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)

c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)