a) \(A=7+2\sqrt{10}\)

\(2A=14+4\sqrt{10}\)

\(2A=10+4\sqrt{10}+4\)

\(2A=\left(\sqrt{10}+2\right)^2\)

\(A=\frac{\left(\sqrt{10}+2\right)^2}{2}\)

b) \(B=11-2\sqrt{28}=11-4\sqrt{7}\)

\(B=7-4\sqrt{7}+4\)

\(B=\left(\sqrt{7}-2\right)^2\)

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

d) \(D=7+4\sqrt{3}=3+4\sqrt{3}+4=\left(\sqrt{3}+2\right)^2\)

a: \(4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

b; \(7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)

c: \(13-4\sqrt{3}=\left(2\sqrt{3}-1\right)^2\)

mk chỉnh lại đề, đúng thì bạn tham khảo

\(\sqrt{25+4\sqrt{10}+4\sqrt{15}+2\sqrt{6}}\)

\(=\sqrt{2+3+20+2.\sqrt{2}.\sqrt{3}+2.\sqrt{3}.2\sqrt{5}+2.2\sqrt{5}.\sqrt{2}}\)

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

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

\(11-6\sqrt{2}=\left(3-\sqrt{2}\right)^2\)

\(6+4\sqrt{2}=\left(2+\sqrt{2}\right)^2\)

\(12-2\sqrt{35}\)

\(=\left(\sqrt{5}\right)^2+\left(\sqrt{7}\right)^2-2\sqrt{35}\)

\(=\left(\sqrt{5}+\sqrt{7}\right)^2\)

\(7+\sqrt{40}\)

\(=\left(\sqrt{5}\right)^2+\left(\sqrt{2}\right)^2+2\sqrt{10}\)

\(=\left(\sqrt{5}+\sqrt{2}\right)^2\)

a) \(12-2\sqrt{35}=\left(\sqrt{5}\right)^2-2\sqrt{5.7}+\left(\sqrt{7}\right)^2=\left(\sqrt{5}-\sqrt{7}\right)^2\)

b) \(7+\sqrt{40}=7+\sqrt{4.10}=7+2\sqrt{10}=\left(\sqrt{5}\right)^2+2\sqrt{5.2}+\left(\sqrt{2}\right)^2=\left(\sqrt{5}+\sqrt{2}\right)^2\)