a, \(\sqrt{2}\)+ \(\sqrt{7-2\sqrt{10}}\)= \(\sqrt{2}\)+  \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)

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

                                                    = \(\sqrt{5}\)    

b,   1 -  \(\sqrt{7+2\sqrt{6}}\)= 1-  \(\sqrt{\left(\sqrt{6}+1\right)}\)=   1 - \(\sqrt{6}\)- 1 =  \(\sqrt{6}\)

#mã mã#

a,\(x\sqrt{y}+y\sqrt{x}=\sqrt{x}\sqrt{y}.\left(\sqrt{x}+\sqrt{y}\right).\)

c,\(\sqrt{a}-a^2=\sqrt{a}.\left(1-a\sqrt{a}\right)\)

d,\(x-5\sqrt{x}+6=x-3\sqrt{x}-2\sqrt{x}+6\)

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

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

=>\(VT=< \sqrt{2}+\sqrt{3}\)

xảy ra dấu = khi và chỉ khi x=2

thiếu 1 nghiệm nx bn nhé

Đề đâu hả bạn ~???

Đề đâu rồi bạn ?

\(a,\)\(\left(\sqrt{7}-\sqrt{5}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{7}-\sqrt{5}\right)\)

\(=\left[\left(\sqrt{2}-\sqrt{5}\right)+\sqrt{7}\right]\left[\left(\sqrt{2}-\sqrt{5}\right)-\sqrt{7}\right]\)

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

\(=2-2\sqrt{10}-5-7\)

\(=-10-2\sqrt{10}\)

\(b,B=\sqrt{2}.\sqrt{2+\sqrt{3}}\)

\(\Rightarrow B^2=|2|.|2+\sqrt{3}|\)

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

\(=4+2\sqrt{3}\)

\(=3+2\sqrt{3}+1\)

\(=\sqrt{3}^2+2\sqrt{3}+\sqrt{1}^2\)

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

\(\Rightarrow B=\sqrt{3}+\sqrt{1}=\sqrt{3}+1\)