a) x<9

b)x\(\in\)R

c) -3\(\le\)x\(\le\)3

d) x>1 hoặc x<-2

e) x\(\ge\)0 ; x\(\ne\)9

a) \(\sqrt{7}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)

Vậy \(\sqrt{7}+\sqrt{15}< 7\)

b) \(\sqrt{17}+\sqrt{5}+1>\sqrt{16}+\sqrt{4}+1=4+2+1=7\)\(=\sqrt{49}>\sqrt{45}\)

c) \(\dfrac{23-2\sqrt{19}}{3}< \dfrac{23-2\sqrt{16}}{3}=\dfrac{23-2.4}{3}=5=\sqrt{25}< \sqrt{27}\)

d) Giả sử \(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\Leftrightarrow\left(\sqrt{3\sqrt{2}}\right)^2>\left(\sqrt{2\sqrt{3}}\right)^2\)

\(\Leftrightarrow3\sqrt{2}>2\sqrt{3}\Leftrightarrow\sqrt{18}>\sqrt{12}\Leftrightarrow18>12\)

BĐT cuối cùng đúng nên \(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)