Ta có: \(30< 36\)

=> \(\sqrt{30}< \sqrt{36}=6\)

=> \(\sqrt{30+\sqrt{30}}< \sqrt{30+6}=6\)

=> \(\sqrt{30+\sqrt{30+\sqrt{30}}}< \sqrt{30+6}=6\)

Cứ tiếp tực như vậy ta sẽ so sánh đc:

\(\sqrt{30+\sqrt{30+\sqrt{30+...+\sqrt{30}}}}< 6\)

1.

Ta có: \(E=\sqrt{37-6\sqrt{30}}=\sqrt{\left(3\sqrt{3}-\sqrt{10}\right)^2}=\left|3\sqrt{3}-\sqrt{10}\right|=3\sqrt{3}-\sqrt{10}\)

\(F=\sqrt{51-6\sqrt{30}}=\sqrt{\left(3\sqrt{5}-\sqrt{6}\right)^2}=\left|3\sqrt{5}-\sqrt{6}\right|=3\sqrt{5}-\sqrt{6}\)

\(G=\sqrt{59-6\sqrt{30}}=\sqrt{\left(3\sqrt{6}-\sqrt{5}\right)^2}=\left|3\sqrt{6}-\sqrt{5}\right|=3\sqrt{6}-\sqrt{5}\)

\(H=\sqrt{17-2\sqrt{30}}=\sqrt{\left(\sqrt{15}-\sqrt{2}\right)^2}=\left|\sqrt{15}-\sqrt{2}\right|=\sqrt{15}-\sqrt{2}\)

\(E=\sqrt{37-6\sqrt{30}}\\ =\sqrt{\left(3\sqrt{3}-\sqrt{10}\right)^2}\\ =\left|3\sqrt{3}-\sqrt{10}\right|\\ =3\sqrt{3}-\sqrt{10}\)

\(F=\sqrt{51-6\sqrt{30}}\\ =\sqrt{\left(3\sqrt{5}-\sqrt{6}\right)^2}\\ =\left|3\sqrt{5}-\sqrt{6}\right|\\ =3\sqrt{5}-\sqrt{6}\)

\(G=\sqrt{59-6\sqrt{30}}\\ =\sqrt{\left(3\sqrt{6}-\sqrt{5}\right)^2}\\ =\left|3\sqrt{6}-\sqrt{5}\right|\\ =3\sqrt{6}-\sqrt{5}\)

\(H=\sqrt{17-2\sqrt{30}}\\ =\sqrt{\left(\sqrt{15}-\sqrt{2}\right)^2}\\ =\left|\sqrt{15}-\sqrt{2}\right|=\sqrt{15}-\sqrt{2}\)

\(E=\sqrt{37-6\sqrt{30}}=3\sqrt{3}-\sqrt{10}\)

\(F=\sqrt{51-6\sqrt{30}}=3\sqrt{5}-\sqrt{6}\)

\(G=\sqrt{59-6\sqrt{30}}=3\sqrt{6}-\sqrt{5}\)

\(H=\sqrt{17-2\sqrt{30}}=\sqrt{15}-\sqrt{2}\)

\(\sqrt{2}+\sqrt{6}+\sqrt{12}+...+\sqrt{110}\)\(=\sqrt{1.2}+\sqrt{2.3}+\sqrt{3.4}+...+\sqrt{10.11}\)

\(< \frac{1+2}{2}+\frac{2+3}{2}+\frac{3+4}{2}+...+\frac{10+11}{2}\)\(=\frac{1}{2}\left[\left(1+2+3+...+10\right)+\left(2+3+4+...+11\right)\right]\)\(=\frac{1}{2}\left(\frac{11.10}{2}+\frac{13.10}{2}\right)=\frac{1}{2}\left(55+65\right)=60\)

Vậy \(\sqrt{2}+\sqrt{6}+\sqrt{12}+...+\sqrt{110}< 60.\)

chị ơi toán lớp 7 hả

\(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{56}+\sqrt{72}+\sqrt{90}+\sqrt{110}\) < 60 nha.

1.

 Ta có: \(A=\sqrt{31-2\sqrt{30}}=\sqrt{\left(\sqrt{30}-1\right)^2}=\left|\sqrt{30}-1\right|=\sqrt{30}-1\)

\(B=\sqrt{11-2\sqrt{30}}=\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}=\left|\sqrt{6}-\sqrt{5}\right|=\sqrt{6}-\sqrt{5}\)

\(C=\sqrt{13-2\sqrt{30}}=\sqrt{\left(\sqrt{10}-\sqrt{3}\right)^2}=\left|\sqrt{10}-\sqrt{3}\right|=\sqrt{10}-\sqrt{3}\)

\(D=\sqrt{39-6\sqrt{30}}=\sqrt{\left(\sqrt{30}-3\right)^2}=\left|\sqrt{30}-3\right|=\sqrt{30}-3\)

\(A=\sqrt{31-2\sqrt{30}}=\sqrt{30}-1\)

\(B=\sqrt{11-2\sqrt{30}}=\sqrt{6}-\sqrt{5}\)

\(C=\sqrt{13-2\sqrt{30}}=\sqrt{10}-\sqrt{3}\)

\(D=\sqrt{39-6\sqrt{30}}=\sqrt{30}-3\)

\(A=\sqrt{6}+\sqrt{12}+\sqrt{30}+\sqrt{56}\)

\(B^2=\left(\sqrt{6}+\sqrt{30}\right)^2=36+2\sqrt{180}>36+26=62\)

B>7;\(\sqrt{30}>5;\sqrt{56}>7\)

A>7+5+7=19

A>19

ghi de sai ban oi

\(A=\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}\)

\(A< \sqrt{2,25}+\sqrt{6,25}+\sqrt{12,25}+\sqrt{20,25}+\sqrt{30,25}+\sqrt{42,25}=24=B\)

Vậy \(A< B\)

Chúc bạn học tốt ~ 

A/đặt A=\(\sqrt{30+12\sqrt{6}}+\sqrt{30-12\sqrt{6}}\)

\(\Leftrightarrow A^2=60+\sqrt{\left(30+12\sqrt{6}\right)\left(30-12\sqrt{6}\right)}\)

=\(60+6=66\)

\(\Leftrightarrow A=\sqrt{66}\)

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

\(\Leftrightarrow B^2=8-\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}=5\)

\(\Leftrightarrow B=\sqrt{5}\)

\(\Leftrightarrow B-\sqrt{2}=\sqrt{5}-\sqrt{2}\)

hay \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\) \(-\sqrt{2}=\sqrt{5}-\sqrt{2}\)

Có: \(\left(\sqrt{30+12\sqrt{6}}-\sqrt{30-12\sqrt{6}}\right)^2\)

\(=30+12\sqrt{6}-2\sqrt{\left(30+12\sqrt{6}\right)\left(30-12\sqrt{6}\right)}+30-12\sqrt{6}\)

\(=60-2\sqrt{30^2-12^2\cdot6}\)          (hằng đẳng thức số 3)

\(=60-2\sqrt{36}=60-2\cdot6=60-12=48\)

=>\(\sqrt{30+12\sqrt{6}}-\sqrt{30-12\sqrt{6}}=\sqrt{48}=4\sqrt{3}\)