Ta có :

\(\sqrt{225}-\left(\dfrac{1}{\sqrt{13}}-1\right)=15-\dfrac{1}{\sqrt{13}}+1=16-\dfrac{1}{\sqrt{13}}\)

\(\sqrt{289}-\left(\dfrac{1}{\sqrt{14}}+1\right)=17-\dfrac{1}{\sqrt{14}}-1=16-\dfrac{1}{\sqrt{14}}\)

Vì 13 < 14 \(\Rightarrow\sqrt{13}< \sqrt{14}\)

\(\Rightarrow\dfrac{1}{\sqrt{13}}>\dfrac{1}{\sqrt{14}}\)

\(\Rightarrow16-\dfrac{1}{\sqrt{13}}< 16-\dfrac{1}{\sqrt{14}}\)

\(\Rightarrow\sqrt{225}-\left(\dfrac{1}{\sqrt{13}}-1\right)< \sqrt{289}-\left(\dfrac{1}{\sqrt{14}}+1\right)\)

Ta có: \(\sqrt{225}-\left(\dfrac{1}{\sqrt{13}}-1\right)\)

\(=15-\dfrac{1}{\sqrt{13}}+1\)

\(=\left(15+1\right)-\dfrac{1}{\sqrt{13}}\)

\(=16-\dfrac{1}{\sqrt{13}}\)

Và: \(\sqrt{289}-\left(\dfrac{1}{\sqrt{14}}+1\right)\)

\(=17-\dfrac{1}{\sqrt{14}}-1\)

\(=\left(17-1\right)-\dfrac{1}{\sqrt{14}}\)

\(=16-\dfrac{1}{\sqrt{14}}\)

Vì \(13< 14\Rightarrow\sqrt{13}< \sqrt{14}\Rightarrow\dfrac{1}{\sqrt{13}}>\dfrac{1}{\sqrt{14}}\Rightarrow-\dfrac{1}{\sqrt{13}}< -\dfrac{1}{\sqrt{14}}\Rightarrow16-\dfrac{1}{\sqrt{13}}< 16-\dfrac{1}{\sqrt{14}}\)

Hay \(\sqrt{225}-\left(\dfrac{1}{\sqrt{13}}-1\right)< \sqrt{289}-\left(\dfrac{1}{\sqrt{14}}+1\right)\)

Chúc bn học tốt

√225−(1√13 −1) < √289−(1√14 +1).

√225−(1√13 −1) < √289−(1√14 +1).

a, \(\sqrt{\dfrac{289}{225}}=\sqrt{\dfrac{17^2}{15^2}}=\dfrac{17}{15}\)

b, \(\sqrt{2\dfrac{14}{25}}=\sqrt{\dfrac{64}{25}}=\sqrt{\dfrac{8^2}{5^2}}=\dfrac{8}{5}\)

c, \(\sqrt{\dfrac{0,25}{9}}=\sqrt{\dfrac{0,5^2}{3^2}}=\dfrac{0,5}{3}\)

d, \(\sqrt{\dfrac{8,1}{1,6}}=\sqrt{\dfrac{0,1}{0,1}.\dfrac{81}{16}}=\sqrt{1.\dfrac{81}{16}}=\dfrac{9}{4}\)

Chúc bạn học tốt!!!

a) \(\sqrt{\dfrac{289}{225}}\)

\(=\dfrac{\sqrt{289}}{\sqrt{225}}\)

\(=\dfrac{\sqrt{17^2}}{\sqrt{15^2}}\)

\(=\dfrac{17}{15}\)

b) \(\sqrt{2\dfrac{14}{15}}\)

\(=\sqrt{\dfrac{44}{15}}\)

\(=\dfrac{\sqrt{44}}{\sqrt{15}}\)

\(=\dfrac{2\sqrt{11}}{\sqrt{15}}\)

\(=\dfrac{2\sqrt{165}}{15}\)

c) \(\sqrt{\dfrac{0,25}{9}}\)

\(=\sqrt{\dfrac{1}{\dfrac{4}{9}}}\)

\(=\dfrac{\dfrac{1}{2}}{3}\)

\(=\dfrac{1}{6}\)

d) \(\sqrt{\dfrac{8,1}{1,6}}\)

\(=\sqrt{5,0625}\)

\(=\sqrt{\dfrac{81}{16}}\)

\(=\dfrac{9}{4}\)

Giải:

Ta có tính chất tổng quát:

\(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)^2k-k^2\left(k+1\right)}\)

\(=\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)k\left(k+1-k\right)}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)

Áp dụng vào biểu thức

\(\Rightarrow A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{224}}-\frac{1}{\sqrt{225}}\)

\(=1-\frac{1}{\sqrt{225}}\)

a) Ta có: \(-\dfrac{3}{2}\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2\cdot\left(1+\sqrt{5}\right)^2}\)

\(=\dfrac{-3}{2}\left(\sqrt{5}-2\right)+4\cdot\left(\sqrt{5}+1\right)\)

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

\(=\dfrac{5}{2}\sqrt{5}+7\)

b) Ta có: \(\left(1+\dfrac{1}{\tan^225^0}\right)\cdot\sin^225^0-\tan55^0\cdot\tan35^0\)

\(=\dfrac{\tan^225^0+1}{\tan^225^0}\cdot\sin25^0-1\)

\(=\left(\dfrac{\sin^225^0}{\cos^225^0}+1\right)\cdot\dfrac{\cos^225^0}{\sin^225^0}\cdot\sin25^0-1\)

\(=\dfrac{\sin^225^0+\cos^225^0}{\cos^225^0}\cdot\dfrac{\cos^225^0}{\sin25^0}-1\)

\(=\dfrac{1}{\sin25^0}-1\)

\(=\dfrac{1-\sin25^0}{\sin25^0}\)

a, \(\sqrt{\frac{289}{25}}=\frac{\sqrt{289}}{\sqrt{25}}=\frac{17}{5}\)

b, \(\sqrt{2\frac{14}{25}}=\sqrt{\frac{64}{25}}=\frac{8}{5}\)

c, \(\sqrt{\frac{0,25}{9}}=\frac{\sqrt{0,25}}{\sqrt{9}}=\frac{0,5}{3}=\frac{1}{2}.\frac{1}{3}=\frac{1}{6}\)

d, \(\sqrt{\frac{8,1}{16}}\)đề có sai ko cô ? 

a) căn 289 / 225 = 17/15

b) căn 64/ 25 = 8/5

c) căn 0,25 / 9 = 1/6

d) căn 8,1 / 1,6 = 9/4

289225=289225=1715.

1425=2.25+1425=6425=6425=85.

0,259=0,259=0,53=530=16.

8,11,6=8116=8116=94.

a) \(\sqrt{\dfrac{59}{25}+\dfrac{6}{5}\sqrt{2}}=\sqrt{2+2.\dfrac{3}{5}\sqrt{2}+\dfrac{9}{25}}=\sqrt{\left(\sqrt{2}+\dfrac{3}{5}\right)^2}\)

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

b) \(\sqrt{\dfrac{129}{16}+\sqrt{2}}=\sqrt{8+2.2\sqrt{2}.\dfrac{1}{4}+\dfrac{1}{16}}\)

= \(\sqrt{\left(2\sqrt{2}+\dfrac{1}{4}\right)^2}\) = / \(2\sqrt{2}+\dfrac{1}{4}\) / = \(2\sqrt{2}+\dfrac{1}{4}\)

c) Tương tự , mình bận rồi , nếu chưa biết tẹo mk làm cho.

c) \(\sqrt{\dfrac{289+4\sqrt{72}}{16}}=\sqrt{\dfrac{289}{16}+\dfrac{1}{4}\sqrt{72}}=\sqrt{\dfrac{289}{16}+\dfrac{1}{4}.6\sqrt{2}}=\sqrt{18+2.\dfrac{1}{4}.3\sqrt{2}+\dfrac{1}{16}}=\sqrt{\left(3\sqrt{2}+\dfrac{1}{4}\right)^2}\) = / \(3\sqrt{2}+\dfrac{1}{4}\) / = \(3\sqrt{2}+\dfrac{1}{4}\)