Bài 10:
a: \(\sqrt{16+9}=\sqrt{25}=5;\sqrt{16}+\sqrt{9}=4+3=7\)
mà 5<7
nên \(\sqrt{16+9}< \sqrt{16}+\sqrt{9}\)
b: \(\sqrt{25-9}=\sqrt{14};\sqrt{25}-\sqrt{9}=5-3=2=\sqrt{4}\)
mà \(\sqrt{14}>\sqrt{4}\)
nên \(\sqrt{25-9}>\sqrt{25}-\sqrt{9}\)
Bài 11:
a: \(\sqrt{35}< \sqrt{36}=6;\sqrt{99}< \sqrt{100}=10\)
Do đó: \(\sqrt{35}+\sqrt{99}< 6+10\)
=>\(\sqrt{35}+\sqrt{99}< 16\)
b: \(\sqrt{26}>\sqrt{25}=5;\sqrt{50}>\sqrt{49}=7\)
Do đó: \(\sqrt{26}+\sqrt{50}>5+7=12\)
a) Giả sử \(\sqrt{1\dfrac{1}{2}}>\sqrt{\dfrac{9}{8}}\)
\(\Leftrightarrow\dfrac{3}{2}>\dfrac{9}{8}\left(đúng\right)\) \(\left(\dfrac{3}{2}=\dfrac{9}{6}>\dfrac{9}{8}\right)\)
Vậy \(\sqrt{1\dfrac{1}{2}}>\sqrt{\dfrac{9}{8}}\)
\(\)b) Giả sử \(\sqrt{\dfrac{14}{39}}>\sqrt{\dfrac{2}{13}}\Leftrightarrow\dfrac{14}{39}>\dfrac{2}{13}\left(đúng\right)\) \(\left(\dfrac{14}{39}>\dfrac{6}{39}=\dfrac{2}{13}\right)\)
Vậy \(\sqrt{\dfrac{14}{39}}>\sqrt{\dfrac{2}{13}}\)
Bài 8 :
a) Giả sử \(\sqrt{1\dfrac{3}{4}}>\sqrt{1.8}\Leftrightarrow\dfrac{7}{4}>8\Leftrightarrow7>32\left(sai\right)\)
Vậy \(\sqrt{1\dfrac{3}{4}}< \sqrt{1.8}\)
b) Giả sử \(-\sqrt{2\dfrac{4}{7}}>-\sqrt{2.5}\Leftrightarrow\sqrt{2\dfrac{4}{7}}< \sqrt{2.5}\Leftrightarrow\dfrac{18}{7}< 10\Leftrightarrow18< 70\left(đúng\right)\)
Vậy \(-\sqrt{2\dfrac{4}{7}}>-\sqrt{2.5}\)
Bài 9 :
a) Giả sử \(\sqrt{5.\left(\dfrac{2}{5}\right)^2}>\sqrt{1-\dfrac{2}{5}}\)
\(\Leftrightarrow5.\left(\dfrac{2}{5}\right)^2>1-\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{4}{5}>\dfrac{3}{5}\left(đúng\right)\)
Vậy \(\sqrt{5.\left(\dfrac{2}{5}\right)^2}>\sqrt{1-\dfrac{2}{5}}\)
b) Giả sử \(\sqrt{6:\dfrac{4}{5}}>\sqrt{1\dfrac{2}{3}}\)
\(\Leftrightarrow6:\dfrac{4}{5}>1\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{15}{2}>\dfrac{5}{3}\Leftrightarrow15.3>5.2\Leftrightarrow45>10\left(đúng\right)\)
Vậy \(\sqrt{6:\dfrac{4}{5}}>\sqrt{1\dfrac{2}{3}}\)
