Question

упростите вырожение
$\sqrt{75 } + \sqrt{48 } - \sqrt{100}$
$3 \sqrt{8} - \sqrt{50} + 2 \sqrt{18}$

$\sqrt{16} n + \sqrt{25n} - \sqrt{9n}$
$\sqrt{3a} - 2 \sqrt{12a} + 3 \sqrt{75a}$
​

Answer 1

$\sqrt{75}+\sqrt{48}-\sqrt{100}=\sqrt{25\cdot3}+\sqrt{16\cdot3}-10=5\sqrt3+4\sqrt3-10=9\sqrt3-10$

$3\sqrt8-\sqrt{50}+2\sqrt{18}=3\sqrt{4\cdot2}-\sqrt{25\cdot2}+2\sqrt{2\cdot9}=3\cdot2\sqrt2-5\sqrt2+2\cdot3\sqrt2=\\\\=6\sqrt2-5\sqrt2+6\sqrt2=7\sqrt2$

$\sqrt{16n}+\sqrt{25n}-\sqrt{9n}=4\sqrt{n}+5\sqrt{n}-3\sqrt{n}=6\sqrt{n}$

или

$\sqrt{16}n+\sqrt{25n}-\sqrt{9}n=4n+5\sqrt{n}-3\sqrt{n}=4n+2\sqrt{n}$

$\sqrt{3a}-2\sqrt{12a}+3\sqrt{75a}=\sqrt{3a}-2\sqrt{4\cdot3a}+3\sqrt{25\cdot3a}=\\\\=\sqrt{3a}-2\cdot2\sqrt{3a}+3\cdot5\sqrt{3a}=\sqrt{3a}(1-4+15)=12\sqrt{3a}$