Question

Tính

a)\(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)

b)\(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\) với \(a\ge0\)

c)\(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}-\sqrt{60}\)

d)\(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right).\sqrt{11}+3\sqrt{32}\)

Answer 1

Lời giải:

a)

$\sqrt{98}-\sqrt{72}+0.5\sqrt{8}=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}$

$=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}$

b)

$\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}$

$=4\sqrt{a}+4\sqrt{10}.\sqrt{a}-9\sqrt{10}.\sqrt{a}$

$=(4+4\sqrt{10}-9\sqrt{10})\sqrt{a}=(4-5\sqrt{10}).\sqrt{a}$

c)

$(2\sqrt{3}+\sqrt{5})\sqrt{3}-\sqrt{60}=2.3+\sqrt{15}-2\sqrt{15}$

$=6-\sqrt{15}$

d)

$(\sqrt{99}-\sqrt{18}-\sqrt{11})\sqrt{11}+3\sqrt{32}$

$=\sqrt{99}.\sqrt{11}-\sqrt{18}.\sqrt{11}-11+3\sqrt{32}$

$=\sqrt{9}.\sqrt{11}.\sqrt{11}-3\sqrt{2}.\sqrt{11}-11+12\sqrt{2}$

$=3.11+\sqrt{2}(12-3\sqrt{11})-11$

$=22+\sqrt{2}(12-3\sqrt{11})$