Question

$a) \sqrt{144}+( \sqrt{12}+2 \sqrt{72} -6 \sqrt{18)} \\ b) \sqrt{242}( \sqrt{28}- \sqrt{7}-2 \sqrt{7} ) \\ c) \sqrt{2 \frac{1}{4} }+ \sqrt{0,(4)}+ \sqrt \frac{256}{36} \\ d) ( \sqrt{0,01} - \sqrt{0,0016} ) * 100 \\ e) \sqrt{12 }( \sqrt{108}-2 \sqrt{48} )- \sqrt{18}( \sqrt{288}- \sqrt{7} ) \\ f) \sqrt{18}(3 \sqrt{50}+ \sqrt{162})- \sqrt{12}( \sqrt{432} - \sqrt{192})$       Va rogg ajutati-ma cu asteaa! 

Answer 1

a)$\sqrt{144} +( \sqrt{12} +2 \sqrt{72} -6 \sqrt{18} )=12+(2 \sqrt{3} +12\sqrt{2} -18 \sqrt{2} )=$$12+ 2\sqrt{3} -6\sqrt{2}$

b)$\sqrt{242} *( \sqrt{28} -\sqrt{7} -2 \sqrt{7} )=11\sqrt{2}*(2 \sqrt{7} -\sqrt{7} - 2\sqrt{7} )=11\sqrt{2}*(-\sqrt{7})=11\sqrt{14}$

c)$\sqrt{2 \frac{1}{4} } + \sqrt{0,(4)} + \sqrt{ \frac{256}{36} } =\sqrt{ \frac{9}{4} }+\sqrt{ \frac{4}{9} }+ \frac{16}{6} = \frac{3}{2} + \frac{2}{3} + \frac{16}{6} = \frac{3*3+2*2+16}{6}$$= \frac{29}{6}$

d)$(\sqrt{0,01} - \sqrt{0,0016} )*100=(\sqrt{ \frac{1}{100} } - \sqrt{ \frac{16}{10000} } )*100=(\frac{1}{10}-\frac{4}{100})*100$$= \frac{10-4}{100}*100= \frac{6}{100}*100=6$

e) $\sqrt{12} ( \sqrt{108} -2 \sqrt{48} )- \sqrt{18} ( \sqrt{288} - \sqrt{7} )=$$2 \sqrt{3} (6 \sqrt{3}-4 \sqrt{3})-3 \sqrt{2}(12 \sqrt{2} - \sqrt{7} )=2 \sqrt{3} *2 \sqrt{3} -36 \sqrt{4} +3 \sqrt{14} =$$4\sqrt{9} -36*2+3 \sqrt{14} =4*3-72+3 \sqrt{14}=-60+3 \sqrt{14}$

f)$\sqrt{18} ( 3\sqrt{50} + \sqrt{162} )- \sqrt{12} ( \sqrt{432} - \sqrt{192} )=$$3\sqrt{2} ( 15\sqrt{2} + 9\sqrt{2} )- 2\sqrt{3} (12 \sqrt{3} -8 \sqrt{3} )=3\sqrt{2} *24\sqrt{2} -2\sqrt{3} *4\sqrt{3} =$$72\sqrt{4} -8\sqrt{9}=72*2-8*3=144-24=120$