Question

va rog ajutatima, pana maine la 13:30
la exercitiul 14 punctul d este egal cu
$(\sqrt{60} \times 3) \div 4 + (3 \times \sqrt{20} ) \div \sqrt{27}$
​doar cele cu T

Answer 1

Răspuns:

Explicație pas cu pas:

12)

a)

$\sqrt{5}+(\frac{1}{\sqrt{5} }+\frac{1}{\sqrt{2} })*\sqrt{10}-(\frac{1}{\sqrt{8}}-\frac{1}{\sqrt{20} })*4\sqrt{10}-5\sqrt{2}=$

$\sqrt{5}+\frac{\sqrt{10}}{\sqrt{5} }+\frac{\sqrt{10}}{\sqrt{2} }-(\frac{4\sqrt{10}}{\sqrt{8}}-\frac{4\sqrt{10}}{\sqrt{20} })-5\sqrt{2}=$

$\sqrt{5}+\frac{\sqrt{5}*\sqrt{2}}{\sqrt{5} }+\frac{\sqrt{5}*\sqrt{2}}{\sqrt{2} }$$-(\frac{4\sqrt{2}*\sqrt{5}}{2\sqrt{2}}-\frac{4\sqrt{5}*\sqrt{2}}{2\sqrt{5} })-5\sqrt{2}=$

$\sqrt{5}+\sqrt{2}+\sqrt{5}-2\sqrt{5}+2\sqrt{2}-5\sqrt{2} =-2\sqrt{2}$

b)

$4\sqrt{7}+(\frac{2}{\sqrt{7} }+\frac{1}{\sqrt{5}})*\sqrt{35}-(\frac{1}{\sqrt{125}}-\frac{1}{\sqrt{175}})*25\sqrt{35}-7\sqrt{5}=$

$4\sqrt{7}+(\frac{\sqrt{35}}{\sqrt{7} }+\frac{\sqrt{35}}{\sqrt{5}})-(\frac{25\sqrt{35}}{\sqrt{125}}-\frac{25\sqrt{35}}{\sqrt{175}})-7\sqrt{5}=$

$4\sqrt{7}+(\frac{\sqrt{7}*\sqrt{5}}{\sqrt{7} }+\frac{\sqrt{7}*\sqrt{5}}{\sqrt{5}})-(\frac{25\sqrt{7}*\sqrt{5}}{5\sqrt{5}}-\frac{25\sqrt{7}*\sqrt{5}}{5\sqrt{7}})-7\sqrt{5}=$

$4\sqrt{7}+\sqrt{5}+\sqrt{7}-5\sqrt{7}+5\sqrt{5}-7\sqrt{5}=-\sqrt{5}$

13)

a)

$(\sqrt{0,(6)}+\frac{2\sqrt{6} }{3} )*\sqrt{6}+(\sqrt{0,(3)}+\frac{6}{3\sqrt{3} } )*\sqrt{3}=$

$(\frac{\sqrt{6}}{\sqrt{9}}+\frac{2\sqrt{6} }{3} )*\sqrt{6}+(\frac{\sqrt{3}}{\sqrt{9}}}+\frac{6}{3\sqrt{3} } )*\sqrt{3}=$

$\frac{\sqrt{6}*\sqrt{6}}{\sqrt{9}}+\frac{2\sqrt{6}*\sqrt{6} }{3}+\frac{\sqrt{3}*\sqrt{3}}{\sqrt{9}}}+\frac{6*\sqrt{3}}{3\sqrt{3} }=$

$\frac{6}{3}+\frac{2*6}{3}+\frac{3}{3}+\frac{6}{3} =2+4+1+2=9$

b)

$(\sqrt{0,(3)}+\frac{3}{2\sqrt{3}}+\frac{\sqrt{3}}{2} )*\sqrt{3}-(\sqrt{0,(6)}+\frac{\sqrt{24}}{3})*\sqrt{6}=$

$\frac{\sqrt{3}*\sqrt{3}}{\sqrt{9}}+\frac{3*\sqrt{3}}{2\sqrt{3}}+\frac{\sqrt{3}*\sqrt{3}}{2}-\frac{\sqrt{6}*\sqrt{6}}{\sqrt{9}}}-\frac{2\sqrt{6}*\sqrt{6}}{3}=$

$\frac{3}{3}+\frac{3}{2}+\frac{3}{2} -\frac{6}{3}-\frac{2*6}{3} =1+\frac{6}{2}-2-4=1+3-2-4=-2$

14)

c)

$\frac{x}{\sqrt{24}}-\frac{\sqrt{150}}{4}=\frac{3\sqrt{96}}{16}=>\frac{x}{2\sqrt{6}}-\frac{5\sqrt{6}}{4}=\frac{3*4\sqrt{6}}{16} =>\frac{x}{2\sqrt{6}}=\frac{3\sqrt{6}}{4}+\frac{5\sqrt{6}}4}$

$\frac{x}{2\sqrt{6}}=\frac{8\sqrt{6}}{4}=>\frac{x}{2\sqrt{6}}=2\sqrt{6}=>x=2\sqrt{6}*2\sqrt{6}=>x=4*6=24$

d)

$\frac{x}{\sqrt{12}}+\frac{\sqrt{12}}{\sqrt{5}}+\frac{4\sqrt{135}}{9}=\frac{3\sqrt{60}}{4}+\frac{3\sqrt{20}}{\sqrt{27}}$

$\frac{x}{2\sqrt{3}}+\frac{2\sqrt{3}}{\sqrt{5}}+\frac{4*3\sqrt{15}}{9}=\frac{3*2\sqrt{15}}{4}+\frac{3*2\sqrt{5}}{3\sqrt{3}}$

$\frac{x}{2\sqrt{3}}+\frac{2\sqrt{3}}{\sqrt{5}}+\frac{4\sqrt{15}}{3}=\frac{3\sqrt{15}}{2}+\frac{2\sqrt{5}}{\sqrt{3}}$

$\frac{x}{2\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{3}}=\frac{3\sqrt{15}}{2}-\frac{2\sqrt{3}}{\sqrt{5}}-\frac{4\sqrt{15}}{3}$

$\frac{x-2\sqrt{5}}{2\sqrt{3}}=\frac{3*3\sqrt{15}}{2*3}-\frac{2\sqrt{3}}{\sqrt{5}}-\frac{4*2\sqrt{15}}{3*2}$

$\frac{x-2\sqrt{5}}{2\sqrt{3}}=\frac{9\sqrt{15}}{6}-\frac{2\sqrt{3}}{\sqrt{5}}-\frac{8\sqrt{15}}{6}$

$\frac{x-2\sqrt{5}}{2\sqrt{3}}=\frac{\sqrt{15}}{6}-\frac{2\sqrt{3}}{\sqrt{5}}$

$\frac{x-2\sqrt{5}}{2\sqrt{3}}=\frac{\sqrt{5}*\sqrt{3}*\sqrt{5}}{6*\sqrt{5}}-\frac{2\sqrt{3}*6}{\sqrt{5}*6}$

$\frac{x-2\sqrt{5}}{2\sqrt{3}}=\frac{5\sqrt{3}}{6\sqrt{5}}-\frac{12\sqrt{3}}{6\sqrt{5}}$

$\frac{x-2\sqrt{5}}{2\sqrt{3}}=\frac{-7\sqrt{3}}{6\sqrt{5}}$

${(x-2\sqrt{5})6\sqrt{5}=-7\sqrt{3}*2\sqrt{3}=>6x\sqrt{5}-2\sqrt{5}*6\sqrt{5}=-42=>\\\\6x\sqrt{5}-60=-42$$6x\sqrt{5}-60=-42=>6x\sqrt{5}=60-42=>6x\sqrt{5}=18=>x\sqrt{5}=\frac{18}{6}=>x\sqrt{5}=3=>x=\frac{3}{\sqrt{5}}=>x=\frac{3*\sqrt{5}}{\sqrt{5}*\sqrt{5}}=>x=\frac{3\sqrt{5}}{5}$