Bună!
x,y- 2 numere reale pozitive => x,y∈R
[tex]\frac{75}{100} *x=\frac{4}{3} *y[/tex]
[tex]\frac{x+y}{2} -\sqrt{x*y} =d[/tex]
a) d=?
[tex]x=\frac{2}{3} \\=> \frac{75}{100}*\frac{2}{3} =\frac{4}{3} *y\\\frac{25}{50} =\frac{4y}{3} =\frac{1}{2} \\8y=3 => y=\frac{3}{8}[/tex]
[tex]=> d=\frac{\frac{2}{3}+\frac{3}{8} }{2} -\sqrt{\frac{2}{3} *\frac{3}{8} } =\frac{\frac{25}{24} }{2} -\sqrt{\frac{1}{4} } =\frac{25}{24} *2-\frac{1}{2} =\frac{24}{17} -\frac{1}{2} =\frac{48-17}{34} =\frac{31}{34}[/tex]
b) d=0,5
x,y=?
[tex]\frac{x+y}{2} -\sqrt{x*y} =0,5=\frac{5}{10} \\10x+10y-20\sqrt{x*y} =\frac{10}{20} =1 \\-20\sqrt{x*y} =1 -10(x+y)\\\\400xy=(1-10x-10y)^{2}= (1-10x-10y)(1-10x-10y)\\400xy=1-10x-10y-10x+100x^{2} +100xy-10y+100xy+100y^{2} \\-100x^{2} -100y^{2} -200xy+400xy+10x+10y-1=0\\-100(x^{2} +y^{2} )+200xy+10(x+y)=1\\-100(x^{2} -2xy+y^{2} )+10(x+y)=1\\-100(x-y)^{2} +10(x+y)-1=0[/tex][tex]-10[10(x-y)^{2} -(x+y)]=1\\-10(10x^{2} -20xy+10y^{2} -x-y)=1[/tex]
La b) imi da multe bătăi de cap... nu ajung la un rezultat...
sper sa te fi ajutat
