Question

1.Сократить дробь
1)$frac{3y^{1/4}}{y ^{1/2}-5y ^{1/4} }$
2)$frac{a-b}{a ^{0.5}+b ^{0.5} }$
2.Найти значение выражения
1)$frac{x-9 x^{1/2} }{ x^{3/4}+3 x^{1/2} }$ , при х= 0,0081
2)$frac{y-49 y^{0.5} }{y^{0.75}+7 y^{0.5} }$ . при y= 0.2401
3.Упростить выражение
1)$frac{ x^{0.5} }{ x^{0.5}-5 } - frac{5}{x^{0.5}+5} + frac{x}{25-x}$
2)$( frac{5a^{0.5}+b^{0.5} }{a^{0.5}-5b^{0.5} } + frac{5 a^{0.5}- b^{0.5} }{ a^{0.5}+b^{0.5} })* frac{a-25b}{a+b}$

Answer 1

$frac{3y^{1/4}}{y ^{1/2}-5y ^{1/4} } =frac{3y^{1/4}}{y ^{1/4}(y ^{1/4}-5) }=frac{3}{y ^{1/4}-5 }$

$frac{a-b}{a ^{0.5}+b ^{0.5} } = frac{(a ^{0.5}+b ^{0.5})(a ^{0.5}-b ^{0.5})}{a ^{0.5}+b ^{0.5} } =a ^{0.5}-b ^{0.5}$

$frac{x-9 x^{1/2} }{ x^{3/4}+3 x^{1/2} } = frac{x^{1/2}(x^{1/4}-3)(x^{1/4}+3) }{ x^{1/2}(x^{1/4}+3 ) } = \ =x^{1/4}-3= sqrt[4]{x} -3= sqrt[4]{0,0081} -3=0,3-3=-2,7$

$frac{y-49 y^{0.5} }{y^{0.75}+7 y^{0.5} } = frac{y^{0.5}(y^{0.25}-7)(y^{0.25}+7) }{y^{0.5}(y^{0.25}+7 ) } = \ =y^{0.25}-7= sqrt[4]{y} -7=sqrt[4]{0.2401} -7=0.7-7=-6.3$

$frac{ x^{0.5} }{ x^{0.5}-5 } - frac{5}{x^{0.5}+5} + frac{x}{25-x} = frac{ x^{0.5}(x^{0.5}+5)-5(x^{0.5}-5)-x }{ x-25 } = \ = frac{ x+5x^{0.5}-5x^{0.5}+25-x }{ x-25 } =frac{ 25 }{ x-25 }$

$( frac{5a^{0.5}+b^{0.5} }{a^{0.5}-5b^{0.5} } + frac{5 a^{0.5}- b^{0.5} }{ a^{0.5}+b^{0.5} }) frac{a-25b}{a+b} = \ = frac{((5a^{0.5}+b^{0.5})(a^{0.5}+b^{0.5})+(5 a^{0.5}- b^{0.5} ) (a^{0.5}-5b^{0.5}))(a-25b)}{(a^{0.5}-5b^{0.5})(a^{0.5}+b^{0.5})(a+b)}= \ = frac{(5a+5a^{0.5}b^{0.5}+a^{0.5}b^{0.5}+b+5a-25a^{0.5}b^{0.5}-a^{0.5}b^{0.5}+5b)(a-25b)}{(a^{0.5}-5b^{0.5})(a^{0.5}+b^{0.5})(a+b)}= \ = frac{(b+15a-20a^{0.5}b^{0.5})(a-25b)}{(a^{0.5}-5b^{0.5})(a^{0.5}+b^{0.5})(a+b)}$

Answer 2

это из задачника Сканави?