Question

1.сократить дробь 1) 2) 2.найти значение выражения 1) , при х= 0,0081 2) . при y= 0.2401 3. выражение 1) 2)

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?