Question

помогите даю много балов и зделайтн пж 1 вартант​

Answer 1

Ответ:

Объяснение:

1. а) $\frac{25}{x^{2} }*\frac{x^{7} }{40} = 25 *\frac{x^{5} }{40} = 5*\frac{x^{5} }{8} = \frac{5x^{5} }{8}$

2. а)$(\frac{2x^{4} }{y^{5}z^{3}})^{3} = \frac{(2x^{4})^{3} }{(y^{5}z^{3})^{3}} =\frac{8x^{12} }{y^{15} z^{9} }$

3. а) $\frac{x+1}{x} - \frac{1}{x+2}*\frac{x^{2}-4}{x} = \frac{x+1}{x} - \frac{1}{x+2}*\frac{(x-2)(x+2)}{x} = \frac{x+1}{x} - \frac{x-2}{1} = \frac{x+1-(x-2)}{x} =\frac{x+1-x+2}{x} = \frac{3}{x}$

б)$\frac{4}{x+5} *\frac{x^{2}-25 }{2} = \frac{2}{x+5}*(x-5)*(x+5) = 2(x-5)=2x-10$

в)$\frac{a^{2} - x^{2}}{c^{2}-d^{2}} *\frac{c+d}{a-x}+\frac{x}{d-c}= \frac{(a - x)}{(c-d)}*\frac{(a + x)}{(c+d)} *\frac{c+d}{a-x}+\frac{x}{-(c-d)} = \frac{a+x}{c-d}-\frac{x}{c-d} = \frac{a+ x -x}{c-d} =\frac{a}{c-d}$

4. а)$\frac{9-4y^{2} }{4y^{2}-6y+9} : \frac{3-2y}{8y^{3}+27} = \frac{(3-2y)*(3+2y) }{4y^{2}-6y+9} * \frac{8y^{3}+27}{3-2y} = \fra\frac{x}{y} c{3-2y}{4y^{2}-6y+9} *(2y+3) * (4y^{2}-6y + 9) = (3+2y) *(2y+3) = (3+2y)^{2} = (3+2*(-\frac{1}{2}) )^{2} = (3-1)^{2}= 2^{2} = 4$