С решением пожалуйста В1 1 не надо а в В2 2

Question 1

Question 2

Вариант 1

2.

$\frac{5-3x}{25-x^{2}} +\frac{2x}{25-x^{2}} =\frac{5-3x+2x}{25-x^{2}} =\frac{5-x}{(5-x)(5+x)} =\frac{1}{5+x}$

$\frac{1}{5-1.5} =\frac{1}{3.5} =1:\frac{35}{10} =\frac{10}{35} =\frac{2}{7}$

3.

а) $\frac{2x+1}{12x^{2}y} +\frac{2-3y}{18xy^{2}} =\frac{3y(2x+1)}{36x^{2}y^{2}} +\frac{2x(2-3y)}{36x^{2}y^{2}} =\frac{6xy+3y+4x-6xy}{36x^{2}y^{2}} =\frac{3y+4x}{36x^{2}y^{2}}$

б) $\frac{a+4}{a} -\frac{a+6}{a+2} =\frac{(a+4)(a+2)}{a(a+2)} -\frac{a(a+6)}{a(a+2)} =\frac{a^{2}+6a+8-a^{2}-6a}{a(a+2)} =\frac{8}{a(a+2)}$

в) $\frac{a+1}{2a(a-1)} -\frac{a-1}{2a(a+1)} =\frac{(a+1)(a+1)}{2a(a+1)(a-1)} -\frac{(a-1)(a-1)}{2a(a+1)(a-1)} =\frac{a^{2}+2a+1-a^{2}+2a-1}{2a(a^{2}-1)} =\frac{4a}{2a(a^{2}-1)} =\frac{2}{a^{2}-1}$

г) $\frac{x+2}{2x-4} -\frac{3x-2}{x^{2}-2x} =\frac{x+2}{2(x-2)} -\frac{3x-2}{x(x-2)} =\frac{x(x+2)}{2x(x-2)} -\frac{2(3x-2)}{2x(x-2)} =\frac{x^{2}+2x-6x+4}{2x(x-2)} =\frac{x^{2}-4x+4}{2x(x-2)} =\frac{(x-2)^{2}}{2x(x-2)} =\frac{x-2}{2x}$

Вариант 2

1.

а) $\frac{a+4}{4a} *\frac{8a^{2}}{a^{2}-16}=\frac{(a+4)*8a^{2}}{4a(a-4)(a+4)}= \frac{2a}{a-4}$

б) $(\frac{3x^{2}*y^{-3}}{z})^{2}:\frac{(3x)^{3}z^{-2}}{y^{5}} =\frac{9x^{4}y^{-6}}{z^{2}} :\frac{27x^{3}z^{-2}}{y^{5}} =\frac{9x^{4}}{z^{2}y^{6}} :\frac{27x^{3}}{y^{5}z^{2}} =\frac{9x^{4}*y^{5}z^{2}}{z^{2}y^{6}*27x^{3}} =\frac{x}{3y}$

3.

$x+81x^{-1}=18\\x+\frac{81}{x} =18\\x^{2}+81=18x\\x^{2}-18x+81=0\\(x-9)^{2}=0\\x-9=0\\x=9$

4.

$(\frac{b+1}{b-1}-\frac{b}{b+1} ):\frac{3b+1}{2b-2} =\frac{(b+1)^{2}-b(b-1)}{(b-1)(b+1)} *\frac{2(b-1)}{3b+1} =\frac{b^{2}+2b+1-b^{2}+b}{(b-1)(b+1)} *\frac{2(b-1)}{3b+1} =\frac{(3b+1)*2(b-1)}{(b-1)(b+1)*(3b+1)} =\frac{2}{b+1}$

Question 3

спасибо