Question

Решите несколько не трудных задач...................... можете перевернуть фотку

Answer 1

2.

1)

$\frac{26a^{5}b^{8}}{39a^{7}b^{4}} =\frac{2b^{4}}{3a^{2}}$

2)

$\frac{10mn-25n}{5mn} =\frac{5n(2m-5)}{5mn} =\frac{2m-5}{m}$

3)

$\frac{x^{2}-16}{2x+8} =\frac{(x-4)(x+4)}{2(x+4)} =\frac{x-4}{2}$

4)

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

3.

1)

$\frac{3-2y}{y^{2}} -\frac{y-12}{6y} =\frac{6(3-2y)}{6y^{2}} -\frac{y(y-12)}{6y^{2}} =\frac{18-12y-y^{2}+12y}{6y^{2}} =\frac{18-y^{2}}{6y^{2}}$

2)

$\frac{20}{a^{2}+5a} -\frac{4}{a} =\frac{20}{a(a+5)} -\frac{4}{a} =\frac{20}{a(a+5)}-\frac{4(a+5)}{a(a+5)}=\frac{20-4a-20}{a(a+5)} =-\frac{4a}{a(a+5)} =-\frac{4}{a+5}$

3)

$\frac{y}{y-10} -\frac{y^{2}}{y^{2}-100} =\frac{y}{y-10} -\frac{y^{2}}{(y-10)(y+10)} =\frac{y(y+10)}{(y-10)(y+10)} -\frac{y^{2}}{(y-10)(y+10)} =\frac{y^{2}+10y-y^{2}}{(y-10)(y+10)} =\\\\=\frac{10y}{y^{2}-100}$

4)

$\frac{12c^{2}}{2c-3} -6c=\frac{12c^{2}}{2c-3}-\frac{6c(2c-3)}{2c-3} =\frac{12c^{2}-12c^{2}+18c}{2c-3} =\frac{18c}{2c-3}$

4.

1)

$\frac{a-15}{4a-20} -\frac{a-5}{4a+20} +\frac{30}{a^{2}-25} =\frac{a-15}{4(a-5)} -\frac{a-5}{4(a+5)} +\frac{30}{(a-5)(a+5)} =\frac{(a-15)(a+5)}{4(a-5)(a+5)} -\frac{(a-5)(a-5)}{a(a-5)(a+5)} +\\\\+\frac{30*4}{4(a-5)(a+5)} =\frac{a^{2}-10a-75}{4(a-5)(a+5)} -\frac{a^{2}-10a+25}{4(a-5)(a+5)} +\frac{120}{4(a-5)(a+5)} =\frac{a^{2}-10a-75-a^{2}+10a-25+120}{4(a-5)(a+5)} =\\\\=\frac{20}{4(a-5)(a+5)} =\frac{5}{a^{2}-25}$2)

$\frac{8a^{3}+100a}{a^{3}+125} -\frac{4a^{2}}{a^{2}-5a+25} =\frac{8a^{3}+100a}{(a+5)(a^{2}-5a+25)} -\frac{4a^{2}(a+5)}{(a+5)(a^{2}-5a+25)} =\frac{8a^{3}+100a-4a^{3}-20a^{2}}{(a+5)(a^{2}-5a+25)} =\\\\=\frac{4a^{3}-20a^{2}+100a}{(a+5)(a^{2}-5a+25)} =\frac{4a(a^{2}-5a+25)}{(a+5)(a^{2}-5a+25)} =\frac{4a}{a+5}$