1. Өрнекті ықшамдаңдар: а) $\frac{3-2b}{a+b}\cdot\frac{9b+9a}{4b^2-9}-\frac{6b}{2b+3}$; ә) $\frac{16a^2-25c^2}{ac+c^2}:\frac{5c-4a}{3c}+\frac{21a+24c}{a+c}$; б) $\left(\frac{b+2}{b^2-3b+9}-\frac{6}{b^3+27}\right):\frac{3b+15}{2b^2-6b+18}$; в) $\left(\frac{a-1}{a^2+2a+4}-\frac{2}{a^3-8}\right):\frac{3-a}{a^2+2a+4}$; г) $\frac{2a+1}{a-7}+\left(\frac{a}{a+8}+\frac{a}{7-a}\right):\frac{a}{8+a}$; ғ) $\left(\frac{b}{2b-3}-\frac{b}{3+2b}\right):\frac{b}{9+12b+4b^2}+\frac{45-6b}{3-2b}$; д) $25\sqrt b-0.5\sqrt{4b}+100\sqrt{0.16b}$.

Question

Вопрос: $Фото задания: 1. Өрнекті ықшамдаңдар: а) $\frac{3-2b}{a+b}\cdot\frac{9b+9a}{4b^2-9}-\frac{6b}{2b+3}$; ә)…$ Фото задания

1. Өрнекті ықшамдаңдар: а) $\frac{3-2b}{a+b}\cdot\frac{9b+9a}{4b^2-9}-\frac{6b}{2b+3}$; ә) $\frac{16a^2-25c^2}{ac+c^2}:\frac{5c-4a}{3c}+\frac{21a+24c}{a+c}$; б) $\left(\frac{b+2}{b^2-3b+9}-\frac{6}{b^3+27}\right):\frac{3b+15}{2b^2-6b+18}$; в) $\left(\frac{a-1}{a^2+2a+4}-\frac{2}{a^3-8}\right):\frac{3-a}{a^2+2a+4}$; г) $\frac{2a+1}{a-7}+\left(\frac{a}{a+8}+\frac{a}{7-a}\right):\frac{a}{8+a}$; ғ) $\left(\frac{b}{2b-3}-\frac{b}{3+2b}\right):\frac{b}{9+12b+4b^2}+\frac{45-6b}{3-2b}$; д) $25\sqrt b-0.5\sqrt{4b}+100\sqrt{0.16b}$.

Accepted Answer

$4b^2-9=(2b-3)(2b+3)$, $9b+9a=9(a+b)$.
\[\frac{3-2b}{a+b}\cdot\frac{9(a+b)}{(2b-3)(2b+3)}-\frac{6b}{2b+3}=-\frac9{2b+3}-\frac{6b}{2b+3}=-\frac{3(2b+3)}{2b+3}=-3.\]
Жауабы: $-3$.
$16a^2-25c^2=(4a-5c)(4a+5c)$, $ac+c^2=c(a+c)$.
\[\frac{(4a-5c)(4a+5c)}{c(a+c)}\cdot\frac{3c}{5c-4a}+\frac{3(7a+8c)}{a+c}\]
Since $5c-4a=-(4a-5c)$, the first term is $-\frac{3(4a+5c)}{a+c}$. Therefore
\[\frac{-12a-15c+21a+24c}{a+c}=\frac{9a+9c}{a+c}=9.\]
Жауабы: $9$.
$b^3+27=(b+3)(b^2-3b+9)$, $2b^2-6b+18=2(b^2-3b+9)$.
\[\frac{b+2}{b^2-3b+9}-\frac6{(b+3)(b^2-3b+9)}=\frac{(b+2)(b+3)-6}{(b+3)(b^2-3b+9)}=\frac{b(b+5)}{(b+3)(b^2-3b+9)}.\]
Деление на $\frac{3(b+5)}{2(b^2-3b+9)}$:
\[\frac{b(b+5)}{(b+3)(b^2-3b+9)}\cdot\frac{2(b^2-3b+9)}{3(b+5)}=\frac{2b}{3(b+3)}.\]
Жауабы: $\frac{2b}{3(b+3)}$.
$a^3-8=(a-2)(a^2+2a+4)$.
\[\frac{a-1}{a^2+2a+4}-\frac2{(a-2)(a^2+2a+4)}=\frac{(a-1)(a-2)-2}{(a-2)(a^2+2a+4)}=\frac{a(a-3)}{(a-2)(a^2+2a+4)}.\]
Деление на $\frac{3-a}{a^2+2a+4}=-\frac{a-3}{a^2+2a+4}$:
\[\frac{a(a-3)}{(a-2)(a^2+2a+4)}\cdot\frac{a^2+2a+4}{3-a}=-\frac{a}{a-2}.\]
Жауабы: $-\frac{a}{a-2}$.
\[\frac{a}{a+8}+\frac{a}{7-a}=\frac{a(7-a)+a(a+8)}{(a+8)(7-a)}=\frac{15a}{(a+8)(7-a)}.\]
Сонда
\[\frac{15a}{(a+8)(7-a)}:\frac{a}{a+8}=\frac{15}{7-a}.\]
\[\frac{2a+1}{a-7}+\frac{15}{7-a}=\frac{2a+1}{a-7}-\frac{15}{a-7}=\frac{2a-14}{a-7}=2.\]
Жауабы: $2$.
\[\frac{b}{2b-3}-\frac{b}{2b+3}=\frac{6b}{(2b-3)(2b+3)}=\frac{6b}{4b^2-9}.\]
$9+12b+4b^2=(2b+3)^2$, сондықтан
\[\frac{6b}{(2b-3)(2b+3)}:\frac{b}{(2b+3)^2}=\frac{6(2b+3)}{2b-3}.\]
Ал $\frac{45-6b}{3-2b}=\frac{6b-45}{2b-3}$. Демек,
\[\frac{6(2b+3)+6b-45}{2b-3}=\frac{18b-27}{2b-3}=9.\]
Жауабы: $9$.
$\sqrt{4b}=2\sqrt b$, $\sqrt{0.16b}=0.4\sqrt b$.
\[25\sqrt b-0.5\cdot2\sqrt b+100\cdot0.4\sqrt b=(25-1+40)\sqrt b=64\sqrt b.\]
Жауабы: $64\sqrt b$.

Ответ: