ГДЗ Алгебра 8 клас (Самостійні та діагностичні роботи) Істер О.С. 2025 рік

САМОСТІЙНА РОБОТА №2 [2М]

Додавання і віднімання дробів

ВАРІАНТ 3

Завдання 1 Знайдіть суму.

$\frac{7m^2}{a}$ + $\frac{4m^2}{a}$ = $\frac{7m^2+4m^2}{a}$ = $\frac{11m^2}{a}$

А $\frac{11m^2}{a}$     Б $\frac{3m^2}{a}$     В $\frac{11m^2}{2a}$    Г $\frac{11m^4}{a}$

Завдання 2 Виконайте дію.

1) $\frac{2x+y}{8}$ – $\frac{y\text{–}4x}{8}$ = $\frac{2x+y\text{–}\left(y\text{–}4x\right)}{8}$ =

= $\frac{2x+y\text{–}y+4x}{8}$ = $\frac{2x+4x}{8}$ = $\frac{6x}{8}$ = $\frac{3x}{4}$

2) $\frac{m+3}{5m}$ + $\frac{8\text{–}3t}{15t}$ = $\frac{3t(m+3)}{5m\text{•}3t}$ + $\frac{m\left(8\text{–}3t\right)}{m\text{•}15t}$ =

= $\frac{3t(m+3)+m\left(8\text{–}3t\right)}{15\mathrm{mt}}$ = $\frac{3\mathrm{mt}+6t+8m\text{–}3\mathrm{mt}}{15\mathrm{mt}}$ = $\frac{6t+8m}{15\mathrm{mt}}$

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

= $\frac{20\text{–}5(a+4)}{a(a+4)}$ = $\frac{20\text{–}5a\text{–}20}{a(a+4)}$ = $\frac{\text{—}5a}{a(a+4)}$ = —$\frac{5}{a+4}$

Завдання 3 Знайдіть значення виразу, якщо m = 2025, p = $\frac{1}{6}$.

$\frac{m^2+36p^2}{m\text{–}6p}$ + $\frac{12\mathrm{mp}}{6p\text{–}m}$ = $\frac{m^2+36p^2}{m\text{–}6p}$ + $\frac{12\mathrm{mp}}{\left(\text{—}1\right)\left(m\text{–}6p\right)}$ =

= $\frac{m^2+36p^2}{m\text{–}6p}$ – $\frac{12\mathrm{mp}}{m\text{–}6p}$ = $\frac{m^2+36p^2\text{–}12\mathrm{mp}}{m\text{–}6p}$=

= $\frac{m^2\text{–}2m\text{•}6p+{\left(6p\right)}^2}{m\text{–}6p}$ = $\frac{{\left(m\text{–}6p\right)}^2}{m\text{–}6p}$ = m – 6p

Якщо m = 2025, p = $\frac{1}{6}$, то

m – 6p = 2025 – 6 • $\frac{1}{6}$ = 2025 – 1 = 2024

Завдання 4 Доведіть тотожність.

$\frac{9a+1\text{,}5b}{9a^2\text{–}1\text{,}5\mathrm{ab}}$ – $\frac{9a–1\text{,}5b}{9a^2\text{+}1\text{,}5\mathrm{ab}}$ + $\frac{36a}{9a^2\text{–}0\text{,}25b^2}$ = $\frac{24}{6a\text{–}b}$

$\frac{9a+1\text{,}5b}{9a^2\text{–}1\text{,}5\mathrm{ab}}$ – $\frac{9a–1\text{,}5b}{9a^2\text{+}1\text{,}5\mathrm{ab}}$ + $\frac{36a}{9a^2\text{–}0\text{,}25b^2}$ = $\frac{3(3a+0\text{,}5b)}{3a\left(3a\text{–}0\text{,}5b\right)}$ – $\frac{3\left(3a\text{–}0\text{,}5b\right)}{3a(3a+0\text{,}5b)}$ + $\frac{36a}{{\left(3a\right)}^2\text{–}{\left(0\text{,}5b\right)}^2}$ = $\frac{3a+0\text{,}5b}{a\left(3a\text{–}0\text{,}5b\right)}$ – $\frac{3a\text{–}0\text{,}5b}{a(3a+0\text{,}5b)}$ + $\frac{36a}{\left(3a\text{–}0\text{,}5b\right)(3a+0\text{,}5b)}$ = $\frac{(3a+0\text{,}5b)(3a+0\text{,}5b)}{a\left(3a\text{–}0\text{,}5b\right)(3a+0\text{,}5b)}$ – $\frac{\left(3a\text{–}0\text{,}5b\right)\left(3a\text{–}0\text{,}5b\right)}{a(3a+0\text{,}5b)\left(3a\text{–}0\text{,}5b\right)}$ + $\frac{36a\text{•}a}{(3a+0\text{,}5b)\left(3a\text{–}0\text{,}5b\right)a}$ = $\frac{{(3a+0\text{,}5b)}^2\text{–}{\left(3a\text{–}0\text{,}5b\right)}^2+36a^2}{a(3a+0\text{,}5b)\left(3a\text{–}0\text{,}5b\right)}$ = $\frac{\left(\right(3a+0\text{,}5b\left)\text{–}\right(3a\text{–}0\text{,}5b\left)\right)\left(\right(3a+0\text{,}5b)+(3a\text{–}0\text{,}5b\left)\right)+36a^2}{a(3a+0\text{,}5b)\left(3a\text{–}0\text{,}5b\right)}$= $\frac{(3a+0\text{,}5b\text{–}3a+0\text{,}5b)(3a+0\text{,}5b+3a\text{–}0\text{,}5b)+36a^2}{a(3a+0\text{,}5b)\left(3a\text{–}0\text{,}5b\right)}$=  $\frac{6ba+36a^2}{a(3a+0\text{,}5b)\left(3a\text{–}0\text{,}5b\right)}$ = $\frac{12a(0\text{,}5b+3a)}{a(3a+0\text{,}5b)\left(3a\text{–}0\text{,}5b\right)}$ = $\frac{12}{3a\text{–}0\text{,}5b}$ = $\frac{24}{2\left(3a\text{–}0\text{,}5b\right)}$ = $\frac{24}{6a\text{–}b}$