Question

решите пожалуйста ааа а прошу

Answer 1

4. Решить уравнение.

$(3x^2 + 9x - 5) - (7x^2 - 6x + 2) = 7 - 2x - 4x^2;$

$3x^2 + 9x - 5 - 7x^2 + 6x - 2 = 7 - 2x - 4x^2;$

$(3x^2 - 7x^2) + (9x + 6x) - 5 - 2 = 7 - 2x - 4x^2;$

$-4x^2 + 15x - 7 = 7 - 2x - 4x^2;$

$15x - 7 = 7 - 2x;$

$15x + 2x = 7 + 7;$

$17x = 14;$

$x = \frac{14}{17}.$

Ответ: $\frac{14}{17}$.

5. Вычислить.

1) $\frac{216^5*36^3}{6^{20}} = \frac{(6^3)^5*(6^2)^3}{6^{20}} = \frac{6^{15}*6^6}{6^{20}}  = \frac{6^{15+6}}{6^{20}} = \frac{6^{21}}{6^{20}} = 6^{21-20} = 6^1 = 6.$

2) $\frac{18^{14}}{6^{12}*3^{14}} = \frac{(2*3^2)^{14}}{(2*3)^{12}*3^{14}} = \frac{2^{14}*(3^2)^{14}}{2^{12}*3^{12}*3^{14}} = \frac{2^{14}*3^{28}}{2^{12}*3^{12+14}} = \frac{2^{14}*3^{28}}{2^{12}*3^{26}} =\\= 2^{14-12}*3^{28-26} = 2^2*3^2 = 4*9 = 36.$

3) $(\frac{6}{11})^9*(1\frac{5}{6})^7 = (\frac{6}{11})^9*(\frac{11}{6})^7 = (\frac{6}{11})^9*(\frac{6}{11})^{-7 } = (\frac{6}{11})^{9+(-7)} =\\= (\frac{6}{11})^{9-7} = (\frac{6}{11})^2 = \frac{6^2}{11^2} = \frac{36}{121}.$

Ответ: 6; 36; $\frac{36}{121}$.

6. Найти (*).

$(5x^2 - 3xy - y^2) - (*) = x^2 + 3xy;$

$(*) = (5x^2 - 3xy - y^2) - (x^2 + 3xy);$

$(*) = 5x^2 - 3xy - y^2 - x^2 - 3xy;$

$(*) = 4x^2 - 6xy - y^2.$

Ответ: 4x² - 6xy - y².