Suppose that a friend missed class and never learned what \( 25^{\frac{1}{2}} \) means. a. Use exponent rules your friend would already know to calculate \( 25^{\frac{1}{2}} \cdot 25^{\text {a }} \)

268. Решите уравнение: \( \begin{array}{l}\text { a) }(y+2)\left(y^{2}-2 y+4\right)-y(y-3)(y+3)=44 \text {; } \\ \text { б) }(m-3)\left(m^{2}+3 m+9\right)-m(m+2)(m-2)=13 \\ \text { (B) }(3 a+1)\left(9 a^{2}-3 a+1\right)-3 a\left(9 a^{2}+4\right)=25 \text {; } \\ \text { (A) }(2 n-1)\left(4 n^{2}+2 n+1\right)-4 n\left(2 n^{2}-5\right)=21 \text {. }\end{array} \)

\( \log _ { 2 } ^ { 2 } x - 15 \log _ { 5 } x - 4 \leq 0 \)

Вычислить следующие определители: а) \( \left|\begin{array}{cc}1 & -2 \\ -2 & 4\end{array}\right| \); б) \( \left|\begin{array}{ccc}1 & 2 & 0 \\ 0 & 0 & 3 \\ 5 & 0 & -1\end{array}\right| \), 2. Найти точку пересечения прямых \( l_{1}: 3 x-2 y=0 \) и \( l_{2}: y=-x+5 \). Построить прямые.

268. Решите уравнение: \( \begin{array}{l}\text { а) }(y+2)\left(y^{2}-2 y+4\right)-y(y-3)(y+3)=44 \text {; } \\ \text { б) }(m-3)\left(m^{2}+3 m+9\right)-m(m+2)(m-2)=13 \text {; } \\ \text { B) }(3 a+1)\left(9 a^{2}-3 a+1\right)-3 a\left(9 a^{2}+4\right)=25 \text {; }(2 n-1)\left(4 n^{2}+2 n+1\right)-4 n\left(2 n^{2}-5\right)=21 \text {. }\end{array} \).

7. \( a, b \) ve \( c \) gerçel sayıları için \( a>0>b \cdot a \) \( b>c \) olduğuna göre, aşağıdakilerden hangisi kesinlil pozitiftir? \( \begin{array}{lll}\text { A) } a \cdot c & \text { B) } a \cdot b \cdot c & \text { C) } c \\ \text { D) } a+c & \text { E) } b^{2} \cdot c\end{array} \)