Question

Алгебра решите пж
Очень надо зарание спасибо

Answer 1

Ответ:

$1)\ \ 8(5y+3)^2+9(3y-1)^2=8(25y^2+30y+9)+9(9y^2-6y+1)=\\\\=281y^2+186y+81\\\\\\2)\ \ (2x-5)^2-2(7x-1)^2=4x^2-20x+25-2(49x^2-14x+1)=\\\\=-94x^2+8x+23\\\\\\3)\ \ (4y^2+3)^2+(9-4y^2)^2-2(4y^2+3)(4y^2-9)=\\\\=16y^4+24y^2+9+81-72y^2+16y^4-2(16y^4-24y^2-27)=144$

$4)\ \ (a^2-6ab+9b^2)(a^2+6ab+9b^2)-(a^2-9b^2)^2=\\\\=(a-3b)^2(a+3b)^2-(a-3b)^2(a+3b)^2=0$

$1)\ \left\{\begin{array}{l}\dfrac{2x}{3}=2+\dfrac{y}{2}\ |\cdot 6\\\ \dfrac{2x}{3}+y=8\ |\cdot 3\end{array}\right\ \ \left\{\begin{array}{l}4x=12+3y\\2x+3y=24\end{array}\right\ \ \left\{\begin{array}{l}4x-3y=12\\2x+3y=24\ |\cdot (-2)\end{array}\right\ \oplus \\\\\\\left\{\begin{array}{l}4x=3y+12\\-9y=-36\end{array}\right\ \ \left\{\begin{array}{l}4x=12+12\\y=4\end{array}\right\ \ \left\{\begin{array}{l}4x=24\\y=4\end{array}\right\ \ \left\{\begin{array}{l}x=6\\y=4\end{array}\right\ \ \ (6;4)$

$2)\ \ \left\{\begin{array}{l}\dfrac{2}{x}-\dfrac{1}{y}=5\\\ \dfrac{2}{x}+\dfrac{1}{y}=7\end{array}\right\ \oplus \ominus \ \left\{\begin{array}{l}2\cdot \dfrac{2}{x}=12\\-2\cdot \dfrac{1}{y}=-2\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{4}{x}=12\\-\dfrac{2}{y}=-2\end{array}\right\\\\\\\left\{\begin{array}{l}x=\dfrac{4}{12}\\y=1\end{array}\right\ \ \left\{\begin{array}{l}x=\dfrac{1}{3}\\y=1\end{array}\right\ \ \ \ \ \ \ \ \ \Big(\ \dfrac{1}{3}\ ;\ 1\ \Big)$