Y=7-2x
X^2-7+2x=1
=> x^2+2x-8=0
D=4+32=36
X1= (-2+6)/2=2
X2=(-2-6)/2=-4
y1=7-2*2=3
y2=7-2*(-4)=15
Ответ:(2;3)^(-4;15)
Формула суммы бесконечно убывающей геом. прогресии:
![S= \frac{b_1}{1-q}](https://tex.z-dn.net/?f=S%3D+%5Cfrac%7Bb_1%7D%7B1-q%7D+)
.
![b_4=b_1q^3=\frac{9}{8}\\\\q= \frac{\sqrt3}{2} \; ,\; \; \; \; b_1\cdot (\frac{\sqrt3}{2})^3=\frac{9}{8}\; ,\; \; \; b_1\cdot \frac{3\sqrt3}{8}=\frac{9}{8}\\\\b_1=\frac{9}{8}:\frac{3\sqrt3}{8}=\frac{9}{3\sqrt3}=\frac{3}{\sqrt3}=\sqrt3\\\\S= \frac{\sqrt3}{1-\frac{\sqrt3}{2}} =\frac{2\sqrt3}{2-\sqrt3}=\frac{2\sqrt3\cdot (2+\sqrt3)}{4-3}=2\sqrt3\cdot (2+\sqrt3)=4\sqrt3+6](https://tex.z-dn.net/?f=b_4%3Db_1q%5E3%3D%5Cfrac%7B9%7D%7B8%7D%5C%5C%5C%5Cq%3D+%5Cfrac%7B%5Csqrt3%7D%7B2%7D+%5C%3B+%2C%5C%3B+%5C%3B+%5C%3B+%5C%3B+b_1%5Ccdot+%28%5Cfrac%7B%5Csqrt3%7D%7B2%7D%29%5E3%3D%5Cfrac%7B9%7D%7B8%7D%5C%3B+%2C%5C%3B+%5C%3B+%5C%3B+b_1%5Ccdot+%5Cfrac%7B3%5Csqrt3%7D%7B8%7D%3D%5Cfrac%7B9%7D%7B8%7D%5C%5C%5C%5Cb_1%3D%5Cfrac%7B9%7D%7B8%7D%3A%5Cfrac%7B3%5Csqrt3%7D%7B8%7D%3D%5Cfrac%7B9%7D%7B3%5Csqrt3%7D%3D%5Cfrac%7B3%7D%7B%5Csqrt3%7D%3D%5Csqrt3%5C%5C%5C%5CS%3D+%5Cfrac%7B%5Csqrt3%7D%7B1-%5Cfrac%7B%5Csqrt3%7D%7B2%7D%7D+%3D%5Cfrac%7B2%5Csqrt3%7D%7B2-%5Csqrt3%7D%3D%5Cfrac%7B2%5Csqrt3%5Ccdot+%282%2B%5Csqrt3%29%7D%7B4-3%7D%3D2%5Csqrt3%5Ccdot+%282%2B%5Csqrt3%29%3D4%5Csqrt3%2B6)
7х - 2у = -3 |*2
3х + 4у = 23
14х - 4у = -6
+
3х + 4у = 23
_____
17х = 17
х = 1
х = 1
3*1 + 4у = 23
х = 1
4у = 20
х = 1
у = 5
ОТВЕТ: (1;5)
13)
x - л. было в баке изначально
(1-0,3)x=0,7x - л. осталось в баке после 1-й поездки
0,1*0,7x=0,07x л. расходовал на 2-ю поездку
0,3x+0,07x=37
0,37x=37
x=100
14) x - одна из частей
750-x - вторая из частей
0,04x+0,12(750-x)=0,056*750
0,04x+90-0,12x=42
-0,08x=-48
x=600
750-600=150