Question

Решите ТРИ ЗАДАЧИ ТРИ

Answer 1

5. a)

 $3x^2+8x-11=0\\D=8^2-4*3*(-11)=64+132=196\\x_1=\frac{-8-14}{6} =\frac{-22}{6} =-3\frac{2}{3} \\x_2=\frac{-8+14}{6} =\frac{6}{6} =1$

5. б)

$2x^2=32x\\2x^2-32x=0\\2x(x-16)=0\\x_1=0\\x_2=16$

5. в)

$64x^2-49=0\\64x^2=49\\x^2=\frac{49}{64}\\x_1=\sqrt{\frac{49}{64}} =\frac{7}{8} \\x_2=-\sqrt{\frac{49}{64}} =-\frac{7}{8}$

5. г)

$x^2-7x+10=0\\D=7^2-4*10=49-40=9\\x_1=\frac{7-3}{2}=2 \\x_2=\frac{7+3}{2}=5$

6.

$v_1=v_2+15\\t_1=t_2-\frac{2}{3} \\s_1=s_2\\v_1t_1=v_2t_2\\(v_2+15)(t_2-\frac{2}{3} )=v_2t_2\\v_2=\frac{s_2}{t_2}=\frac{300}{t_2} \\(\frac{300}{t_2}+15)(t_2-\frac{2}{3} )=\frac{300}{t_2}t_2/*t_2\\(300+15t_2)(t_2-\frac{2}{3} )=300t_2\\300t_2-200+15t_2^2-10t_2=300t_2\\15t_2^2-10t_2-200=0/:5\\3t^2_2-2t_2-40=0\\D=4+4*3*40=484\\t_{2(1)}=\frac{2-22}{6} < 0\\t_{2(2)}=\frac{2+22}{6} =4 \\t_1=3\frac{1}{3} \\v_1=300:3\frac{1}{3} =300:\frac{10}{3} =300*\frac{3}{10} =90\\v_2=300:4=75$

7.

$1+\frac{3}{x+4} =\frac{3}{x}\\x \neq 0\\x\neq -4\\\frac{x+4}{x+4} +\frac{3}{x+4} =\frac{3}{x}\\\frac{x+4+3}{x+4} =\frac{3}{x} \\\frac{x+7}{x+4} =\frac{3}{x}\\(x+7)x=(x+4)3\\x^2+7x=3x+12\\x^2+7x-3x-12=0\\x^2+4x-12=0\\D=4^2-4*(-12)=16+48=64\\x_1=\frac{-4-8}{2} =-6\\x_2=\frac{-4+8}{2} =2$