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rentbox
@rentbox
August 2022
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Решите В2 и В4 пожалуйста, очень надо
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sedinalana
2
ABCD параллелограмм,BD=6,AC=10,<AOB=60
S=BD*AC/2*sin<AOB=6*10/2*√3/2=15√3
AO=OC=5
DO=OD=3
<AOD=180-<AOB=180-60=120
AD²=AO²+OD²-2*AO*OD*cos<AOD
AD²=9+25-2*3*5*(-1/2)=34+15=49
AD=7
S=AD*h
h=S/AD=15√3/7
Ответ 15√3/7
4
2x/(x²-5x-2)+3x/(x²+5x-2)=-5/8
2/(x-5-2/x)+3/(x+5-2/x)=-5/8
x-2/x=a
2/(a-5)+3/(a+5)=-5/8
16(a+5)+24(a-5)+5(a²-25)=0,a≠5;a≠-5
16a+80+24a-120+5x²-125=0
5a²+40a-165=0
a²+8a-33=0
D=64+132=196>0
a1+a2=-8 U a1*a2=-33
a1=-11
x-2/x=-11
x²+11x-2=0,x≠0
D=121+8=129>0
x1+x2=-11
a2=3
x-2/x=3
x²-3x-2=0
D=9+8=17>0
x3+x4=3
(x1=x2+x3+x4)/4=(-11+3)/4=-8/4=-2
Ответ -2
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Answers & Comments
ABCD параллелограмм,BD=6,AC=10,<AOB=60
S=BD*AC/2*sin<AOB=6*10/2*√3/2=15√3
AO=OC=5
DO=OD=3
<AOD=180-<AOB=180-60=120
AD²=AO²+OD²-2*AO*OD*cos<AOD
AD²=9+25-2*3*5*(-1/2)=34+15=49
AD=7
S=AD*h
h=S/AD=15√3/7
Ответ 15√3/7
4
2x/(x²-5x-2)+3x/(x²+5x-2)=-5/8
2/(x-5-2/x)+3/(x+5-2/x)=-5/8
x-2/x=a
2/(a-5)+3/(a+5)=-5/8
16(a+5)+24(a-5)+5(a²-25)=0,a≠5;a≠-5
16a+80+24a-120+5x²-125=0
5a²+40a-165=0
a²+8a-33=0
D=64+132=196>0
a1+a2=-8 U a1*a2=-33
a1=-11
x-2/x=-11
x²+11x-2=0,x≠0
D=121+8=129>0
x1+x2=-11
a2=3
x-2/x=3
x²-3x-2=0
D=9+8=17>0
x3+x4=3
(x1=x2+x3+x4)/4=(-11+3)/4=-8/4=-2
Ответ -2