(x/x²+2x+4+ x²+8/x³-8 - 1/x-2) /это дробь - вопрос №17883434248812 от karinakrasyuk 24.05.2020 09:49

Question

Алгебра: (x/x²+2x+4+ x²+8/x³-8 - 1/x-2) /это дробь​

karinakrasyuk · Answer

8:(1+8:(1-8:(1+4:(1-4:(1-8х =1993-1
8:(1+8:(1-8:(1+4:(1-4:(1-8х ) =1992
1+8:(1-8:(1+4:(1-4:(1-8х =1/249
8:(1-8:(1+4:(1-4:(1-8х))) =1/249-249/249
8:(8:(1-8:(1+4:(1-4:(1-8х))) =-248/249
1-8:(1+4:(1-4:(1-8х)) =8 * (-249/248)
1-8:(1+4:(1-4:(1-8х)) =-249/31
8:(1+4:(1-4:(1-8х)) =31/31+249/31
8:(1+4:(1-4:(1-8х)) =280/31
1+4:(1-4:(1-8х)) =8*(31/280)
1+4:(1-4:(1-8х)) =31/35
4:(1-4:(1-8х)) =31/35-35/35
4:(1-4:(1-8х)) =-4/35
1-4:(1-8х) =4 * (-35/4)
1-4:(1-8х) =-35
4:(1-8х) =1+35
4:(1-8х) =36
1-8х=1/9
8х=9/9-1/9
8х=8/9
х=1/9
проверка1-8*1/9=1/9 4:1/9=36 1-36=-35 4:(-35)=-4/35 1+(-4/35)=31/35 8:31/35=280/31 1-280/31=-249/31
8/(-249/31)=-248/249 1+(-248/249)=1/249 8:1/249=1992 1+1992=1993 1993=1993

(x/x²+2x+4+ x²+8/x³-8 - 1/x-2) /это дробь​

(x/x²+2x+4+ x²+8/x³-8 - 1/x-2)
/это дробь