(х+1)(х-1)(х-3)(х-5)+7=0

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Ответ:

(x+1)(x+3)⋅(x−1)(x+5)+7=0(x2+4x+3)(x2+4x−5)+7=0t=x2+4x−5→x2+4x+3=t+8t⋅(t+8)+7=0t2+8t+7=0t1=−1,t2=−7a)x2+4x−5=−1,x2+4x−4=0x1,2=−2±8=−2(1±2)b)x2+4x−5=−7,x2+4x=2=0x3,4=−2±2=−2(2±1)