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В решении.
Объяснение:
При каком значении параметра a уравнение x²+ax+a-1=0 разложится как (x-7)(x+1)?
x²+ax+a-1 = (x-7)(x+1)
Раскрыть скобки:
x²+ax+a-1 = х²+х-7х-7
Привести подобные члены:
ах+а-1-х+7х+7=0
Разложить на множители:
(ах+а)-(1+х)+7(х+1)=0
а(х+1)-(1+х)+7(х+1)=0
(х+1)(а-1+7)=0
(х+1)(а+6)=0
х+1=0
х= -1;
а+6=0
а= -6.
Проверка:
Подставить вычисленное значение а в уравнение и решить его:
x²+ax+a-1=0
х²-6х-6-1=0
х²-6х-7=0
D=b²-4ac =36+28=64 √D= 8
х₁=(-b-√D)/2a
х₁=(6-8)/2
х₁= -2/2
х₁= -1;
х₂=(-b+√D)/2a
х₂=(6+8)/2
х₂=14/2
х₂=7;
ответ: х²-6х-7=(х-7)(х+1) при а= -6.
m-n/2mn+2/m-n=(m-n)²+2·2mn/2mn= (m-n)²+2mn/2mn(m-n)=m²-2mn+n²+4mn/2mn(m-n)=
=m²+2mn+n²/2mn(m-n)=(m+n)²/2mn(m-n)
p+3/p+4-p-3/p-4 = (p+3)/(p+4) - (p-3)/(p-4)=(p+3)(p-4)-(p+4)(p-3)/(p+4)(p-4)=
p²+3p-4p-12-p²-4p+3p+12/p²-4=-2p/p²-4
7p+q/p^2-pq+p+7q/q^2-pq=(7p+q)/p(p-q)+(p+7q)/q(q-p)=(7p+q)/p(p-q)-(p+7q)/q(p-q)=(7p+q)·q-(p+7q)·p/pq(p-q)=7pq+q²-p²-7pq/pq(p-q)=(q²-p²)/pq(p-q)=-(p+q)/pq(p-q)