Răspuns:
Explicație pas cu pas:
I
1) (1-2x)(x-3)=-7 x-2x²-3+6x+7=0 -2x²+7x+7=0
2x²-7x-7=0 x1,2=-7±√49+72/4 x1=-7+11/4=1 x2=-7-11/4=-18/4=-9/2
2) 2[16-(4x+3)²]=0
2(4-4x-3)(4+4x+3)=2(1-4x)(7+4x)=0
x1=1/4 x2=-7/4
3) x²+x-20+3(x²-x-6)=x²+4x+4+2x+18
x-20+3x²-3x-18-6x-22=0 3x²-8x-60=0
x1,2=4±√16+180/3 x1=4+14/3=6 x2=4-14/3=-10/3
II
1) 3x²+11x+8=0 x1,2=-11±√121-96/6=-11±5/6
x1=-11-5/6=-8/3 x2=-1
3x²+11x+8=(3x+8)(x+1)
2x²-x-3=0 x1,2=1±√1+25/4
x1=1+5/4=6/4=3/2 x2=-1
2x²-x-3=(x+1)(2x-3)
F=(3x²+11x+8)/(2x²-x-3)=(3x+8)(x+1)/(x+1)(2x-3)=(3x+8)/(2x-3)
2) F=[(x+2)-3]²/(x²+6x-7)=(x-1)²/(x-1)(x+7)=(x-1)/(x+7)
III
E=(x²-5x)(x²-5x+10)+24=
=x^4-5x³-5x³+10x²-50x+24=x^4-10x³+25x²-14x²-50x+24=
14x²+50x-24=2(7x²+25x-12)=
x1,2=-25±√625-336/14=-25±17/14
x1=-25-17/14=-3 x2=-25+17/14=-8/14=-4/7
E=(x²-5x)²-2(x+3)(7x+4)
