Question

sa se rezolve ecuatiile:
a)(x-1)²=9
b)3-(x-5)²=1
c)(3-x)(3+x)-6=2
d)5-(x-2)(x+2)=3
e)$x^{2}$ -2x+1=7
f)4$x^{2}$ -4x+1=5
Va rooog mult,e urgent <3

Answer 1

$\displaystyle a).(x-1)^2=9 \\ x^2-2 \cdot x \cdot 1+1^2=9 \\ x^2-2x+1=9 \\ x^2-2x+1-9=0 \\ x^2-2x-8=0 \\ a=1,b=-2,c=-8 \\ \Delta=b^2-4ac=2^2-4 \cdot 1 \cdot (-8)=4+32=36\ \textgreater \ 0 \\ x_1= \frac{2+ \sqrt{36} }{2 \cdot 1} = \frac{2+6}{2} = \frac{8}{2} =4 \\ \\ x_2= \frac{2- \sqrt{36} }{2 \cdot 1} = \frac{2-6}{2} = \frac{-4}{2} =-2$

$\displaystyle b).3-(x-5)^2=1 \\ 3-(x^2-2 \cdot x \cdot 5+5^2)=1 \\ 3-(x^2-10x+25)=1 \\ 3-x^2+10x-25=1 \\ -x^2+10x=1-3+25 \\ -x^2+10x=23 \\ -x^2+10x-23=0 \\ a=-1,b=10,c=-23 \\ \Delta=b^2-4ac=10^2-4 \cdot (-1)\cdot (-23)=100-92=8\ \textgreater \ 0 \\ x_1= \frac{-10+ \sqrt{8} }{2 \cdot (-1)} = \frac{-10+2 \sqrt{2} }{-2} = \frac{-2(5- \sqrt{2} )}{-2}=5- \sqrt{2} \\ \\ x_2= \frac{-10- \sqrt{8} }{2 \cdot (-1)} = \frac{-10-2 \sqrt{2} }{-2} = \frac{-2(5+ \sqrt{2} )}{-2} =5+ \sqrt{2}$

$\displaystyle c).(3-x)(3+x)-6=2 \\ 9-x^2-6=2 \\ -x^2=2-9+6 \\ -x^2=-1 \\ -x^2+1=0 \\ a=-1,b=0,c=1 \\ \Delta=b^2-4ac=0-4 \cdot (-1)\cdot 1=0+4=4\ \textgreater \ 0 \\ x_1= \frac{0+ \sqrt{4} }{2 \cdot (-1)} = \frac{0+2}{-2} = \frac{2}{-2} =-1 \\ \\ x_2= \frac{0- \sqrt{4} }{2 \cdot (-1)} = \frac{0-2}{-2} = \frac{-2}{-2} =1$

$\displaystyle d).5-(x-2)(x+2)=3 \\ 5-x^2+4=3 \\ -x^2=3-5-4 \\ -x^2=-6 \\ -x^2+6=0 \\ a=-1,b=0,c=6 \\ \Delta=b^2-4ac=0-4 \cdot (-1) \cdot 6=0+24=24\ \textgreater \ 0 \\ x_1= \frac{0+ \sqrt{24} }{2 \cdot (-1)} = \frac{0+2 \sqrt{6} }{-2} = \frac{2 \sqrt{6} }{-2} =- \sqrt{6} \\ \\ x_2= \frac{0- \sqrt{24} }{2 \cdot (-1)} = \frac{0-2 \sqrt{6} }{-2} = \frac{-2 \sqrt{6} }{-2} = \sqrt{6}$

$\displaystyle e).x^2-2x+1=7 \\ x^2-2x+1-7=0 \\ x^2-2x-6=0 \\ a=1,b=-2,c=-6 \\ \Delta=b^2-4ac=2^2-4 \cdot 1 \cdot (-6)=4+24=28\ \textgreater \ 0 \\ x_1= \frac{2+ \sqrt{28} }{2 \cdot 1} = \frac{2+2 \sqrt{7} }{2} = \frac{2(1+ \sqrt{7} )}{2} =1+ \sqrt{7} \\ \\ x_2= \frac{2- \sqrt{28} }{2 \cdot 1} = \frac{2-2 \sqrt{7} }{2} = \frac{2(1- \sqrt{7} )}{2} =1- \sqrt{7}$

$\displaystyle f).4x^2-4x+1=5 \\ 4x^2-4x+1-5=0 \\ 4x^2-4x-4=0 \\ a=4,b=-4,c=-4 \\ \Delta=b^2-4ac=4^2-4 \cdot 4 \cdot (-4)=16+64=80\ \textgreater \ 0 \\ x_1= \frac{4+ \sqrt{80} }{2 \cdot 4} = \frac{4+4 \sqrt{5} }{8} = \frac{4(1+ \sqrt{5} )}{8} = \frac{1+ \sqrt{5} }{2} \\ \\ x_2= \frac{4- \sqrt{80} }{2 \cdot 4} = \frac{4-4 \sqrt{5} }{8} = \frac{4(1- \sqrt{5} )}{8} = \frac{1- \sqrt{5} }{2}$