Question

алгебра 9 класс

Answer 1

a) (x^2-y^2)/(x^3-y^3)*((x-y)^2/3xy+1)-1/y=

=((x+y)(x-y))/((x^2+xy+y^2)*(x-y))*((x-y)^2/3xy)-1/y=

=(x+y)/((x^2+xy+y^2))*((x-y)^2/3xy)-1/y=

=(x+y)/(x^2+xy+y^2)*(3xy/3xy+(x-y)^2/3xy)-1/y=

=(x+y)/(x^2+xy+y^2)*(x^2+3xy-2xy+y^2)/3xy-1/y=

=(x+y)/(x^2+xy+y^2)*(x^2+xy+y^2)/3xy-1/y=

=((x+y)(x^2+xy+y^2)/(3xy(x^2+xy+y^2)-1/y=

=((x+y)-3x)/3xy=-(2x-y)/3xy

 

 

б) x^2+4y^2=4xy

3x-y=5

 

y=3x-5

 

x^2+4(3x-5)^2=4x(3x-5)

x^2+4(9x^2-30x+25)=12x^2-20x

x^2+36x^2-120x+100=12x^2-20x

25x^2-100x+100=0

x^2-4x+4=0

D=b^2-4ac=16-16=0

x=(-b±sqrt(D))/2=4/2=2

y=3x-5=6-5=5-6=1

то есть

x=2, y=1

 

 

b) (x-1)/(x+1)<=3/2x

2x(x-1)<=3(x+1)

2x^2-2x-3x-3<=0

2x^2-5x-3<=0

2x^2-5x-3=0

D=b^2-4ac=25+24=49

x1,2=(-b±sqrt(D)/2a

x1=(5-7)/4=-0,5

x2=(5+7/4=3

То есть

x от -0,5 до 3