Question

корень из(4х2+9х+5)-корень(2х2+х-1)=корень(х2-1)

Answer 1

${4x^{2} + 9x +5} - sqrt{2x^{2} + x -1} = sqrt{x^{2} -1}$  

ОДЗ:

4x^2+9x+5>=0

D=81-80=1

x1=-1

x2=-1.25

(=-00; -1.25] U [-1 +00)

2x^2+x-1>=0

D=1+8=9

(2x-1)(x+1)>=0

x=(-00 -1] U [1/2 +00)

x^2-1>=0

(х-1)(х+1)>=0

x=(-00 -1] U [1 +00)

Общее ОДЗ:

х={-1} U [1 +00)

$sqrt{4x^{2} + 9x +5} = sqrt{2x^{2} + x -1} + sqrt{x^{2} -1}$    

$4x^{2} + 9x +5= x^{2} - 1 + 2sqrt{(x^{2}-1)*(2x^{2}+x-1)} +2x^{2}+x-1$

4x^{2} +9x+5=3x^{2}-2+2$sqrt{2x^{4}+x^{3}-3x^{2}-x+1} +x$

-2$sqrt{2x^{4}+x^{3}-3x^{2}-x+1} =3x^{2}-2+x-4x^{2}-9x-5$

-2$sqrt{2x^{4}+x^{3}-3x^{2}-x+1} =-x^{2} -7 - 8x$

4(2x^{4}+x^{3}-3x^{2}-x+1)=x^{4}+49+64x^{2}+14x^{2}+16x^{3}+11x

8x^{4}+4x^{3}-12x^{2}-4x+4=x^{4}+49+78x^{2}+16x^{3}+112x

7x^{4}-12x^{3}-90x^{2}-116x-45=0

7x^{4}+7x^{3}-19x^{3}-19x^{2}-71x^{2}-71x-45x-45=0

7x^{3}*(x+1)-19x^{2}*(x+1)-71x*(x+1)-45(x+1)=0

(x+1)(7x^{3}-19x^{2}-71x-45)=0

(x+1)(7x^{3}+7x^{2}-26x^{2}-26x-45x-45)=0

(x+1)(7x^{2}*(x+1)-26x*(x+1)-45(x+1)=0

(x+1)^{2}*(7x^{2}-26x-45)=0

(x+1)^{2}*(7x^{2}+9x-35x-45)=0

(x+1)^{2}*(x(7x+9)-5(7x+9))=0

(x+1)^{2}*(7x+9)*(x-5)=0

(x+1)^{2}=0

x=-1

7x+9=0

x=- $frac{9}{7}$ - не удовлетворяет ОДЗ

x-5=0

x=5

Ответ:

x=5

x=-1

Answer 2

$sqrt{(4x+5)(x+1)} -sqrt{(2x-1)(x+1)} =sqrt{(x-1)(x+1)} \\ (4x+5)(x+1)=(2x-1)(x+1) + (x-1)(x+1) + 2|x+1|sqrt{(2x-1)(x-1)}\\1)x<-1\(x+1)(4x+5-2x+1-x+1+2sqrt{(2x-1)(x-1)})=0\x+1 = 0\ x=-1\\ x+7 =-2sqrt{(2x-1)(x-1)})\x^2-26x+45=0\x_1=-frac{9}{7} notin ODZ\ x_2=5notin ODZ\\ 2)x+7=2sqrt{(2x-1)(x-1)} \ x^2+14x+49=8x^2-12x+4\ 7x^2-26x-45=0\ x_1=-frac{9}{7} notin ODZ\ x_2=5\\ OTBET: x=-1;x=5$