Sửa đề: \(19 + 3x + 4\sqrt{-x^2 - x + 6} = 10\sqrt{2 - x} + 10\sqrt{x + 3}\) (1)

ĐKXĐ: -3<=x<=2

Đặt \(u=\sqrt{2-x};v=\sqrt{x+3}\) (Điều kiện: u>0; v>0)

\(u^2+v^2=2-x+3+x=5\)

\(uv=\sqrt{\left(2-x\right)\left(x+3\right)}=\sqrt{2x+6-x^2-3x}=\sqrt{-x^2-x+6}\)

\(v^2=x+3\)

=>\(x=v^2-3\)

=>\(3x=3v^2-9\)

(1) sẽ trở thành: \(19 + (3v^2 - 9) + 4uv = 10u + 10v\)

=>\(10+3v^2+4uv-10u-10v=0\)

=>\(u(4v-10)=-3v^2+10v-10=-3\left(v^2-\frac{10}{3}v+\frac{10}{3}\right)=-3\left(v^2-2\cdot v\cdot\frac53+\frac{25}{9}+\frac59\right)=-3\left(v-\frac53\right)^2-\frac53<0\) ∀v

=>u(4v-10)<0

=>4v-10<0

=>4v<10

=>v<5/2

=>\(u = \frac{3v^2 - 10v + 10}{10 - 4v}\)

=>\(5 - v^2 = \left( \frac{3v^2 - 10v + 10}{10 - 4v} \right)^2\)

=>\((5 - v^2)(10 - 4v)^2 = (3v^2 - 10v + 10)^2\)

=>\((5 - v^2)(16v^2 - 80v + 100) = 9v^4 - 60v^3 + 160v^2 - 200v + 100\)

=>\(-16v^4 + 80v^3 - 20v^2 - 400v + 500 = 9v^4 - 60v^3 + 160v^2 - 200v + 100\)

=>\(25v^4 - 140v^3 + 180v^2 + 200v - 400 = 0\)

=>\(5v^4 - 28v^3 + 36v^2 + 40v - 80 = 0\)

=>\((v - 2)(5v^3 - 18v^2 + 40) = 0\)

=>v-2=0

=>v=2

=>x+3=4

=>x=1(nhận)

bang 8

chu so tan cung la 8

-5²y⁵-x²y⁵+2x²y⁵ = ( -5-1+2 ) x²y⁵ 

                             = -4x²y⁵

a)\(\frac{x-1}{x-5}=\frac{6}{7}\) điều kiện : x khác 5

<=>7x-7=6x-30<=> x=-23

b) \(\frac{12-7x}{-13}=\frac{4-3x}{-5}\)

<=> -60+35x=-52+39x

<=> 4x=-8

<=> x=-2

( X - 39 ) x 4 : 9 = 12

( X - 39 ) x 4  = 12 x 9

( X - 39 ) x 4 = 108

 X - 39  = 108 : 4

 X - 39  = 27

   X = 27 + 39 

   X = 66

\(\left(x-39\right).4:9=12\)

\(\left(x-39\right).4=12.9\)

\(\left(x-39\right).4=108\)

\(x-39=108:4\)

 \(x-39=27\)

        \(x=27+39\)

        \(x=66\)