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)

tìm...x....à?????????????

  (x²+x)²+4(x²+x)-12=0

(x²+x)(x²+x)+4(x²+x)   = 12

(x²+x)  [(x²+x)+4]        =12

x(x+1) [x(x+1)+4]         =12

...????

đặt \(x^2+x\) = t

ta có : t ² +4t -12 = 0

\(\Leftrightarrow\) t²+6t-2t-12=0

\(\Leftrightarrow\)t(t+6)-2(t+6)=0

\(\Leftrightarrow\)(t+6)(t-2)=0

<=> thay t = x²+x

đoạn sau tự làm nhé !!!

<=>(x^2+x+1)x^2+x-11=(x-1)(x+2)(x^2+x+5)

=>x=1

=>x+2=0

=>x=-2

áp dụng denta

<=>x^2+x+5=0

1^2-4(1.5)=-19

vì -19<0 =>\(\Delta<0\) ko có nghiệm thực

=>x=-2 hoặc 1

Ta có: \(\dfrac{\left(x+3\right)\left(x-3\right)}{3}+2=x\left(1-x\right)\)

\(\Leftrightarrow\dfrac{x^2-9}{3}+\dfrac{6}{3}=\dfrac{3x\left(1-x\right)}{3}\)

\(\Leftrightarrow x^2-9+6=3x-3x^2\)

\(\Leftrightarrow x^2-3-3x+3x^2=0\)

\(\Leftrightarrow4x^2-3x-3=0\)

\(\Delta=9-4\cdot4\cdot\left(-3\right)=9+48=57\)

Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là 

\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{57}}{8}\\x_2=\dfrac{3+\sqrt{57}}{8}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3-\sqrt{57}}{8};\dfrac{3+\sqrt{57}}{8}\right\}\)

Ta có : 

\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)

\(\Leftrightarrow\)\(\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2028}{6}-3\right)=0\)

\(\Leftrightarrow\)\(\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)

\(\Leftrightarrow\)\(\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)

Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\)

Nên 

\(x+2010=0\)

\(\Rightarrow\)\(x=-2010\)

Vậy \(x=-2010\)

Chúc bạn học tốt ~ 

`48/[x+4]+48/[x-4]=5`           `ĐK: x \ne +-4`

`<=>[48(x-4)+48(x+4)]/[(x-4)(x+4)]=[5(x+4)(x-4)]/[(x-4)(x+4)]`

   `=>48x-192+48x+192=5x^2-80`

`<=>5x^2-96x-80=0`

`<=>5x^2-100+4x-80=0`

`<=>5x(x-20)+4(x-20)=0`

`<=>(x-20)(5x+4)=0`

`<=>` $\left[\begin{matrix} x=20\\ x=\dfrac{-4}{5}\end{matrix}\right.$   (t/m)

Vậy `S={-4/5;20}`

ĐK : \(x\ne\pm4\)

\(\Leftrightarrow\cdot\dfrac{48\left(x+4\right)+48\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}=\dfrac{5\left(x+4\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\)

\(\Leftrightarrow48x+192+48x-192==5x^2-80\)

\(\Leftrightarrow96x=5x^2-80\)

\(\Leftrightarrow5x^2-96x-80=0\)

\(\Leftrightarrow5x^2+4x-100-80=0\)

\(\Leftrightarrow4\left(x-20\right)+5x\left(x-20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-20=0\\5x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-\dfrac{4}{5}\end{matrix}\right.\)