Ta có: \(2x^2+3x+\sqrt{2x^2+3x+9}=33\)

   \(\Leftrightarrow\left(2x^2+3x-27\right)+\left(\sqrt{2x^2+3x+9}-6\right)=0\)

   \(\Leftrightarrow\left(2x+9\right)\left(x-3\right)+\dfrac{2x^2+3x-27}{\sqrt{2x^2+3x+9}+6}=0\)

   \(\Leftrightarrow\left(2x+9\right)\left(x-3\right)+\dfrac{\left(2x+9\right)\left(x-3\right)}{\sqrt{2x^2+3x+9}+6}=0\)

   \(\Leftrightarrow\left(2x+9\right)\left(x-3\right)\left(1+\dfrac{1}{\sqrt{2x^2+3x+9}+6}\right)=0\)

   \(\Leftrightarrow\left[{}\begin{matrix}2x+9=0\\x-3=0\\1+\dfrac{1}{\sqrt{2x^2+3x+9}+6}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{2}\\x=3\\1+\dfrac{1}{\sqrt{2x^2+3x+9}+6}=0\left(1\right)\end{matrix}\right.\)

Giải (1) ta có:

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

     \(\Leftrightarrow1=-\sqrt{2x^2+3x+9}-6\)

     \(\Leftrightarrow7=-\sqrt{2x^2+3x+9}\)

     \(\Leftrightarrow49=2x^2+3x+9\)

      \(\Leftrightarrow2x^2+3x-40=0\)

Ta có:Δ=3²-4.2.(-40)=329

Vì Δ>0 nên phương trình có 2 nghiệm phân biệt là:

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-3+\sqrt{329}}{4}\\x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-3-\sqrt{329}}{4}\end{matrix}\right.\)

Vậy phương trình có 4 nghiệm là ....

Bạn coi lại đề xem có sai không chứ nghiệm giải ra xấu cực. Và phương trình không rút gọn hết nghe cũng rất vô lý.

a: \(x\left(x-1\right)+2x^2-2=0\)

=>\(x\left(x-1\right)+2\left(x-1\right)\left(x+1\right)=0\)

=>\(\left(x-1\right)\left(x+2x+2\right)=0\)

=>(x-1)(3x+2)=0

=>\(\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)

b: \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)

=>\(\left(3x+1\right)\left(3x-1\right)-\left(3x+1\right)\left(2x-3\right)=0\)

=>\(\left(3x+1\right)\left(3x-1-2x+3\right)=0\)

=>(3x+1)(x+2)=0

=>\(\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-2\end{matrix}\right.\)

a: 

=>

=>

=>(x-1)(3x+2)=0

=>⎡⎣x=−13x=−2

b)đk:\(x\ge\dfrac{1}{2}\)

Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)

\(x\sqrt{2x-1}=\sqrt{\left(2x^2-x\right)x}\le\dfrac{2x^2-x+x}{2}=x^2\)

=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\) 

Dấu = xảy ra\(\Leftrightarrow x=1\)

Vậy....

c) đk: \(x\ge0\)

\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)

\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)

pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)

\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...

 

a)ĐKXĐ: x≥-1/3; x≤6

<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)

(vì x≥-1/3 nên3x+1≥0 )

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x-3\ge0\\2x^2-3x+1\ge0\\x^2+2x-3\le2x^2-3x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge1\\x\le\dfrac{1}{2}\end{matrix}\right.\\x^2-5x+4\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge4\\x\le1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x\le-3\\x\ge4\end{matrix}\right.\)