\(2\left(3x+5\right)\sqrt{x^2+9}=3x^2+2x+30\)

\(\Leftrightarrow\sqrt{x^2+9}=\frac{3x^2+2x+30}{2\left(3x+5\right)}\)

\(\Leftrightarrow\sqrt{x^2+9}-3=\frac{3x^2+2x+30}{2\left(3x+5\right)}-3\)

\(\Leftrightarrow\frac{x^2+9-9}{\sqrt{x^2+9}+3}-\frac{3x^2-16x}{6x+10}=0\)

\(\Leftrightarrow\frac{x^2}{\sqrt{x^2+9}+3}-\frac{x\left(3x-16\right)}{6x+10}=0\)

\(\Leftrightarrow x\left(\frac{x}{\sqrt{x^2+9}+3}-\frac{3x-16}{6x+10}\right)=0\)

Pt trong ngoặc vô nghiệm suy ra x=0

e.

\(\left\{{}\begin{matrix}2x-3y+5=0\\3x+5y-21=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}10x-15y=-25\\9x+15y=63\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}19x=38\\3x+5y=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{21-3x}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

f.

\(\left\{{}\begin{matrix}x-y\sqrt{2}=0\\2x\sqrt{2}+y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y\sqrt{2}=0\\4x+y\sqrt{2}=5\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x=5\sqrt{2}\\2x\sqrt{2}+y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{2}\\y=5-2x\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{2}\\y=1\end{matrix}\right.\)

a.

\(\Leftrightarrow\left\{{}\begin{matrix}5x=-25\\3x-5y=-30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{3x+30}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=3\end{matrix}\right.\)

b.

\(\Leftrightarrow\left\{{}\begin{matrix}8x-6y=-10\\9x+6y=-24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}17x=-34\\9x+6y=-24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=\dfrac{-24-9x}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)

Đặt \(2x^2-3x+1=t\Rightarrow2x^2-3x-9=t-10\)

Phương trình trở thành:

\(t\left(t-10\right)=-9\Leftrightarrow t^2-10t+9=0\Rightarrow\left[{}\begin{matrix}t=1\\t=9\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x=0\\2x^2-3x-8=0\end{matrix}\right.\)

\(\Leftrightarrow...\) (bấm máy)

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 ....

Đặt \(\sqrt{x^2+9}=a\) ( \(a\ge9\) ) => \(x^2+9=a^2\)

Đặt \(3x+5=b\) => \(2x+3=\dfrac{2}{3}a-\dfrac{1}{3}\)

Ta có; \(2\left(3x+5\right)\sqrt{x^2+9}=3x^2+2x+30\)

<=> \(2ab=3a^2+\left(\dfrac{2}{3}b-\dfrac{1}{3}\right)\)

<=> \(6ab=9a^2+2b-1\)

<=> \(\left(9a^2-1\right)-\left(6ab-2b\right)=0\)

<=> \(\left(3a-1\right)\left(3a+1\right)-2b\left(3a-1\right)=0\)

<=> \(\left(3a-1\right)\left(3a+1-2b\right)=0\)

<=> \(\left[{}\begin{matrix}3a=1\left(1\right)\\3a-2b=-1\left(2\right)\end{matrix}\right.\)

(1) => \(3\sqrt{x^2+9}=1\) => Vô nghiệm ( vì \(\sqrt{x^2+9}\ge9\) )

(2) => \(3\sqrt{x^2+9}-2\left(3x+5\right)=-1\)

=> \(x=0\) (TM)

P/s: Mk nghĩ vì bn khá giỏi nên mk sẽ lm hơi tắt!

\(2\left(3x+5\right)\sqrt{x^2+9}=3x^2+2x+30\)

\(\Leftrightarrow2\left(3x+5\right)\sqrt{x^2+9}-30=3x^2+2x\)

\(\Leftrightarrow\dfrac{4\left(3x+5\right)^2\left(x^2+9\right)-900}{2\left(3x+5\right)\sqrt{x^2+9}+30}=x\left(3x+2\right)\)

\(\Leftrightarrow\dfrac{36x^4+120x^3+424x^2+1080x}{2\left(3x+5\right)\sqrt{x^2+9}+30}-x\left(3x+2\right)=0\)

\(\Leftrightarrow\dfrac{4x\left(9x^3+30x^2+106x+270\right)}{2\left(3x+5\right)\sqrt{x^2+9}+30}-x\left(3x+2\right)=0\)

\(\Leftrightarrow x\left(\dfrac{4\left(9x^3+30x^2+106x+270\right)}{2\left(3x+5\right)\sqrt{x^2+9}+30}-\left(3x+2\right)\right)=0\)

Dễ thấy: \(\dfrac{4\left(9x^3+30x^2+106x+270\right)}{2\left(3x+5\right)\sqrt{x^2+9}+30}-\left(3x+2\right)>0\)

\(\Rightarrow x=0\)

ĐK: ....>=0 (đúng với mọi x thuộc R)

Đặt \(\sqrt{2x^2+3x+5}=a;\text{ }\sqrt{2x^2-3x+5}=b\)

\(a^2-b^2=2x^2+3x+5-\left(2x^2-3x+5\right)=6x\)

phương trình đã cho thành \(a+b=\frac{1}{2}\left(a^2-b^2\right)\Leftrightarrow\left(a+b\right)\left(a-b\right)=2\left(a+b\right)\)

\(\Leftrightarrow\left(a+b\right)\left(a-b-2\right)=0\)

\(\Leftrightarrow a-b=2\text{ (do }a,b>0\text{)}\)

Mà \(a+b=3x\)

\(\Rightarrow a+b+a-b=2+3x\Leftrightarrow2a=2+3x\)

\(\Rightarrow2\sqrt{2x^2+3x+5}=2+3x\)

\(\Leftrightarrow4\left(2x^2+3x+5\right)=\left(2+3x\right)^2\text{ và }2+3x\ge0\)

\(\Leftrightarrow x^2=16\text{ và }x\ge-\frac{2}{3}\)

\(\Leftrightarrow x=4\)

Kết luận: x = 4.