\(pt\Leftrightarrow3\left(x^2+7x+7\right)-3+2\sqrt{x^2+7x+7}=0\)

\(\Leftrightarrow3t^2-3+2t=0\left(t=\sqrt{x^2+7x+7}\ge0\right)\)

\(\Leftrightarrow t=\frac{-2\pm\sqrt{40}}{6}\)\(\Rightarrow\sqrt{x^2+7x+7}=\frac{-2\pm\sqrt{40}}{6}\)

Giải tiếp nhé, nghiệm xấu thật

\(3^2+21x+18+2\sqrt{x^2+7x+7}=0\)

\(\Rightarrow3\left(x^2+7x+7\right)-3+2\sqrt{x^2+7x+7}\)

\(\Rightarrow3\left(x^2+7x+7\right)+2\sqrt{x^2+7x+7}-5=0\)

Đặt: \(a=\sqrt{x^2+7x+7}\left(a\ge0\right)\)ta được PT:

\(3a^2+2a-5=0\)

\(\Rightarrow\left(a-1\right)\left(3a+5\right)=0\)

\(\Rightarrow a=1\)(nhận)

     \(a=-\frac{5}{3}\)(loại)

\(a=1\Rightarrow\sqrt{x^2+7x+7}=1\Rightarrow x^2+7x+7=1\Rightarrow x^2+7x+6=0\)

\(\Rightarrow\left(x+6\right)\left(x+1\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=-6\end{cases}}\)

Vậy: \(x=-1;x=-6\)

\(a.3x^2+21x+18+2\sqrt{x^2+7x+7}=2\)

\(\Leftrightarrow3\left(x^2+7x+6\right)+2\sqrt{x^2+7x+7}=2\circledast\)

Đặt : \(x^2+7x+7=t\left(t\ge0\right)\) , ta có :

\(\circledast\Leftrightarrow3\left(t-1\right)+2\sqrt{t}=2\)

\(\Leftrightarrow3t+2\sqrt{t}-5=0\)

\(\Leftrightarrow3\sqrt{t}\left(\sqrt{t}-1\right)+5\left(\sqrt{t}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{t}-1=0\\3\sqrt{t}+5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=1\left(TM\right)\\vô-nghiệm\end{matrix}\right.\)

Với : \(t=1\) , thì : \(x^2+7x+7=1\Leftrightarrow x^2+x+6x+6=0\)

\(\Leftrightarrow x\left(x+1\right)+6\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)

KL...........

\(b.2x^2-8x-3\sqrt{x^2-4x-5}=12\circledast\)

ĐKXĐ : \(\left[{}\begin{matrix}x\ge5\\x\le-1\end{matrix}\right.\)

\(\circledast\Leftrightarrow2x^2-8x-12-3\sqrt{x^2-4x-5}=0\)

\(\Leftrightarrow2\left(x^2-4x-3\right)-3\sqrt{x^2-4x-5}=0\)

Đặt : \(x^2-4x-5=t\left(t\ge0\right)\) , ta có :

\(2\left(t+2\right)-3\sqrt{t}=0\)

\(\Leftrightarrow2t-3\sqrt{t}+4=0\)

\(\Leftrightarrow2\left(t-2.\dfrac{3}{4}\sqrt{t}+\dfrac{9}{16}\right)+4-\dfrac{9}{8}=0\)

\(\Leftrightarrow\left(\sqrt{t}-\dfrac{3}{4}\right)^2=\dfrac{23}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{t}-\dfrac{3}{4}=\dfrac{\sqrt{23}}{4}\\\sqrt{t}-\dfrac{3}{4}=-\dfrac{\sqrt{23}}{4}\end{matrix}\right.\)

Tới đây dễ rồi , bạn tự làm nốt nhé...:)

☛ Câu hỏi của Tô Thu Huyền - Toán lớp 9 | Học trực tuyến

TXĐ: \(x\le\dfrac{-7}{2};x\ge6;x=1\)

\(\sqrt{\left(x-1\right)\left(2x+7\right)}+\sqrt{\left(x-1\right)\left(3x-18\right)}=\sqrt{\left(x-1\right)\left(7x+1\right)}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{2x+7}+\sqrt{3x-18}=\sqrt{7x+1}\end{matrix}\right.\)

Pt1: \(\sqrt{x-1}=0\Rightarrow x=1\)

Pt2: \(\sqrt{2x+7}+\sqrt{3x-18}=\sqrt{7x+1}\)

\(\Leftrightarrow5x-11+2\sqrt{\left(2x+7\right)\left(3x-18\right)}=7x+1\)

\(\Leftrightarrow\sqrt{\left(2x+7\right)\left(3x-18\right)}=x+6\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+6\ge0\\\left(2x+7\right)\left(3x-18\right)=\left(x+6\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-6\\5x^2-27x-162=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{-18}{5}\end{matrix}\right.\)

Vậy pt có 3 nghiệm: \(\left[{}\begin{matrix}x=1\\x=9\\x=\dfrac{-18}{5}\end{matrix}\right.\)

Lời giải:
ĐK: $x\geq \frac{-18}{7}$

PT $\Leftrightarrow x^2+3x-4-3(\sqrt{x+3}-2)-(\sqrt{7x+18}-5)=0$

$\Leftrightarrow (x-1)(x+4)-3.\frac{x-1}{\sqrt{x+3}+2}-\frac{7(x-1)}{\sqrt{7x+18}+5}=0$

$\Leftrightarrow (x-1)\left(x+4-\frac{3}{\sqrt{x+3}+2}-\frac{7}{\sqrt{7x+18}+5}\right)=0$

Xét các TH:

Nếu $x-1=0\Rightarrow x=1$ (thỏa mãn)

Nếu $x+4-\frac{3}{\sqrt{x+3}+2}-\frac{7}{\sqrt{7x+18}+5}=0$

$\Leftrightarrow (x+2)+1-\frac{3}{\sqrt{x+3}+2}+1-\frac{7}{\sqrt{7x+18}+5}=0$

$\Leftrightarrow x+2+\frac{\sqrt{x+3}-1}{\sqrt{x+3}+2}+\frac{\sqrt{7x+18}-2}{\sqrt{7x+18}+5}=0$

\(\Leftrightarrow (x+2)+\frac{x+2}{(\sqrt{x+3}+1)(\sqrt{x+3}+2)}+\frac{7(x+2)}{(\sqrt{7x+18}+2)(\sqrt{7x+18}+5)}=0\)

\(\Leftrightarrow (x+2)\left( 1+\frac{1}{(\sqrt{x+3}+1)(\sqrt{x+3}+2)}+\frac{7}{(\sqrt{7x+18}+2)(\sqrt{7x+18}+5)}\right)=0\)

Dễ thấy biểu thức trong ngoặc lớn luôn dương nên $x+2=0\Leftrightarrow x=-2$

Vậy $x=-2$ hoặc $x=1$