GPT: \(\left(x-2\right)\left(\sqrt{3x+1}-1\right)=3x\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\sqrt{x^2-3x+2}=t\ge0\)
\(\Rightarrow log_3\left(t+2\right)+5^{t^2-1}-2=0\)
Nhận thấy \(t=1\) là 1 nghiệm của pt
Xét hàm \(f\left(t\right)=log_3\left(t+2\right)+5^{t^2-1}-2\)
\(f'\left(t\right)=\dfrac{1}{\left(t+2\right)ln3}+2t.5^{t^2-1}.ln5>0\) ; \(\forall t\ge0\)
\(\Rightarrow f\left(t\right)\) đồng biến \(\Rightarrow f\left(t\right)\) có tối đa 1 nghiệm
\(\Rightarrow t=1\) là nghiệm duy nhất
\(\Rightarrow\sqrt{x^2-3x+2}=1\)
\(\Rightarrow...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x^2+x+1=\left(3x+1\right)\sqrt{x^2+1}\) (ĐKXĐ : \(x>-\frac{1}{3}\) )
\(\Leftrightarrow3x^2-2x=\left(3x+1\right)\sqrt{x^2+1}-\left(3x+1\right)\)
\(\Leftrightarrow3x^2-2x=\left(3x+1\right)\left(\sqrt{x^2+1}-1\right)\)
\(\Leftrightarrow x\left(3x-2\right)=\left(3x+1\right)\left(\frac{x^2+1-1}{\sqrt{x^2+1}+1}\right)\)
\(\Leftrightarrow x\left(3x-2\right)=x\left(3x+1\right)\left(\frac{1}{\sqrt{x^2+1}+1}\right)\)
\(\Leftrightarrow x\left(3x-2-\frac{3x+1}{\sqrt{x^2+1}+1}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x-2-\frac{3x+1}{\sqrt{x^2+1}+1}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x\approx1,2818\end{cases}}\)
Thử lại, ta có x = 0 thoả mãn nghiệm phương trình.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐK:x\in R\)
Đặt \(\sqrt{x^2+3}=t\left(t\ge0\right)\)
\(PT\Leftrightarrow2t^2-\left(7x+1\right)t+3x^2+3x=0\\ \Delta=\left(7x+1\right)^2-4\cdot2\left(3x^2+3x\right)=25x^2-10x+1=\left(5x-1\right)^2\ge0\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{7x+1-5x+1}{4}\\t=\dfrac{7x+1+5x-1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{2x+2}{4}=\dfrac{x+1}{2}\\t=\dfrac{12x}{4}=3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=\dfrac{x+1}{2}\\\sqrt{x^2+3}=3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+3=\dfrac{x^2+2x+1}{4}\\x^2+3=9x^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x^2-2x+11=0\\x^2=\dfrac{3}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\Delta=4-132< 0\\\left[{}\begin{matrix}x=\dfrac{\sqrt{6}}{4}\\x=-\dfrac{\sqrt{6}}{4}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{\sqrt{6}}{4};\dfrac{\sqrt{6}}{4}\right\}\)
xét x=1 không phải là nghiệm của pt
xét x khác 1
\(\left(x-2\right)\left(\sqrt{3x+1}-1\right)=3x\Leftrightarrow\sqrt{3x+1}=\frac{3x}{x-2}+1=\frac{4x-2}{x-2}\)\(;x\ne2\)
\(\Rightarrow3x+1=\frac{16x^2-16x+4}{x^2-4x+4}\Rightarrow\left(3x+1\right)\left(x^2-4x+4\right)=16x^2-16x+4\)
\(\Leftrightarrow3x^3-11x^2+8x+4=16x^2-16x+4\)
\(\Leftrightarrow3x^3-27x^2+24x=0\)
\(\Leftrightarrow x\left(3x^2-27x+24\right)=0\)
\(\Leftrightarrow x.\left(x-8\right)\left(x-1\right)=0\)