Phương trình chứa căn

\(\dfrac{-17}{15}\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-3}\\b=\sqrt{9-2x}\end{matrix}\right.\left(a,b\ge0\right)\Leftrightarrow ab=\sqrt{-2x^2+15x-27};a^2+b^2=-x+6\)

\(PT\Leftrightarrow a+b=2ab+a^2+b^2+2\\ \Leftrightarrow\left(a+b\right)^2-\left(a+b\right)-2=0\\ \Leftrightarrow\left(a+b\right)^2+\left(a+b\right)-2\left(a+b\right)-2=0\\ \Leftrightarrow\left(a+b\right)\left(a+b+1\right)-2\left(a+b+1\right)=0\\ \Leftrightarrow\left(a+b-2\right)\left(a+b+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a+b=2\left(n\right)\\a+b=-1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow a+b=2\\ \Leftrightarrow\sqrt{x-3}=2-\sqrt{9-2x}\left(3\le x\le\dfrac{9}{2}\right)\\ \Leftrightarrow x-3=11-2x-4\sqrt{9-2x}\\ \Leftrightarrow x+14=4\sqrt{9-2x}\\ \Leftrightarrow x^2+28x+196=144-32x\\ \Leftrightarrow x^2+60x+52=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-30+4\sqrt{53}\left(l\right)\\x=-30-4\sqrt{53}\left(l\right)\end{matrix}\right.\)

Vậy PT vô nghiệm

Điều kiện: \(x^2-3x+2\ge0\)\(\Leftrightarrow x\ge2\)

 \(\sqrt{x^2-3x+2}=x^2-3x-4\)

\(\Leftrightarrow x^2-3x+2-2\cdot\dfrac{1}{2}\sqrt{x^2-3x+2}+\dfrac{1}{4}=\dfrac{25}{4}\)

\(\Leftrightarrow\left(\sqrt{x^2-3x+2}-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-3x+2}-\dfrac{1}{2}=\dfrac{5}{2}\\\sqrt{x^2-3x+2}-\dfrac{1}{2}=\dfrac{-5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-3x+2}=3\\\sqrt{x^2-3x+2}=-2\end{matrix}\right.\)

Đối chiếu điều kiện ta được \(\sqrt{x^2-3x+2}=3\)

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

\(\Leftrightarrow x^2-3x-7=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{37}}{2}\\x=\dfrac{3-\sqrt{37}}{2}\end{matrix}\right.\)

Đối chiếu điều kiện \(\Rightarrow x=\dfrac{3+\sqrt{37}}{2}\)

Lời giải:
ĐKXĐ: $x\in\mathbb{R}$

Đặt $\sqrt{x^2+x+1}=a; \sqrt{x^2-x+1}=b(a,b\geq 0)$. PT trở thành:
$a=a^2-b^2+b$

$\Leftrightarrow (a-b)(a+b)-(a-b)=0$

$\Leftrightarrow (a-b)(a+b-1)=0$

$\Rightarrow a=b$ hoặc $a+b=1$

Nếu $a=b\Leftrightarrow a^2=b^2\Leftrightarrow x^2+x+1=x^2-x+1$

$\Leftrightarrow x=0$

Nếu $a+b=1$

$\Leftrightarrow \sqrt{x^2+x+1}+\sqrt{x^2-x+1}=1$

$\Leftrightarrow \sqrt{x^2+x+1}=1-\sqrt{x^2-x+1}$

$\Rightarrow x^2+x+1=x^2-x+2-2\sqrt{x^2-x+1}$

$\Leftrightarrow 1-2x=2\sqrt{x^2-x+1}$

$\Rightarrow (1-2x)^2=4(x^2-x+1)$

$\Leftrightarrow -3=0$ (vô lý)

Vậy pt có nghiệm $x=0$

1. Đề bài ko chính xác

ĐKXĐ: \(\left[{}\begin{matrix}x\ge0\\x\le-2\end{matrix}\right.\)

- Với \(x\ge0\Rightarrow2x\sqrt{x^2+2x}\ge2x\sqrt{x^2}=2x^2\ge x^2>x^2-1\)

\(\Rightarrow\) Pt vô nghiệm

- Với \(x\le-2\Rightarrow\left\{{}\begin{matrix}x^2-1\ge3>0\\2x\sqrt{x^2-1}< 0\end{matrix}\right.\) \(\Rightarrow x^2-1>2x\sqrt{x^2+2x}\)

\(\Rightarrow\) Pt vô nghiệm

Hay pt luôn vô nghiệm

2.

\(\Leftrightarrow2x^2+3x+9-3\sqrt{2x^2+3x+9}-6=0\)

Đặt \(\sqrt{2x^2+3x+9}=u>0\)

\(\Rightarrow u^2-5u-6=0\Rightarrow\left[{}\begin{matrix}u=-1\left(loại\right)\\u=6\end{matrix}\right.\)

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

\(\Leftrightarrow2x^2+3x+9=36\)

\(\Leftrightarrow2x^2+3x-27=0\)

Bấm máy

3.

ĐKXĐ: ...

\(\left(x+4\right)\left(x+1\right)-3\sqrt{x^2+5x+2}=6\)

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

Đặt \(\sqrt{x^2+5x+2}=u\ge0\)

\(\Rightarrow u^2-3u-4=0\Rightarrow\left[{}\begin{matrix}u=-1\left(loại\right)\\u=4\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+5x+2}=4\)

\(\Leftrightarrow x^2+5x+2=16\)

\(\Leftrightarrow x^2+5x-14=0\)

Bấm máy