PT tương đương \(2x-6=3\sqrt{x-2}-\sqrt{x+6}\)

Bình phương hai vế \(4x^2-34x+48=6\sqrt{\left(x-2\right)\left(x+6\right)}\)

Tiếp tục bình phương được phương trình tương đương \(\left(x-3\right)^2\left(x^2-11x+19\right)=0\)

P/s: Tham khảo nha!

(x-3)^2*(x^2-11x+19)=0

ĐKXĐ: \(-\frac{3}{2}\le x\le12\)

\(\Leftrightarrow x^2-2x\sqrt{2x+3}+2x+3+12-x-6\sqrt{12-x}+9=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+3}\right)^2+\left(\sqrt{12-x}-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{2x+3}=0\\\sqrt{12-x}-3=0\end{matrix}\right.\) \(\Rightarrow x=3\)

a) \(x^2-6x+26=6\sqrt{2x+1}\) (ĐKXĐ : \(x\ge-\frac{1}{2}\) )

\(\Leftrightarrow x^2-6x+26-6\sqrt{2x+1}=0\)

\(\Leftrightarrow\left(x^2-6x+8\right)-\left(6\sqrt{2x+1}-18\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)-6\left(\sqrt{2x+1}-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)-6\left(\frac{2x+1-9}{\sqrt{2x+1}+3}\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)-\frac{12\left(x-4\right)}{\sqrt{2x+1}+3}=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-2-\frac{12}{\sqrt{2x+1}+3}\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\x-2-\frac{12}{\sqrt{2x+1}+3}=0\end{array}\right.\)

Với x - 4 = 0 => x = 4 (TMĐK)

Với \(x-2-\frac{12}{\sqrt{2x+1}+3}=0\Rightarrow x=4\left(TM\right)\)

Vậy phương trình có nghiệm x = 4

b) \(x+\sqrt{2x-1}=3+\sqrt{x+2}\) ( ĐKXĐ : \(x\ge\frac{1}{2}\))

\(x+\sqrt{2x-1}-3-\sqrt{x+2}=0\)

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

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

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

Vì \(x\ge\frac{1}{2}\) nên  \(\frac{2}{\sqrt{2x-1}+\sqrt{5}}-\frac{1}{\sqrt{x+2}+\sqrt{5}}+1>0\) . Do đó x-3 = 0 => x = 3 (TMĐK)

Vậy phương trình có nghiệm x = 3

em                                                                                                                                                                                                            ko

biết

@Cold Wind

<=>\(x^3+x+6=2\left(x+1\right)\sqrt{3+2x-x^2}=2x\sqrt{3+2x-x^2}+2\sqrt{3+2x-x^2}\)<=>\(x^3+x+6-4x-4=2x\left[\sqrt{3+2x-x^2}-2\right]+\left[\sqrt{3+2x-x^2}-2\right]\)

<=>\(\left(x^3-x-2x+2\right)\left(\sqrt{3+2x-x^2}+2\right)=2x\left[3+2x-x^2-4\right]+\left[3+2x-x^2-4\right]\)\(\left(\sqrt{3+2x-x^2}+2\right)\left[x\left(x^2-1\right)-2\left(x-1\right)\right]=-2\left(x+1\right)\left(x-1\right)^2\)

x =1 là nghiệm

<=>\(\left(\sqrt{3+2x-x^2}+2\right)\left[x\left(x+1\right)-2\right]=-2\left(x+1\right)\left(x-1\right)\)

\(\left[{}\begin{matrix}-1\le x< 1\Rightarrow VP< 0;VT\ge0\rightarrow Vonghiem\\1< x\le3\Rightarrow VP>0;VT< 0\rightarrow Vonghiem\end{matrix}\right.\)

x =1 là duy nhất

x=1

Ta có \(y^2-2y+3=y^2-2y+1+2=\left(y-1\right)^2+2\ge2\)

\(\dfrac{6}{x^2+2x+4}=\dfrac{6}{x^2+2x+1+3}=\dfrac{6}{\left(x+1\right)^2+3}\le2\)

Vậy \(y^2-2y+3=\dfrac{6}{x^2+2x+4}=2\Leftrightarrow\)\(\left\{{}\begin{matrix}y^2-2y+3=2\\\dfrac{6}{x^2+2x+4}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left(y-1\right)^2+2=2\\\dfrac{6}{\left(x+1\right)^2+3}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=-1\end{matrix}\right.\)

\(y^2-2y+3=\left(y-1\right)^2+2\ge2\)

\(\dfrac{6}{x^2+2x+4}=\dfrac{6}{\left(x+1\right)^2+3}\le2\)

So ez