Question

[Hoàng Hải Đăng]

Giải pt

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

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

Answer 1

[Hoàng Lê Bảo Ngọc]

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