Question

[Nguyễn Linh Chi]

Một vài bài phương trình:

Answer 1

[Kiệt Nguyễn]

\(ĐKXĐ:x\ge2\)

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

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

\(+\sqrt{x-2}\)

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

\(-\sqrt{x-2}=0\)

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

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

\(TH1:\sqrt{x-2}-\sqrt{x+3}=0\Leftrightarrow\sqrt{x-2}=\sqrt{x+3}\)

\(\Leftrightarrow x-2=x+3\left(L\right)\)

\(TH2:\sqrt{x-1}-1=0\Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\)(t/m đk)

Vậy x = 2

Answer 2

[Kiệt Nguyễn]

\(\sqrt{x^2+12}+5=3x+\sqrt{x^2+5}\)

\(\Leftrightarrow\sqrt{x^2+12}-\sqrt{x^2+5}=3x-5\)

Dễ thấy \(VT>0\Rightarrow3x-5>0\Leftrightarrow x>\frac{5}{3}\)

\(pt\Leftrightarrow\left(\sqrt{x^2+5}-3\right)-\left(\sqrt{x^2+12}-4\right)+3x-6=0\)

\(\Leftrightarrow\frac{x^2-4}{\sqrt{x^2+5}+3}-\frac{x^2-4}{\sqrt{x^2+12}+4}+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{x+2}{\sqrt{x^2+5}+3}-\frac{x+2}{\sqrt{x^2+12}+4}+3\right)=0\)

Ta có: \(\frac{x+2}{\sqrt{x^2+5}+3}-\frac{x+2}{\sqrt{x^2+12}+4}\)\(=\left(x+2\right)\left(\frac{1}{\sqrt{x^2+5}+3}-\frac{1}{\sqrt{x^2+12}+4}\right)\)

\(=\left(x+2\right).\frac{\sqrt{x^2+12}-\sqrt{x^2+5}+1}{\left(\sqrt{x^2+5}+3\right)\left(\sqrt{x^2+12}+4\right)}>0\forall x>\frac{5}{3}\)

\(\Rightarrow x-2=0\Leftrightarrow x=2\)

Vậy x = 2