Question

Giải hệ phương trình \(\left\{{}\begin{matrix}\sqrt{x-3}-\sqrt[3]{y^2+5y+7}=\sqrt{y-1}\sqrt[3]{x^2+x+1}\\\sqrt{x+4}+2y=9\end{matrix}\right.\)

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

Làm 1 trong 2 phần cũng ok nhé

Answer 1

2. \(pt\Leftrightarrow\sqrt{x^2+12}-\sqrt{x^2+5}=3x-5\Rightarrow3x-5>0\Rightarrow x>\frac{5}{3}\)

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

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

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

+ \(\forall x>\frac{5}{3}\) ta có: \(\left\{{}\begin{matrix}x+2>0\\\sqrt{x^2+12}+4>\sqrt{x^2+5}+3\end{matrix}\right.\)

\(\Rightarrow\frac{x+2}{\sqrt{x^2+12}+4}< \frac{x+2}{\sqrt{x^2+5}+3}\Rightarrow\frac{x+2}{\sqrt{x^2+12}+4}-\frac{x+2}{\sqrt{x^2+5}+3}-3< 0\) nên từ (1) suy ra:

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