Câu hỏi của Hoàng Trung Đức

Question

Căn (x+1) + Căn (X +10)= căn (x+2) + căn (x+5)

Nguyễn Huệ Lam · Accepted Answer

$\sqrt{x+1}+\sqrt{x+10}=\sqrt{x+2}+\sqrt{x+5}.$ ĐK: $x\ge-1$

$PT\Leftrightarrow\sqrt{x+1}-\sqrt{x+2}+\sqrt{x+10}-\sqrt{x+5}=0$

$\Leftrightarrow\sqrt{x+1}-\left(\sqrt{x+2}+1\right)+\left(\sqrt{x+10}+3\right)-\left(\sqrt{x+5}+2\right)=0$

$\Leftrightarrow\frac{x+1}{\sqrt{x+1}}-\frac{x+2-1}{\sqrt{x+2}-1}+\frac{x+10-9}{\sqrt{x+10}-3}-\frac{x+5-4}{\sqrt{x+5}-2}=0$

$\Leftrightarrow\left(x+1\right)\left(\frac{1}{\sqrt{x+1}}-\frac{1}{\sqrt{x+2}-1}+\frac{1}{\sqrt{x+10}-3}+\frac{1}{\sqrt{x+5}-2}\right)=0$

$\Rightarrow\orbr{\begin{cases}x=-1\left(TM\right)\...=0\left(VN\right)\end{cases}}$

Câu trả lời:

$\sqrt{x+1}+\sqrt{x+10}=\sqrt{x+2}+\sqrt{x+5}.$ ĐK: $x\ge-1$

$PT\Leftrightarrow\sqrt{x+1}-\sqrt{x+2}+\sqrt{x+10}-\sqrt{x+5}=0$

$\Leftrightarrow\sqrt{x+1}-\left(\sqrt{x+2}+1\right)+\left(\sqrt{x+10}+3\right)-\left(\sqrt{x+5}+2\right)=0$

$\Leftrightarrow\frac{x+1}{\sqrt{x+1}}-\frac{x+2-1}{\sqrt{x+2}-1}+\frac{x+10-9}{\sqrt{x+10}-3}-\frac{x+5-4}{\sqrt{x+5}-2}=0$

$\Leftrightarrow\left(x+1\right)\left(\frac{1}{\sqrt{x+1}}-\frac{1}{\sqrt{x+2}-1}+\frac{1}{\sqrt{x+10}-3}+\frac{1}{\sqrt{x+5}-2}\right)=0$

$\Rightarrow\orbr{\begin{cases}x=-1\left(TM\right)\...=0\left(VN\right)\end{cases}}$