Câu hỏi của Nguyễn Thế Phúc Anh

Question

Giai pt

$\left(\sqrt{2x+3}+2\right).\left(\sqrt{x+6}-\sqrt{x+1}\right)=5$

Nguyễn Thế Phúc Anh · Accepted Answer

1 like tức thì nàoCâu trả lời: 1 like tức thì nào

Phạm Tuấn Đạt · Answer

$\left(\sqrt{2x+3}+2\right)\left(\sqrt{x+6}-\sqrt{x+1}\right)=5$

$ĐKXĐ:x\ge-1$.Nhận thấy $\sqrt{x+6}-\sqrt{x+1}>0$

$\Leftrightarrow\left(\sqrt{2x+3}+2\right)\frac{\left(\sqrt{x+6}+\sqrt{x+1}\right)\left(\sqrt{x+6}-\sqrt{x+1}\right)}{\sqrt{x+6}-\sqrt{x+1}}=5$

$\Leftrightarrow\left(\sqrt{2x+3}+2\right)\frac{5}{\sqrt{x+6}-\sqrt{x+1}}=5$

$\Leftrightarrow\frac{\sqrt{2x+3}+2}{\sqrt{x+6}-\sqrt{x+1}}=1$

$\Leftrightarrow\sqrt{2x+3}+2-\sqrt{x+6}+\sqrt{x+1}=0$

Th1:$\sqrt{x+1}=2\Leftrightarrow x=3\left(thoaman\right)$

Th2:$\sqrt{x+1}-2\ne0\Leftrightarrow x\ne3$

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

$\Leftrightarrow\frac{x-3}{\sqrt{2x+3}+\sqrt{x+6}}+\frac{x-3}{\sqrt{x+1}-2}=0$

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

Tự lm tiếp nhaCâu trả lời: $\left(\sqrt{2x+3}+2\right)\left(\sqrt{x+6}-\sqrt{x+1}\right)=5$