Question

giải phương trình :
\(\sqrt{x-2}+2\cdot\sqrt{x-3}+\sqrt{x+6+6\cdot\sqrt{x-3}}=4\)

Answer 1

\(\sqrt{x-2}+2\sqrt{x-3}+\sqrt{x+6+6\sqrt{x-3}}=4\)

\(\left(\text{Đ}\text{KXĐ}:x\ge3\right)\)

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

\(\Leftrightarrow\sqrt{x-2}+2\sqrt{x-3}+\sqrt{x-3}+3=4\)

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

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

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

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

Pt \(\dfrac{1}{\sqrt{x-2}+1}+\dfrac{3}{\sqrt{x-3}}\) vô n_o

=> x - 3 = 0

<=> x = 3 (nhận)

Answer 2

sửa lẹ t làm cho