Question

giải phương trình:

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

Answer 1

\(\sqrt[]{x+3}+\sqrt[]{x-1}=2\left(x\ge1\right)\)

\(\Leftrightarrow x+3+x-1+2\sqrt[]{\left(x+3\right)\left(x-1\right)}=4\)

\(\Leftrightarrow2x+2+2\sqrt[]{\left(x+3\right)\left(x-1\right)}=4\)

\(\Leftrightarrow2\sqrt[]{\left(x+3\right)\left(x-1\right)}=4-2\left(x+1\right)\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\\Leftrightarrow\left(x+3\right)\left(x-1\right)=\left(1-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\Leftrightarrow x^2+2x-3=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\Leftrightarrow4x=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\Leftrightarrow x=1\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

Answer 2

Điều kiện xác định: \(x\ge1\)

\(\sqrt{x+3}+\sqrt{x-1}=2\\ \Leftrightarrow x+3+x-1+2\sqrt{\left(x+3\right)\left(x-1\right)}=4\)

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

Để phương trình thỏa mãn thì x\(\le1\)mà \(x\le1\)

\(\Rightarrow x=1\)

Thử lại, ta được: \(\sqrt{1+3}+\sqrt{1-1}=2\left(tm\right)\)

Vậy x=1

Answer 3

Bạn kiểm tra lại đề nha, sao lại \(\sqrt{x-1=2}\) nhỉ.